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Text Truncated. Only the first 1MB is shown below. Download the file for the complete contents. Introduction N.1 ================ Hi! Welcome to "The PC Assembler Helper" and "The PC Assembler Tutor". Both the program and the tutorial are designed to help those who are just starting to learn assembler language as well as those who know some assembler instructions but want to have a firmer grasp of the complete instruction set for the 8086. There are two significant problems to learning assembler language. First, it is difficult to do either input or output at the assembler level. Imagine trying to learn BASIC if you were not allowed the following two instructions: PRINT RESULT INPUT NEW.DATA Without PRINT and INPUT, you might be able to write a program but you would not be able to see the results. You would never be sure that the results were what you wanted. Also, you would not be able to vary the data. It would have to be coded into the program. "The PC Assembler Helper" has taken care of this problem. It provides input and output of all standard 8086/8087 integral data types. These include 1 byte, 2 byte, 4 byte and 8 byte signed and unsigned numbers along with 1 byte and 2 byte hex, ASCII and binary data. Lastly, there is i/o for 10 byte BCD numbers. The interface has been designed so that beginners can use it with a minimum of trouble. The second major problem is that most assembler books regard the 8086 as a black box. There is no way of seeing the workings of the chip itself. What exactly happens when you add two numbers? What about multiplication? Once again, "The PC Assembler Helper" has come up with the solution. It allows you to view all the registers and flags at will. These registers can be independently formatted. If one register holds ASCII data while another has binary information and yet a third has a signed number, then each register can be set to display the appropriate type of data. This, you will find, is invaluable. A third, though less important problem, is assembler overhead. There is a certain structure that must be followed to get the program to assemble and run correctly. This has been provided in template files. All you need to do is copy the appropriate template file and put the code in a predefined location to get your program to run correctly. For simple programs this can cut your work in half. It also minimizes the number of typos. "The PC Assembler Tutor" is built around the Helper. It systematically goes through the 8086 instruction set, having you write small programs to illustrate how each instruction works. At The PC Assembler Tutor 2 ______________________ the end you should have a feeling for all the instructions except a few which involve the 8087 or peripheral hardware. These will be mentioned, but not used. You will be a better programmer if you know how all the instructions work. There are times when one specific instruction is just what you want. If you don't know that it exists or how it works, you won't use it. This way, if you run across a situation where you think that a certain instruction might be useful to you, you can go back to the Tutor to refresh your memory and be able to put the instruction to use almost immediately. If you are a beginner, I feel confident that you will learn faster and more thoroughly than with any other method. If you know some assembler but would like to know more, I'm sure that there is lots that would interest you. AAD SBB XLAT REPNE SCAS Do you know what these are? What about segment overrides? Do you know when to use them and when to avoid them? Do you know ALL the allowable addressing modes? What actually is an ASSUME statement? In order to let you see if you find the material interesting, you may go through chapters 0 - 4 chapters without any obligation. You may also make an archival copy of the disks (you are urged to do so). If you continue after the fourth chapter then please register by sending $9.95 (or $10.60 for Californians) to Nelsoft. The registration form is at the end of this introduction. If I followed the pricing structure of other people I would be charging several times as much. My goal is different. I want everyone who can benefit from the program and tutorial to use them, and I want everyone who uses them to do so legally. Therefore, I have priced them so that everyone can pay for them without any inconvenience. If you use them, you can certainly afford my minimal price. The material is sequential. Chapter 0 should be read before starting on the other chapters and the chapters should be read in order. Appendix 1 contains all the subroutine calls in "The Assembler Helper" and how to access them. Appendix 2 is an alphabetical list of all the 8086 instructions, telling what they do and showing all allowable syntaxes. Appendix 3 gives the speed of all instructions along with a list of which flags are affected (you will learn what this means in the Tutor). It is to your benefit to start at the beginning and work your way through. Chapter 0 contains material that you need to know, so you must read it. The text has been broken up into sections so that no printout is longer than 10 pages or so. All text files have a file extension .DOC. If a chapter is much longer than Introduction 3 ____________ that, it will be divided into parts, indicated by -1, -2, -3 after the chapter number. These files should be printable with the DOS 'print' command. The only imbedded printer command code is form feed for the next page. The text runs about 2500 characters a page, so you can estimate the size of the printout from the size of the text file. Curly brackets in the text denote a footnote.{1} Some of the footnotes are technical and will be understood by only a quarter of the people. If you are one of that quarter, fine. If not, the important thing is not that you understand the outline of the proof, but that you believe that what is being proved is true. The assembler level is for those who have some degree of intellegence. You have an unparalleled opportunity to screw things up at this level. If you got Cs and Ds in high school algebra because you didn't quite understand what was going on, then you probably shouldn't do assembler programming.{2} In addition, I assume that you have done a lot of programming, preferably in either Pascal or C. BASIC is a nice language, but it is missing a certain type of structure which is vital for creating robust code in assembler language. If BASIC is all you know, I would recommend that you learn C first and then come back to assembler. You will be a better programmer for it.{3} Finally, "The Assembler Helper" assumes that it has control of the screen. If you are hooked up to a debugger, there may be a conflict. There is a subroutine in the Helper called "set_timer" which may help minimize this conflict. You need to be in chapter 5 or so before you will be able to use it. See \APPENDIX\APP1.DOC for details. If you are ready to go, please look at the following two pages and then read INTRO2.DOC. It will explain a little about what an assembler is. I hope you enjoy using the Helper and the Tutor as much as I enjoyed writing them. Chuck Nelson ____________________ 1. Like this one. 2. If you got Cs and Ds because you were too busy reading "Tales from the Crypt" and Isaac Asimov, that's something entirely different. 3. On re-reading this I decided that it is true, but pretensious. If you like BASIC and program well in BASIC, then you should learn assembler and continue using BASIC. There are certain inherent difficulties with BASIC, so before you start you should read BAS1.DOC. This is on DISK2. The PC Assembler Tutor 4 ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson All rights reserved Microsoft (R) Macro Assembler and Microsoft (R) Overlay Linker are registered trademarks of Microsoft Corporation. This manual contains screen output of the Macro Assembler and the Overlay Linker. Screen shots (C) 1981-1988 Microsoft Corporation. It also contains excerpts from Macro Assembler .LST files and Overlay Linker .MAP files. Portions of these files are Copyright (C) 1981-1988 Microsoft Corporation. Used with permission of Microsoft Corporation. TRADEMARK ACKNOWLEDGEMENT IBM is a registered trademark of International Business Machines Inc. Intel is a registered trademark of Intel Corporation. Macintosh is a registered trademark of Apple Computer, Inc. Microsoft is a registered trademark of Microsoft Corporation. Motorola is a registered trademark of Motorola, Inc. 8086 is a trademark of Intel Corporation. Codeview is a registered trademark of Microsoft Corporation. QuickC is a registered trademark of Microsoft Corporation. Turbo Pascal, Turbo Assembler and Turbo Debugger are registered trademarks of Borland International. The PC Assembler Helper was designed as a learning tool. It is meant to be used in conjunction with simple assembler programs to display the results of individual assembler instructions. It should not be used with high-level languages nor with programs that modify the screen. HELPMEM.COM, the memory resident version, uses the same interrupts as a debugger. Therefore, if there is a debugger attatched to any program that is being used, HELPMEM.COM should not be loaded into memory. WARRANTY THIS PROGRAM, INSTRUCTION MANUAL, AND REFERENCE MATERIALS ARE SOLD "AS IS", WITHOUT WARRANTY AS TO THEIR PERFORMANCE, MERCHANTABILITY, OR FITNESS FOR ANY PARTICULAR PURPOSE. THE ENTIRE RISK AS TO THE RESULTS AND PERFORMANCE OF THESE PROGRAMS IS ASSUMED BY YOU. ***************************************************************** REGISTRATION Hey, Chuck, I'm no chump! I'm using your programs/manual, and I want to pay my fair share. Please make me a registered user of "The PC Assembler Tutor" and "The PC Assembler Helper". Enclosed is a check for $9.95 (plus 6.5% tax or $10.60 for California residents). Say, that's cheaper than a large pizza! Name_________________________________________________________ Last First Initial Address______________________________________________________ Street Address _______________________________________________________ City, State, and Zip Code I got my copy from ___________________________________________ Make checks payable to NELSOFT and send your registration to: NELSOFT P.O. Box 21389 Oakland, CA 94620 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ REGISTRATION BENEFITS As a registered user of "The PC Assembler Helper" and "The PC Assembler Tutor" you are entitled to: 1) Use asmhelp.obj and helpmem.com for personal use. 2) Make 1 (one) printer copy of "The PC Assembler Tutor". 3) Use all programs in "The PC Assembler Tutor" for personal use. 4) Make an archival copy of the disks. 5) Distribute UNALTERED disks to friends for their perusal. 6) Use any updates to either "The PC Assembler Helper" or "The PC Assembler Tutor" under the same registration conditions. Though copies of the disk may be given away if there is no charge, it is illegal to charge for redistribution of the disk or its contents without permission of the author. Under no circumstances may you distribute printed copies of "The PC Assembler Tutor". If you intend to charge for distributing the disk or its information, please read and sign the following distribution agreement. ***************************************************************** DISTRIBUTION LICENSING AGREEMENT FOR THE PC ASSEMBLER HELPER AND THE PC ASSEMBLER TUTOR Anyone wishing to charge people a fee for giving them a copy of The PC Assembler Helper and/or The PC Assembler Tutor must have the written authorization of the author, without which the distributor is guilty of copyright violation. To receive such authorization, send this completed application, along with a copy of your software library's order form to: NELSOFT P.O. Box 21389 Oakland, CA 94620 If you want a distribution disk with the latest copy of these programs, please include $7.00 to cover the cost of the disks, mailing and handling. (This offer is for bona fide user groups and shareware distributors only). NAME OF ORGANIZATION ___________________________________________ YOUR NAME ______________________________________________________ ADDRESS ________________________________________________________ CITY, STATE ____________________________________________________ TERMS OF DISTRIBUTION 1. The fee charged for each disk may not exceed $7.00. On high-density disks, the fee may not be over $10.00. 2. Your library's catalog or listing must state that this material is not free, but is copyrighted material that is provided to allow the user to evaluate it before paying. 3. The offering and sale of disks containing The PC Assembler Helper and The PC Assembler Tutor will be stopped at any time the author so requests. 4. The Tutor and the Helper must be distributed together. The compressed files and the information document must remain in the subdirectory \PCTUTOR. There may be no additional files in this subdirectory. Both the name and the contents of \PCREADME.DOC must remain unaltered. 5. Problems or complaints will be reported to the author. In return for the right to charge a fee for the distribution of The PC Assembler Helper and The PC Assembler Tutor, I agree to comply with the above terms of distribution. Signed, __________________________________________ __________________ Your Signature Date Introduction N.2 ================ 5 WHAT'S AN ASSEMBLER? What is the difference between a compiler and an assembler? A compiler is a program that takes the source code you have written and turns it into machine language instructions that are usable by the computer. A machine language instruction is a binary number that tells the computer to do one specific thing. This is something very specific like: add 1 to a specific variable. The compiler does this in two steps. First, the compiler takes each line of source code and turns the expression on the line into a number of simple tasks to accomplish what is desired, generating a number of assembler TEXT instructions and data definitions. When the compiler is through with the source code, it then takes these TEXT instructions and assembles them to form an object module. An object module is a file that can be linked with other files to form a larger program. Why two steps instead of one? There are several reasons. This allows a compiler writer to write a first part for any language like Pascal, C, BASIC, etc., and then use the same assembler part. This saves development time in a company that has more than one type of compiler. You can insert a new assembler part without worrying about the text generation part, or you can insert a new text generation part without worrying about the assembler part. Secondly, though an assembler is not a simple program, the compiler's text generator is even more complicated. Putting the two together is like trying to juggle eight balls instead of four balls. This leads to the most vital reason. Not only would a unified text-generator/assembler be more error prone, it would be almost impossible to debug. If you are getting an error due to one type of Pascal instruction, is this because it is being misunderstood by the compiler or because the compiler is giving it the wrong machine codes? In the two part system, the compiler writer can look at the intermediate text code and isolate the problem into one of the two halves. An assembler is a program that takes a TEXT file where each line corresponds to a specific machine instruction or type of data, calculates the address in memory where each piece of data or machine instruction will be, translates each instruction and piece of data into machine readable form, and inserts the addresses of data and labels into machine instructions where appropriate.{1} ____________________ 1. A label is just a name which marks a certain spot in the assembler code. The PC Assembler Tutor 6 ______________________ The text name for a machine instruction is called a mnemonic.{2} It indicates what is being done by the instruction. Which would you rather use for multiplication, 'MUL' or '11110111xx100xxx'? These 'x's indicate a digit whose value depends on where something is in memory. For each mnemonic there is a single machine instruction which performs the operation. This means that the assembler's task is relatively simple. It only needs to allocate space for all the variables and instructions, to translate each mnemonic and data value into its corresponding machine code, and finally put it into a machine usable file. Here you need to know what different forms of file there are. 1) An executable (.EXE) file contains certain information for the operating system when the program is started. This allows the program to be as large as is wanted. 2) A .COM file contains no information for the operating system. When the operating system starts a .COM file it simply puts it in memory and starts it. Files with a .COM extension are limited to a length of 64k bytes. 3) Binary files are files which must be loaded into a .COM or .EXE program before being run. They cannot be used by themselves. They are archaic. They are a crutch for those compilers which don't support .OBJ files, and are disappearing. 4) An object (.OBJ) file is a section of a program. It contains code and variables, but also contains information that can be used to combine it with other object files into a larger program. A linker can convert one or more object files into an executable file. Things have moved along in the past few years. TurboPascal 3.0 generated .COM files. This wasn't because .COM files were superior but because it was too difficult to generate the extra information needed to produce an .OBJ file. Interpreted BASIC required binary files because it did not have the ability to use .OBJ files. The situation now is: If you want to link with the current Turbo Pascal, you should use an .OBJ file. If you want to link with QuickC, you need an .OBJ file. If you want to combine an external subprogram with QuickBASIC, you need an .OBJ file. Get the picture? No assembler makes .EXE files. If you have a single file that you want made onto a stand alone .EXE file, you first make an .OBJ file and then use that single file with LINK. ____________________ 2. You don't pronounce that first 'm'. Introduction 7 ____________ When making a .COM file, the normal route is to make .OBJ files, link them together into an EXE file, and then convert them to a .COM file with a program called EXE2BIN. This allows you to divide the problem into a number of subproblems and put them all together at the end. As you can see from the above, the job for the assembler is to take a text file and convert it into an .OBJ file. The three assemblers that you are most likely to have are MASM, A86, and TurboAssembler. They all produce object files. Since assemblers simply supply the machine code for each instruction, they will all produce the exact same code.{3} This is one of the differences between assemblers and compilers. The text generation phase of the compiler requires creativity. It is the compiler writer's idea of how to solve a certain problem in a specific language. This generated text is copyrighted, and you need a license to distribute a program that includes this generated text. An assembler. on the other hand, is just a drudge. If you had a book with the machine codes and had enough time, you could produce the same file byte for byte that the assembler would produce for a .COM file. There is no creativity involved in the generated code, and there is no license involved in its distribution. There is some difference in the speed of those three assemblers, but I'll have a comment about that after I give you the numbers. These numbers are from the time of hitting the ENTER key to the return of the DOS prompt ('>'). These are on a low speed machine so your numbers should be better, but won't be any worse. A86 TurboASM MASM one page of code 3.2 8.8 10.3 20 pages of code 7.3 12.7 20.4 60 pages of code 12.5 22.9 43.3 All these numbers are in seconds. A86 is the fastest. It loads faster because it itself is a .COM file, and it works faster. But even the slowest (MASM), is fast enough. How long does it take to write 60 pages of code? Probably a week or two. Writing assembler code is normally slower than writing code for a high-level language. Even the slowest finishes the assembly in well under a minute. Think of the time it would take to compile 60 pages of Pascal code. In fact, the normal length of a file will be from 10 to 20 pages, so these are the numbers you need to think of. These will be the smaller .OBJ files which are linked together by the linker. If you have one of these assemblers you don't need anything different. If you need to get one, you can use this information ____________________ 3. Or functionally the same code. Sometimes there are two different instructions that do the same thing, just as 6+1, 5+2 and 4+3 all produce 7. The PC Assembler Tutor 8 ______________________ to help you in your selection. The following prices are as of mid-1990. A86 is available through shareware. It costs $50.00 for the assembler alone, $80.00 with the debugger. Add another $10.00 if you want a printed manual. Both MASM and Turbo Assembler come with assembler, debugger, a number of utility programs and several manuals. They both retail for $150.00, but even a quick glance at BYTE magazine will find you a place that is selling them for $105.00 - $110.00. They come bundled, so you cannot buy the assembler without the debugger (The Microsoft debugger is Codeview and the Borland debugger is Turbo Debugger). Speaking of debuggers, you may be thinking, "Well, I have DEBUG, so why do I want another debugger?" DEBUG has been outdated for several years now. It has been supplanted by symbolic debuggers which associate code with specific lines in the original text file. You give the symbolic debugger the .EXE file along with the text files that produced it, and it shows you your source code as you go along. Here's a section of code we will meet later in the Tutor: ; - - - - - - - - - - - - - - - - - - - - start: push ds ; set up for return sub ax,ax push ax mov ax, DATASTUFF ; load ds mov ds,ax outer_loop: lea ax, multiplicand ; load multiplicand call get_unsigned_8byte call print_unsigned_8byte call get_unsigned ; unsigned word to multiplier mov multiplier, ax lea si, multiplicand ; load pointers lea bx, result ; - - - - - - - - - - - - - - - - - - - - Don't worry about what these instructions do. You'll learn that later. Here is DEBUG's idea of what is going on: ******************** DEBUG SCREEN SHOT ************************ -r AX=0000 BX=0000 CX=2749 DX=0000 SP=0A00 BP=0000 SI=0000 DI=0000 DS=0D7E ES=0D7E SS=0D8E CS=0E7F NV UP DI PL NZ NA PO NC 0E7F:0000 1E PUSH DS -u0E7F:0000 2A 0E7F:0000 1E PUSH DS 0E7F:0001 2BC0 SUB AX,AX 0E7F:0003 50 PUSH AX 0E7F:0004 B82E0E MOV AX,0E2E 0E7F:0007 8ED8 MOV DS,AX 0E7F:0009 8D060800 LEA AX,[0008] Introduction 9 ____________ 0E7F:000D E80C14 CALL 141C 0E7F:0010 E84B11 CALL 115E 0E7F:0013 E8F505 CALL 060B 0E7F:0016 A31000 MOV [0010],AX 0E7F:0019 8D360800 LEA SI,[0008] 0E7F:001D 8D1E1200 LEA BX,[0012] ********************** END DEBUG ******************************* Part of this is understandable, but a lot of it is confusing, and you have lost the concept of what you are trying to do. To see what a symbolic debugger does with the same code, here is the Turbo Debugger's idea of what is happening: ************************* TURBO SCREEN SHOT {4} ***************************** File View Run Breakpoints Data Window Options READY .Module: debugtst File: debugtst.asm 74....................................1 . . start: push ds ; set up for return .Registers......3. . sub ax,ax . ax 5C94 .c=0. . push ax . bx 0000 .z=1. . . cx 0000 .s=0. . mov ax, DATASTUFF ; load ds . dx 0000 .o=0. . mov ds,ax . si 0000 .p=1. . . di 0000 .a=0. . outer_loop: . bp 0000 .i=1. . lea ax, multiplicand ; load multipli. sp 09FA .d=0. . call get_unsigned_8byte . ds 4AD6 . . . call print_unsigned_8byte . es 4A26 . . . call get_unsigned ; unsigned word. ss 4A36 . . . mov multiplier, ax . cs 4B27 . . . . ip 0019 . . . .................. . lea si, multiplicand ; load pointers ............................................................................. .Watches....................................................................2 .multiplier,d 23700 .multiplicand qword 00000042E843515D ............................................................................. F1-Help F2-Bkpt F3-Close F4-Here F5-Zoom F6-Next F7-Trace F8-Step F9-Run ***************************************************************************** DEBUG really doesn't meet our needs. Of course, those of you who still use EDLIN to write 20 page documents should feel free to use DEBUG. Modern language compilers have their own debuggers in their environments, so we only need a debugger for the assembler. Turbo Debugger and Code View do the job very well. D86 is better than DEBUG but it has some problems. ____________________ 4. Turbo Debugger is (C) Copyright 1988-1989 Borland International. The PC Assembler Tutor 10 ______________________ If you actually want a debugger that will symbolically debug everything that supports symbolic debugging, you might want to take a look at the Turbo Debugger. It has more power than you are ever likely to need. "The Assembler Helper", the program which comes with the Tutor, is NOT a debugger. Debuggers are designed to show you what is happening with your code; the Helper is designed to show you what is happening with the 8086. It's a fundamental difference in outlook. If you want to use a debugger while you are doing this tutorial it is possible but the results are not guaranteed. Please see DEBUGGER.DOC in \COMMENTS for some information about the different debuggers and how to use ASMHELP with a debugger. You should not try to do this before you are in chapter 5 or 6. You may have noticed that CHASM, an assembler distributed through shareware, was not listed. There is a reason for this. It can't produce .OBJ files, so it cannot produce the standard files for use with current compilers, including QuickBASIC. CHASM is also unusable with this tutorial because it cannot produce files to link with ASMHELP.OBJ, the i/o interface program. If you are getting a shareware assembler, get A86. It's a quality assembler. This tutorial was originally written for those using MASM. In order to allow those using the Turbo Assembler and A86 to follow along, there is a document for each one in \COMMENTS which explains any differences between what is in the chapters (which use MASM as an example) and what the respective assemblers do. There aren't that many differences. The pathnames are \COMMENTS\TURBO.DOC and \COMMENTS\A86.DOC. 1 TABLE OF CONTENTS Chapter 0.1 - Numbers And Arithmetic . . . . . . . . . . . i Base 10 Machine . . . . . . . . . . . . . . . . . . . . i Negative Numbers . . . . . . . . . . . . . . . . . . . ii 10's Complement . . . . . . . . . . . . . . . . . . . iii Addition . . . . . . . . . . . . . . . . . . . . . . . v Subtraction . . . . . . . . . . . . . . . . . . . . . . v Modular Math . . . . . . . . . . . . . . . . . . . . . vi Sign Extension . . . . . . . . . . . . . . . . . . . . ix Overflow . . . . . . . . . . . . . . . . . . . . . . xii Multiplication . . . . . . . . . . . . . . . . . . . xiii Division . . . . . . . . . . . . . . . . . . . . . . xiv Chapter 0.2 - Bases 2 And 16 . . . . . . . . . . . . . . . xv Base Conversion . . . . . . . . . . . . . . . . . . . . xv Binary Math . . . . . . . . . . . . . . . . . . . . . xvi 2's Complement . . . . . . . . . . . . . . . . . . . xvii Sign Extension . . . . . . . . . . . . . . . . . . xviii Chapter 0.3 - Logic . . . . . . . . . . . . . . . . . . . xxi AND . . . . . . . . . . . . . . . . . . . . . . . . . xxi OR . . . . . . . . . . . . . . . . . . . . . . . . . xxii XOR . . . . . . . . . . . . . . . . . . . . . . . . . xxii NOT . . . . . . . . . . . . . . . . . . . . . . . . xxiii Chapter 0.4 - Memory . . . . . . . . . . . . . . . . . . xxv Segmentation . . . . . . . . . . . . . . . . . . . . xxv Numbers In Memory . . . . . . . . . . . . . . . . . xxvii Chapter 0.5 - Style . . . . . . . . . . . . . . . . . . . xxix Chapter 1 - Some Simple Programs . . . . . . . . . . . . . 1 Label . . . . . . . . . . . . . . . . . . . . . . . . . 3 CALL . . . . . . . . . . . . . . . . . . . . . . . . . 3 JMP . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Chapter 2 - Data . . . . . . . . . . . . . . . . . . . . . 11 DB, DW, DD, DQ, DT, DF . . . . . . . . . . . . . . . . 11 Definition of Constants . . . . . . . . . . . . . . . . 13 Chapter 3 - Asmhelp . . . . . . . . . . . . . . . . . . . . 16 Registers . . . . . . . . . . . . . . . . . . . . . . . 16 Show_regs . . . . . . . . . . . . . . . . . . . . . . . 17 MOV . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Chapter 4 - Show_regs . . . . . . . . . . . . . . . . . . . 24 Show_reg Codes . . . . . . . . . . . . . . . . . . . . 29 Chapter 5 - Addition and Subtraction . . . . . . . . . . . 31 LOOP . . . . . . . . . . . . . . . . . . . . . . . . . 31 The PC Assembler Tutor 2 ______________________ OF, ZF, SF, CF . . . . . . . . . . . . . . . . . . . . 32 ADD . . . . . . . . . . . . . . . . . . . . . . . . . . 33 PUSH . . . . . . . . . . . . . . . . . . . . . . . . . 33 POP . . . . . . . . . . . . . . . . . . . . . . . . . . 34 SUB . . . . . . . . . . . . . . . . . . . . . . . . . . 37 JC, JNC, JO, JNO . . . . . . . . . . . . . . . . . . . 38 INTO . . . . . . . . . . . . . . . . . . . . . . . . . 39 INTO.COM . . . . . . . . . . . . . . . . . . . . . . . 39 Chapter 6 - Multiplication and Division . . . . . . . . . . 41 MUL . . . . . . . . . . . . . . . . . . . . . . . . . . 41 IMUL . . . . . . . . . . . . . . . . . . . . . . . . . 41 DIV . . . . . . . . . . . . . . . . . . . . . . . . . . 44 IDIV . . . . . . . . . . . . . . . . . . . . . . . . . 44 Chapter 7 - Logic . . . . . . . . . . . . . . . . . . . . . 47 AND . . . . . . . . . . . . . . . . . . . . . . . . . . 47 TEST . . . . . . . . . . . . . . . . . . . . . . . . . 49 OR . . . . . . . . . . . . . . . . . . . . . . . . . . 50 XOR . . . . . . . . . . . . . . . . . . . . . . . . . . 50 NEG . . . . . . . . . . . . . . . . . . . . . . . . . . 51 NOT . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Masks . . . . . . . . . . . . . . . . . . . . . . . . . 52 Chapter 8 - Shift and Rotate . . . . . . . . . . . . . . . 56 SAL, SHL . . . . . . . . . . . . . . . . . . . . . . . 56 INC, DEC . . . . . . . . . . . . . . . . . . . . . . . 57 SHR . . . . . . . . . . . . . . . . . . . . . . . . . . 58 SAR . . . . . . . . . . . . . . . . . . . . . . . . . . 59 ROL, ROR . . . . . . . . . . . . . . . . . . . . . . . 60 RCL, RCR . . . . . . . . . . . . . . . . . . . . . . . 61 Chapter 9 - Jumps . . . . . . . . . . . . . . . . . . . . . 68 CMP . . . . . . . . . . . . . . . . . . . . . . . . . . 68 Signed and Unsigned Conditional Jumps . . . . . . . . . 70 Flag Conditional Jumps . . . . . . . . . . . . . . . . 75 JCXZ . . . . . . . . . . . . . . . . . . . . . . . . . 75 Chapter 10 - Templates . . . . . . . . . . . . . . . . . . 78 .LST File . . . . . . . . . . . . . . . . . . . . . . . 78 SEGMENTS . . . . . . . . . . . . . . . . . . . . . . . 84 PUBLIC (SEGMENTS) . . . . . . . . . . . . . . . . . . . 85 CLASS . . . . . . . . . . . . . . . . . . . . . . . . 85ff ENDS . . . . . . . . . . . . . . . . . . . . . . . . . 92 ASSUME . . . . . . . . . . . . . . . . . . . . . . . . 93 Segment Overrides . . . . . . . . . . . . . . . . . . . 93 Subroutines . . . . . . . . . . . . . . . . . . . . . . 96 END . . . . . . . . . . . . . . . . . . . . . . . . . . 97 RET . . . . . . . . . . . . . . . . . . . . . . . . . . 98 EXTRN . . . . . . . . . . . . . . . . . . . . . . . . . 99 STACK . . . . . . . . . . . . . . . . . . . . . . . . 101 Chapter 11 - Addressing Modes . . . . . . . . . . . . . . 104 EQU . . . . . . . . . . . . . . . . . . . . . . . . . 110 All Addressing Modes . . . . . . . . . . . . . . . . 114 OFFSET . . . . . . . . . . . . . . . . . . . . . . . 118 SEG . . . . . . . . . . . . . . . . . . . . . . . . . 118 Table Of Contents 3 _________________ LEA . . . . . . . . . . . . . . . . . . . . . . . . . 118 Chapter 12 - Multiple Word Arithmetic I . . . . . . . . . 122 ADC . . . . . . . . . . . . . . . . . . . . . . . . . 123 CLC . . . . . . . . . . . . . . . . . . . . . . . . . 124 SBB . . . . . . . . . . . . . . . . . . . . . . . . . 126 Chapter 13 - Multiple Word Arithmetic II . . . . . . . . 129 Unsigned Multiplication . . . . . . . . . . . . . . . 129 Unsigned Division . . . . . . . . . . . . . . . . . . 130 Chapter 14 - Zoom . . . . . . . . . . . . . . . . . . . . 134 Chapter 15 - Subroutines . . . . . . . . . . . . . . . . 137 PUSHREGS.MAC . . . . . . . . . . . . . . . . . . . . 137 EXTRN in Subroutines . . . . . . . . . . . . . . . . 142 Passing Data . . . . . . . . . . . . . . . . . . . . 144 Near and Far Procedures . . . . . . . . . . . . . . . 145 The Stack . . . . . . . . . . . . . . . . . . . . . . 148 Types of Returns . . . . . . . . . . . . . . . . . . 153 PUSHREGS . . . . . . . . . . . . . . . . . . . . . . 154 POPREGS . . . . . . . . . . . . . . . . . . . . . . . 154 LDS . . . . . . . . . . . . . . . . . . . . . . . . . 157 LES . . . . . . . . . . . . . . . . . . . . . . . . . 157 Towers of Hanoi . . . . . . . . . . . . . . . . . . . 162 Summary . . . . . . . . . . . . . . . . . . . . . . . 166 Chapter 16 - Long Signed Multiplication And Division . . 170 Long Negation . . . . . . . . . . . . . . . . . . . . 170 Chapter 17 - Interrupts . . . . . . . . . . . . . . . . . 177 INT . . . . . . . . . . . . . . . . . . . . . . . . . 177 NMI . . . . . . . . . . . . . . . . . . . . . . . . . 181 IEF . . . . . . . . . . . . . . . . . . . . . . . . . 181 STI . . . . . . . . . . . . . . . . . . . . . . . . . 181 CLI . . . . . . . . . . . . . . . . . . . . . . . . . 181 INT 3 . . . . . . . . . . . . . . . . . . . . . . . . 182 Chapter 18 - Ports . . . . . . . . . . . . . . . . . . . 185 IN . . . . . . . . . . . . . . . . . . . . . . . . . 185 OUT . . . . . . . . . . . . . . . . . . . . . . . . . 186 Parity . . . . . . . . . . . . . . . . . . . . . . . 186 Chapter 19 - Strings . . . . . . . . . . . . . . . . . . 191 SCAS . . . . . . . . . . . . . . . . . . . . . . . . 191 DF . . . . . . . . . . . . . . . . . . . . . . . . . 191 REP/REPE/REPNE . . . . . . . . . . . . . . . . . . . 195 STOS . . . . . . . . . . . . . . . . . . . . . . . . 196 LODS . . . . . . . . . . . . . . . . . . . . . . . . 198 MOVS . . . . . . . . . . . . . . . . . . . . . . . . 199 CMPS . . . . . . . . . . . . . . . . . . . . . . . . 204 Segment Overrides . . . . . . . . . . . . . . . . . . 207 REP and Overrides . . . . . . . . . . . . . . . . . . 209 Chapter 20 - Control Structures . . . . . . . . . . . . . 212 IF . . . . . . . . . . . . . . . . . . . . . . . . . 212 WHILE . . . . . . . . . . . . . . . . . . . . . . . . 214 The PC Assembler Tutor 4 ______________________ DO-WHILE . . . . . . . . . . . . . . . . . . . . . . 215 BREAK . . . . . . . . . . . . . . . . . . . . . . . . 215 CONTINUE . . . . . . . . . . . . . . . . . . . . . . 215 FOR . . . . . . . . . . . . . . . . . . . . . . . . . 216 SWITCH . . . . . . . . . . . . . . . . . . . . . . . 217 Chapter 21 - .COM Files . . . . . . . . . . . . . . . . . 219 .COM Template . . . . . . . . . . . . . . . . . . . . 219 PSP . . . . . . . . . . . . . . . . . . . . . . . . . 220 ASSUME . . . . . . . . . . . . . . . . . . . . . . . 221 Phase Errors . . . . . . . . . . . . . . . . . . . . 221 Chapter 22 - BCD Numbers . . . . . . . . . . . . . . . . 229 Unpacked BCD . . . . . . . . . . . . . . . . . . . . 229 Packed BCD . . . . . . . . . . . . . . . . . . . . . 230 DAA . . . . . . . . . . . . . . . . . . . . . . . . . 232 DAS . . . . . . . . . . . . . . . . . . . . . . . . . 233 AAA . . . . . . . . . . . . . . . . . . . . . . . . . 240 AAS . . . . . . . . . . . . . . . . . . . . . . . . . 242 AAM . . . . . . . . . . . . . . . . . . . . . . . . . 242 AAD . . . . . . . . . . . . . . . . . . . . . . . . . 243 Unpacking . . . . . . . . . . . . . . . . . . . . . . 245 Packing . . . . . . . . . . . . . . . . . . . . . . . 246 Chapter 23 - XLAT . . . . . . . . . . . . . . . . . . . . 253 EBCDIC Numbers . . . . . . . . . . . . . . . . . . . 253 XLAT . . . . . . . . . . . . . . . . . . . . . . . . 253 Translation Table . . . . . . . . . . . . . . . . . . 253 Chapter 24 - Miscellaneous Instructions . . . . . . . . . 264 XCHG . . . . . . . . . . . . . . . . . . . . . . . . 264 ESC . . . . . . . . . . . . . . . . . . . . . . . . . 264 WAIT . . . . . . . . . . . . . . . . . . . . . . . . 264 FWAIT . . . . . . . . . . . . . . . . . . . . . . . . 264 LOCK . . . . . . . . . . . . . . . . . . . . . . . . 265 LOOPE/LOOPNE . . . . . . . . . . . . . . . . . . . . 265 HALT . . . . . . . . . . . . . . . . . . . . . . . . 266 CMC . . . . . . . . . . . . . . . . . . . . . . . . . 266 LAHF . . . . . . . . . . . . . . . . . . . . . . . . 266 SAHF . . . . . . . . . . . . . . . . . . . . . . . . 267 NOP . . . . . . . . . . . . . . . . . . . . . . . . . 267 Chapter 25 - What Does It All Mean? . . . . . . . . . . . 268 Interrupts . . . . . . . . . . . . . . . . . . . . . 269 Data Bus . . . . . . . . . . . . . . . . . . . . . . 273 Alignment Type . . . . . . . . . . . . . . . . . . . 274 Chapter 26 - Simplifying The Template . . . . . . . . . . 276 INT 21h Function 4Ch . . . . . . . . . . . . . . . . 276 Exit Code . . . . . . . . . . . . . . . . . . . . . . 276 Standardized Segments . . . . . . . . . . . . . . . . 277 _DATA . . . . . . . . . . . . . . . . . . . . . . . . 279 _BSS . . . . . . . . . . . . . . . . . . . . . . . . 279 CONST . . . . . . . . . . . . . . . . . . . . . . . . 280 Literals . . . . . . . . . . . . . . . . . . . . . . 280 STACK . . . . . . . . . . . . . . . . . . . . . . . . 280 Groups . . . . . . . . . . . . . . . . . . . . . . . 280 Table Of Contents 5 _________________ DGROUP . . . . . . . . . . . . . . . . . . . . . . . 281 Groups and OFFSET . . . . . . . . . . . . . . . . . . 284 Standardized Segment Names . . . . . . . . . . . . . 287 Standardized Segment Directives . . . . . . . . . . . 288 .MODEL Names . . . . . . . . . . . . . . . . . . . . 291 Summary . . . . . . . . . . . . . . . . . . . . . . . 293 APPENDIX Appendix I - The PC Assembler Helper . . . . . . . . . . . i Appendix II - The 8086 Instruction Set . . . . . . . . . xiii Appendix III - Instruction Speed And Flags . . . . . . xxvii ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ANCILLARY MATERIAL DEBUGGERS (DEBUGGER.DOC) TASM (TASM.DOC) A86 (A86.DOC) BASIC (BASIC1.DOC, BASIC2-1.DOC, BASIC2-2.DOC) Using Basic . . . . . . . . . . . . . . . . . . . . . . . mmi Variable Typing . . . . . . . . . . . . . . . . . . . mmi Default Type . . . . . . . . . . . . . . . . . . . . mmii Interfacing Basic With Assembler . . . . . . . . . . . mmvii Memory Allocation . . . . . . . . . . . . . . . . . mmvii String Allocation . . . . . . . . . . . . . . . . . mmviii Data Output . . . . . . . . . . . . . . . . . . . . . mmx FIELD . . . . . . . . . . . . . . . . . . . . . . . mmxii LSET/MID$/RSET . . . . . . . . . . . . . . . . . . mmxiii MKI$, MKL$, MKS$, MKD$ . . . . . . . . . . . . . . mmxiii CVI, CVL, CVS, CVD . . . . . . . . . . . . . . . . mmxiii STR$, VAL . . . . . . . . . . . . . . . . . . . . . mmxiv VARPTR . . . . . . . . . . . . . . . . . . . . . . mmxvi PTR86 (Basic 3.0) . . . . . . . . . . . . . . . . . mmxvii BUILDLIB.EXE . . . . . . . . . . . . . . . . . . . mmxvii VARSEG (Basic 4.0) . . . . . . . . . . . . . . . mmxviii Basic Calling Conventions . . . . . . . . . . . . . mmxix INT86 . . . . . . . . . . . . . . . . . . . . . . mxxviii SADD . . . . . . . . . . . . . . . . . . . . . . . mmxxix BLOAD . . . . . . . . . . . . . . . . . . . . . . . mmxxx Summary . . . . . . . . . . . . . . . . . . . . . . mmxxxi Interfacing Basic With Assembler . . . . . . . . . mmvii Ancillary Programs (MISHMASH.DOC) General Block Move . . . . . . . . . . . . . . . . . . 1 Block Multiplication . . . . . . . . . . . . . . . . . 3 Binary Multiplication . . . . . . . . . . . . . . . . . 6 Binary Division . . . . . . . . . . . . . . . . . . . . 9 Chapter 0.1 - Numbers and Arithmetic ==================================== i You don't habitually use the base two system to balance your checkbook, so it would be counterproductive to teach you machine arithmetic on a base two system. What number systems have you had a lot of experience with? The base 10 system springs to mind. I'm going to show you what happens on a base 10 system so you will understand the structure of what happens with computer arithmetic. BASE 10 MACHINE Each place inside the microprocessor that can hold a number is called a REGISTER. Normally there are a dozen or so of these. Our base 10 machine has 4 digit registers. They can represent any number from 0000 to 9999. They are exactly like an industrial counters or the counters on your tape machines.{1} If you add 27 to a register, the microprocessor counts forward 27; if you subtract 153 from a register, the microprocessor counts backwards 153. Every time you add 1 to a register, it increments by 1 - that is 0245, 0246, 0247, 0248. Every time you subtract 1 from a register, it decrements by 1 - that is 3480, 3479, 3478, 3477. Let's do some more incrementing. 9997, 9998, 9999, 0000, 0001, 0002. Whoops! That's a problem. When the register reaches 9999 and we add 1, it changes to 0000, not 10,000. How can we tell the difference between 0000 and 10,000? We can't without a little help from the CPU.{2} Immediately after an arithmetical operation, the CPU knows whether you have gone through 10,000 (9999->0000). The CPU has something called a carry flag. It is internal to the CPU and can have the value 0 or 1. After each arithmetical operation, the CPU sets the CARRY FLAG to 1 if you went through the 9999/0000 boundary, and sets the carry flag to 0 if you didn't.{3} Here are some examples, showing addition, the result, and the carry flag. The carry flag is normally abbreviated by CF. number 1 number 2 result CF 0289 4782 5071 0 4398 2964 7382 0 8177 5826 4003 1 ____________________ 1. Exactly like industrial counters that have several hundred thousand parts, that is. 2. The CPU (central processing unit) is the chip(s) that does all the arithmetic. In the case of the PC, it is the 8086. 3. When you set a flag to 0, it is called CLEARING the flag. The PC Assembler Tutor ii ______________________ 6744 4208 0952 1 Note that you must check the carry flag immediately after the arithmetical operation. If you wait, the CPU will reset it after the next arithmetical operation. Now let's do some decrementing. 0003, 0002, 0001, 0000, 9999, 9998. Golly gosh! Another problem. When we got to 0000, rather than getting -1, -2, we got 9999, 9998. Apparently 9999 stands for -1, 9998 stands for -2. Yes, that's the system on this, on the 8086, and on all computers. (Back to that in a moment.) How do we tell that the number went through 0 ; i.e. 0000->9999? The carry flag comes to the rescue again. If the number goes through the 9999/0000 boundary in either direction, the CPU sets the CF to 1; if it doesn't, the CPU sets the CF to 0. Here's some subtraction, with the result and the carry flag. number 1 number 2 result CF 8473 2752 5721 0 2836 4583 1747 1 0654 9281 8627 1 9281 0654 8627 0 Look at examples 3 and 4. The numbers are reversed. The results are the same but they have different signs. But that is as it should be. When you reverse the order in a subtraction, you get the same absolute value, only a different sign (15 - 7 = 8 but 7 - 15 = -8). Remember, the CF is reliable only immediately after the operation. NEGATIVE NUMBERS The negative numbers go 9999=-1, 9998=-2, 9997=-3, 9996=-4, 9995=-5 etc. A more negative number is denoted by a smaller number in the register; -5 = 10,000 -5 = 9995; -498 = 10,000 -498 = 9502, and in general, -x = 10,000 -x. Here are some negative numbers and their representations on our machine. number machine no number machine no -27 9973 -4652 5348 -8916 1084 -6155 3845 As you will notice, these numbers look exactly the same as the unsigned numbers. They ARE exactly the same as the unsigned numbers. The machine has no way of knowing whether a number in a register is signed or unsigned. Unlike BASIC or PASCAL which will complain whenever you try to use a number in an incorrect way, the machine will let you do it. This is the power and the curse of machine language. You are in complete control. It is your responsibility to keep track of whether a number is signed or unsigned. Which signed numbers should be positive and which negative? This has already been decided for you by the computer, but let's think Chapter 0.1 - Numbers and Arithmetic iii ____________________________________ out what a reasonable solution might be. We could have from 0000 to 8000 positive and from 9999 to 8001 negative, but that would give us 8001 positive numbers and 1999 negative numbers. That seems unbalanced. More importantly, if we take -(3279) the machine will give us 6721, which is a POSITIVE number. We don't want that. For reasons of symmetry, the positive numbers are 0000-4999 and the negative numbers are 9999-5000.{4} Our most negative number is -5000 = 10,000 -5000 = 5000. 10'S COMPLEMENT It's time for a digression. If we are going to be using negative numbers like -(473), changing from an external number to an internal number is going to be a bother: i.e. -473 -> 9527. Going the other way is going to be a pain too: i.e. 9527 -> -473. Well, it would be a problem except that we have some help. 0000 = 10,000 = 9999 +1 - 473 result 9526 +1 = 9527 Let's work this through carefully. On our machine, 0000 and 10000 (9999+1) are the same thing, so 0 - 473 is the same as 9999+1-473 which is the same as 9999-473+1. But when we have all 9s, this is a cinch. We never have to borrow - all we have to do is subtract each digit from 9 and then add 1 to the total. We may have to carry at the end, but that is a lot better than all those borrows. We'll do a few examples: (-4276) 0000 = 10,000 = 9999 +1 -4276 result 5723 +1 = 5724 (-3982) 0000 = 10,000 = 9999 +1 -3982 result 6017 +1 = 6018 (-2400) 0000 = 10,000 = 9999 +1 -2400 result 7599 +1 = 7600 (-1989) 0000 = 10,000 = 9999 +1 ____________________ 4. That way, if we tell the machine that we are working with signed numbers, all it has to do is look at the left digit. If the digit is 5-9, we have a negative number, if it is 0-4, we have a positive number. Note that 0000 is considered to be positive. This is true on all computers. The PC Assembler Tutor iv ______________________ -1989 result 8010 +1 = 8011 This is called 10s complement. Subtract each digit from 9, then add 1 to the total. One thing we should check is whether we get the same number back if we negate the negative result; i.e. does -(-1989)) = 1989? From the last example, we see that -1989 = 8011, so: (-8011) 0000 = 10,000 = 9999 +1 -8011 result 1988 +1 = 1989 It seems to work. In fact, it always works. See the footnote for the proof.{5} You are going to use this from time to time, so you might as well practice some. Here are 10 numbers to put into 10s complement form. The answers are in the footnote. (1) -628, (2) -4194, (3) -9983, (4) -1288, (5) -4058, (6) -6952, (7) -162, (8) -9, (9) -2744, (10) -5000.{6} The computer keeps track of whether a number is positive or negative. After an arithmetical operation, it sets a flag to tell whether the result is positive or negative. This flag has no meaning if you are using unsigned numbers. The computer is saying, "If the last arithmetical operation was with signed numbers, then this is the sign of the result." The flag is called the sign flag (SF). It is 0 if the number is positive and 1 if the number is negative. Let's decrement again and look at both the sign flag and carry flag. NUMBER SIGN CARRY 3 0 0 2 0 0 1 0 0 0 0 0 9999 1 1 ____________________ 5. Let x be any number. Then: -x = ( 10,000 - x) = ( 9999 + 1 - x ) ; -(-x) = ( 10,000 - (-x) ) = ( 9999 + 1 - (-x) ) = ( 9999 + 1 - ( 9999 + 1 - x ) ) = ( 9999 + 1 - 9999 - 1 + x ) = x 6. (1) -628 = 9372 , (2) -4194 = 5806 , (3) -9983 = 0017, (4) -1288 = 8712 , (5) -4058 = 5942 , (6) -6952 = 3048 (7) -162 = 9838 , (8) -9 = 9991 , (9) -2744 = 7256, (10) -5000 = 5000. This last one is a little strange. It changes 5000 into itself. In our system, 5000 is a negative number and it winds up as a negative number. This happens on all computers. If you take the maximum negative number and take its negative, you get the same number back. Chapter 0.1 - Numbers and Arithmetic v ____________________________________ 9998 1 0 9997 1 0 9996 1 0 That worked pretty well. The sign flag changed from 0 to 1 when we went from 0 to 9999 and the carry flag was set to 1 for that one operation so we could see that we had gone through the 9999/0000 boundary. Let's do some more decrementing. NUMBER SIGN CARRY 5003 1 0 5002 1 0 5001 1 0 5000 1 0 4999 0 0 4998 0 0 4997 0 0 4996 0 0 This one didn't work too well. 5000 is our most negative number (-5000) and 4999 is our most positive number; when we crossed the 4999/5000 boundary, the sign changed but there was nothing to tell us that the sign had changed. We need to make another flag. This one is called the overflow flag. We check the carry flag (CF) for the 0000/9999 boundary and we check the overflow flag for the 5000/4999 boundary. The last decrementing example with the overflow flag: NUMBER SIGN CARRY OVERFLOW 5003 1 0 0 5002 1 0 0 5001 1 0 0 5000 1 0 0 4999 0 0 1 4998 0 0 0 4997 0 0 0 4996 0 0 0 This time we can find out that we have gone through the boundary. We'll come back to how the computer sets the overflow flag later, but let's do some addition and subtraction now. UNSIGNED ADDITION AND SUBTRACTION Unsigned addition is done the same way as normally. The computer adds the two numbers. If the result is over 9999, it sets the carry flag and drops the left digit (i.e. 14625 -> 4625, CF = 1, 19137 -> 9137 CF = 1, 10000 -> 0000 CF = 1). The largest possible addition is 9999 + 9999 = 19998. This still has a 1 in the left digit. If the carry flag is set after an addition, the result must be between 10000 and 19998. The PC Assembler Tutor vi ______________________ Since this is unsigned addition, we won't worry about the sign flag or the overflow flag for the moment. Here are some examples of unsigned addition. NUMBER 1 NUMBER 2 RESULT CF 5147 2834 7981 0 6421 8888 5309 1 2910 6544 9454 0 6200 6321 2521 1 Directly after the addition, the computer has complete information about the number. If the carry flag is set, that means that there is an extra 10,000, so the result of the second example is 15309 and the result of the fourth example is 12521. There is no way to store all that information in 4 digits in memory so that extra information will be lost if it is not used immediately. Subtraction is similar. The machine subtracts, and if the answer is below 0000, it sets the carry flag, borrows 10000 and adds it to the result. -3158 -> -3135 + 10000 -> 6842 CF = 1 ; -8197 -> -8197 + 10000 -> 1803 CF = 1. After a subtraction, if the carry flag is set, you know the number is 10000 too big. Once again, the carry flag information must be used immediately or it will be lost. Here are some examples: NUMBER 1 NUMBER 2 RESULT CF 3872 2655 1217 0 9826 5967 3859 0 4561 7143 7418 1 2341 4907 7434 1 If the carry flag is set, the computer borrowed 10000, so example 3 is 7418 - 10000 = -2582 and example 4 is 7434 - 10000 = -2566. MODULAR ARITHMETIC What the computer is doing is modular arithmetic. Modular arithmetic is like a clock. If it is 11 o'clock and you go forward 1 hour it's now 12 o'clock; if it's 11 and you go backwards 1 hour it's now 10. If it's 11 and you go forward 4 hours it's not 15, it's 3. If it's 11 and you go backward 15 hours it's not -4, it's 8. The clock is doing mod 12 arithmetic.{7} (A+B) mod 12 (A-B) mod 12 From the clock's viewpoint, 11 o'clock today, 11 o'clock yesterday and 11 o'clock, June 8, 1754 are all the same thing. If ____________________ 7. To be a perfect analogy 12 o'clock should be 0 o'clock. Chapter 0.1 - Numbers and Arithmetic vii ____________________________________ you go forward 200 hours (that's 12X16 + 8) you will have the same result as going forward 8 hours. If you go backwards 200 hours (that's -(12X16 + 8) = -(12X16) -8) you get the same result as going backwards 8 hours. If you go forward 4 hours from 11 (11+4) mod 12 = 3 you get the same result as going backwards 8 hours (11-8) mod 12 = 3. In fact, these come in pairs. If A + B = 12, then going forward A hours gives the same result as going backwards B hours. Forwards 9 = backwards 3; forwards 7 = backwards 5; forwards 11 = backwards 1. In the mod 12 system, the following things are equivalent: (+72 + 4) (+72 - 8) (+60 + 4) (+60 - 8) (+48 + 4) (+48 - 8) (+36 + 4) (+36 - 8) (+24 + 4) (+24 - 8) (+12 + 4) (+12 - 8) ( 0 + 4) ( 0 - 8) (-12 + 4) (-12 - 8) (-24 + 4) (-24 - 8) (-36 + 4) (-36 - 8) (-48 + 4) (-48 - 8) (-60 + 4) (-60 - 8) They form what is known as an equivalence class mod 12. If you use any one of them for addition or subtraction, you will get the same result (mod 12) as with any other one. Here's some addition:{8} (+48 + 4) + 7 = (48 + 11) mod 12 = 11 (-48 - 8) + 7 = (48 - 1 ) mod 12 = 11 ( 0 - 8) + 7 = ( 0 - 1 ) mod 12 = 11 (-60 + 4) + 7 = (-60 +11) mod 12 = 11 And some subtraction: (+48 + 4) - 2 = (48 + 2 ) mod 12 = 2 (-48 - 8) - 2 = (48 - 10) mod 12 = 2 ( 0 - 8) - 2 = ( 0 - 10) mod 12 = 2 (-60 + 4) - 2 = (-60 + 2) mod 12 = 2 Our pretend computer doesn't cycle every 12 numbers, it cycles every 10,000 numbers - it is a mod 10,000 machine. On our machine, the number 6453 has the following equivalence class: (+30000 + 6453) (+30000 - 3547) (+20000 + 6453) (+20000 - 3547) (+10000 + 6453) (+10000 - 3547) ( 0 + 6453) ( 0 - 3547) (-10000 + 6453) (-10000 - 3547) (-20000 + 6453) (-20000 - 3547) (-30000 + 6453) (-30000 - 3547) ____________________ 8. (-10) mod 12 = 2 ; (-11) mod 12 = 1 The PC Assembler Tutor viii ______________________ Any one of these will act the same as any other one. Notice that 10000 - 3547 is the subtraction that we did to get the representation of -3547 on the machine. -3547 = 9999 + 1 3547 6452 + 1 = 6453 6453 and -3547 act EXACTLY the same on this machine. What this means is that there is no difference in adding signed or unsigned numbers on the machine. The result will be correct if interpreted as an unsigned number; it will also be correct if interpreted as a signed number. 6821 + 3179 = 10000 so -3179 = 6821 and 3179 = -6821 5429 + 4571 = 10000 so -4571 = 5429 and 4571 = -5429 Since -3179 and 6821 act the same on our machine and since -4571 and 5429 act the same, let's do some addition. Take your time so you understand why the signed and unsigned numbers are giving the same results mod 10000: --------------------------------------------------------- 6821 + 497 = 7318 -3179 + 497 = (10000 - 3179) + 497 = 10000 -2682 = -2682 7318 + 2682 = 10000 so -2682 = 7318 ---------------------------------------------------------- 5429 + 876 = 6305 -4571 + 876 = (10000 - 4571) + 876 = 10000 - 3695 = -3695 6305 + 3695 = 10000 so -3695 = 6305 ---------------------------------------------------------- Here's some subtraction: ----------------------------------------------------------- 6821 - 507 = 6314 -3179 - 507 = (10000 - 3179) - 507 = 10000 - 3686 = -3686 6314 + 3686 = 10000 so -3686 = 6314 ---------------------------------------------------------- 5429 - 178 = 5251 -4571 - 178 = (10000 - 4571) - 178 = 10000 - 4749 = -4749 5251 + 4749 = 10000 so -4749 = 5251 ----------------------------------------------------------- It is the same addition or subtraction. Interpreted one way it is Chapter 0.1 - Numbers and Arithmetic ix ____________________________________ signed addition or subtraction; interpreted another way it is unsigned addition or subtraction. The machine could have one operation for signed addition and another operation for unsigned addition, but this would be a waste of computer resources. These operations are exactly the same. This machine, like all computers, has only one integer addition operation and one integer subtraction operation. For each operation, it sets the flags of importance for both signed and unsigned arithmetic. For unsigned addition and subtraction, CF, the carry flag tells whether the 0000/9999 boundary has been crossed. For signed addition and subtraction, SF, the sign flag tells the sign of the result and OF, the overflow flag tells whether the result was too negative or too positive. SIGN EXTENSION Although our base 10 machine is set up for 4 digit numbers, it is possible to use it for numbers of any size by writing the appropriate software. We'll use 12 digit numbers as an example, though they could be of any length. The first problem is converting 4 digit numbers into 12 digit numbers. If the number is an unsigned number, this is no problem (we'll write the number in groups of 4 digits to keep it readable): 4816 -> 0000 0000 4816 9842 -> 0000 0000 9842 127 -> 0000 0000 0127 what if it is a signed number? The first thing we need to know about signed numbers is, what is positive and what is negative? Once again, for reasons of symmetry, we choose positive to be 0000 0000 0000 to 4999 9999 9999 and negative to be 5000 0000 0000 to 9999 9999 9999.{9} This longer number system cycles from 9999 9999 9999 to 0000 0000 0000. Therefore, for longer numbers, 0000 0000 0000 = 1 0000 0000 0000. They are equivalent. 0000 0000 0000 = 9999 9999 9999 + 1. If it is a positive signed number, it is still no problem (recall that in our 4 digit system, a positive number is between 0000 and 4999, a negative signed number is between 5000 and 9999). Here are some positive signed numbers and their conversions: 1974 -> 0000 0000 1974 1 -> 0000 0000 0001 3909 -> 0000 0000 3909 ____________________ 9. Once again, the sign will be decided by the left hand digit. If it is 0-4 it is a positive number; if it is 5-9 it is a negative number. The PC Assembler Tutor x ______________________ If it is a negative number, where did its representation come from in our 4 digit system? -x -> 9999 + 1 -x = 9999 - x + 1. This time it won't be 9999 + 1 but 9999 9999 9999 + 1. Let's have some examples. 4 DIGIT SYSTEM 12 DIGIT SYSTEM -1964 9999 + 1 9999 9999 9999 + 1 -1964 -1964 8035 -> 8036 9999 9999 8035 + 1 -> 9999 9999 8036 -2867 9999 + 1 9999 9999 9999 + 1 -2867 -2867 7132 -> 7133 9999 9999 7132 + 1 -> 9999 9999 7133 -182 9999 + 1 9999 9999 9999 + 1 -182 -182 9817 -> 9818 9999 9999 9817 + 1 -> 9999 9999 9818 As you can see, all you need to do to sign extend a negative number is to put 9s to the left. Can't those 9s on the left become 0s when we add that 1 at the end? No. In order for that to happen, the right four digits must be 9999. But that can only happen if the number to be negated is 0000: 9999 9999 9999 + 1 -0000 9999 9999 9999 + 1 -> 0000 0000 0000 In all other cases, adding 1 does not carry anything out of the right four digits. It is impossible to truncate one of these 12 digit numbers to a 4 digit number without making the results unreliable. Here are two examples: (number) 0000 0168 7451 -> 7451 (now a negative number) (actual value) +168 7451 -2549 (number) 9999 9643 2170 -> 2170 (now a positive number) (actual value) -356 7830 +2170 We now have 12 digit numbers. Is it possible to add them and subtract them? Yes but only 4 digits at a time. When you add with pencil and paper you carry left from each digit. The computer can carry left from each group of 4 digits. We'll do the following addition: 0138 6715 6037 + 2514 2759 7784 Chapter 0.1 - Numbers and Arithmetic xi ____________________________________ Do this with pencil and paper and write down all the carries. The computer is going to do this in 3 parts: 1) 6037 + 7784 2) 6715 + 2759 + carry (if any) 3) 0138 + 2514 + carry (if any) The first addition is our regular addition. It will set the carry flag if the 0000/9999 boundary was crossed (i.e. the result was larger than 9999). In our case CF = 1 since the result is 13821. The register holds 3821. We store 3821. Next, we need to add three things: 6715 + 2759 + CF (=1). There is an instruction like this on all computers. It adds two numbers plus the value of the carry flag. Our first addition was ADD (add two numbers). This time the machine instruction is ADC (add two numbers and the carry). The result of our second addition is 9475. The register holds 9475 and CF = 0. We store 9475. Finally, we need to add three more things: 0138 + 2514 + CF (=0). Once again we use ADC. The result is 2652, CF = 0. We store the 2652. That is the whole result: 2652 9475 3821 If CF = 1 at this point, the number has crossed the 9999,9999,9999/0000,0000,0000 boundary. This will work for signed numbers also. The only difference is that at the very end we don't check CF, we check OF to see if the 4999,9999,9999/5000,0000,0000 boundary has been crossed. Just to give you one more example we'll do a subtraction using the same numbers: 0138 6715 6037 2514 2759 7784 Notice that in order for you to do this with pencil and paper you'll have to put the larger number on top before you subtract. With the machine this is unnecessary. Go ahead and do the subtraction with pencil and paper. The machine can do this 4 digits at a time, so this is a three step process: 1) 6037 - 7784 2) 6715 - 2759 - borrow (if any) 3) 0138 - 2514 - borrow (if any) The first one is a regular subtraction and since the bottom number is larger, the result is 8253, CF = 1. (Perhaps you are puzzled because that's not the result that you got. Don't worry, it all comes out in the wash). Step two subtracts but also subtracts any borrow (We had a borrow because CF = 1). There is a special instruction called SBB (subtract with borrow) that does just that. 6715 - 2759 - 1 = 3955, CF = 0. We store the 3955 and go on to the third part. This also is SBB, but since we had no The PC Assembler Tutor xii ______________________ borrow, we have 0138 - 2514 - 0 = 7624, CF = 1. We store 7624. This is the end result, and since CF = 1, we have crossed the 9999,9999,9999/0000,0000,0000 boundary. This is going to be the representation of a negative number mod 1,0000,0000,0000. With pencil and paper, your result was: -2375 6044 1747 The machine result was: 7624 3955 8253 But CF was 1 at the end, so this represents a negative number. What number does it represent? Let's take its negative to get a positive number with the same absolute value: 9999 9999 9999 + 1 7624 3955 8253 2375 6044 1746 + 1 = 2375 6044 1747 This is the same thing you got with pencil and paper. The reason it looked wierd is that a negative number is always stored as its modular equivalent. If you want to read a negative number, you need to take its negative to get a positive number with the same absolute value. If we had been working with signed numbers, we wouldn't have checked CF at the very end, we would have checked OF to see if the 4999,9999,9999/5000,0000,0000 boundary had been crossed. If OF = 1 at the end, then the result was either too negative or too positive. OVERFLOW How does the machine decide that overflow has occured? First, what exactly is overflow and when is it possible for overflow to occur? Overflow is when the result of a signed addition or subtraction is either larger than the largest positive number or more negative than the most negative number. In the case of the 4 digit machine, larger than +4999 or more negative than -5000. If one number is negative and the other is positive, it is not possible for overflow to occur. Take +32 and -4791 as examples. If we start with the positive number (+32) and add the negative number (-4791), the result can't possibly be too positive. Similarly, if we start with the negative number (-4791) and add the positive number (+32), the result can't be too negative. Therefore, the result can be neither too positive nor too negative. Make sure you understand this before going on. What if both are positive? Then overflow is possible. Here are some examples: Chapter 0.1 - Numbers and Arithmetic xiii ____________________________________ (+3500) + (+4500) = 8000 = -2000 (+2872) + (+2872) = 5744 = -4256 (+1799) + (+4157) = 5956 = -4044 In each case, two positive numbers give a negative result. How about two negative numbers? (7154) + (6000) = 3154 = +3154 (actual value) -2946 -4000 (5387) + (5826) = 1213 = +1213 (actual value) -4613 -4174 (8053) + (6191) = 4244 = +4244 (actual value) -1947 -3809 The numbers underneath are the negative numbers that the numbers above them represent. In these cases, adding two negative numbers gives a positive result. This is what the machine checks for. Before the addition, it checks the signs of the numbers. If the signs are the same, then the result must also be the same sign or overflow has occurred.{10} Thus + and + must have a + result; - and - must have a - result. If not, OF (the overflow flag) is set (OF = 1). Otherwise OF is cleared (OF = 0). MULTIPLICATION Unsigned multiplication is easy. The machine simply multiplies the two numbers. Since the result can be up to 8 digits (the maximum result is 9999 X 9999 = 9998 0001) the machine uses two registers to hold the result. We'll call them R1 and R2. 5436 X 174 R1 0094 R2 5864 2641 X 2003 R1 0528 R2 9923 You need to know which register holds which half of the result, but besides that, everything is straightforward. On this machine R1 holds the left four digits and R2 holds the right four digits. Notice that our machine has changed the modular base from N to N*N (from 1 0000 to 1 0000 0000). What this means is that two things which are modularly equivalent under addition and subtraction are not necessarily equivalent under multiplication and division. 6281 and -3719 will not work the same. ____________________ 10. The machine checks something considerably more obscure because it is easier to implement in semiconductor logic, but what it is actually doing is checking to see if the two numbers being added have the same sign. If they do, the result must be the same sign or overflow has occurred. The PC Assembler Tutor xiv ______________________ The machine can't do signed multiplication. What it actually does is convert the numbers to positive numbers (if necessary), perform unsigned multiplication, and then do sign adjustment of the results (if necessary). It uses 2 registers for the result. SIGNED MULTIPLICATION REGS RESULT (number) (5372) X (3195) R1 8521 = -1478 6460 (actual value) -4628 X +3195 R2 3540 (number) (9164) X (8746) R1 0104 = +104 8344 (actual value) -836 X -1254 R2 8344 (number) (9927) X (0013) R1 9999 = -949 (actual value) -73 X +13 R2 9051 Looking at the last example, if we performed unsigned multiplication on those two numbers, we would have 9927 X 0013 = 0012 9051, a completely different answer from the one we got. Therefore, whenever you do multiplication, you have to tell the machine whether you want unsigned or signed multiplication. DIVISION Unsigned division is easy too. The machine divides one number by the other, puts the quotient in one register and the remainder in another. Once again, the only problem is remembering which register has the quotient and which register has the remainder. For us, the quotient is R1 and the remainder is R2. 6190 / 372 R1 0016 16 remainder 238 R2 0238 9845 / 11 R1 0895 895 remainder 0 R2 0000 As with multiplication, signed division is handled by the machine changing all numbers to positive numbers, performing unsigned division, then putting back the appropriate signs. SIGNED DIVISION REGS RESULT (number) (7192) / (9164) R1 0003 +3 rem. -300 (actual value)-2808 / -836 R2 9700 (number) (3753) / (9115) R1 9996 -4 rem. +213 (actual value)+3753 / -885 R2 0213 Looking at the last example, 3753 / 9115, if that were unsigned multiplication the answer would be 0 remainder 3753, a completely different answer from the signed division. Every time you do a division, you have to state whether you want unsigned or signed division. Chapter 0.2 - Bases 2 and 16 ============================ xv I'm making the assumption that if you are along for the ride you already know something about binary and hex numbers. This is a review only. BASE 2 AND BASE 16 Base 2 (binary) allows only 0s and 1s. Base 16 (hexadecimal) allows 0 - 9, and then makes up the next six numbers by using the letters A - F. A = 10, B=11, C=12, D=13, E=14 and F=15. You can directly translate a hex number to a binary number and a binary number to a hex number. A group of four digits in binary is the same as a single digit in hex. We'll get to that in a moment. The binary digits (BITS) are the powers of 2. The values of the digits (in increasing order) are 1, 2, 4, 8, 16, 32, 64, 128, 256 and so on. 1 + 2 + 4 + 8 = 15, so the first four digits can represent a hex number. This repeats itself every four binary digits. Here are some numbers in binary, hex, and decimal BINARY HEX DECIMAL 0100 4 4 1111 F 15 1010 A 10 0011 3 3 Let's go from binary to hex. Here's a binary number. 0110011010101101 To go from binary to hex, first divide the binary number up into groups of four starting from the right. 0110 0110 1010 1101 Now simply change each group into a hex number. 0110 -> 4 + 2 -> 6 0110 -> 4 + 2 -> 6 1010 -> 8 + 2 -> A 1101 -> 8 + 4 + 1 -> D and we have 66AD as the result. Similarly, to go from hex to binary: D39F change each hex digit into a set of four binary digits: D = 13 -> 8 + 4 + 1 -> 1101 ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor xvi ______________________ 3 -> 2 + 1 -> 0011 9 -> 8 + 1 -> 1001 F = 15 -> 8+4+2+1 -> 1111 and then put them all together: 1101001110011111 Of course, having 16 digits strung out like that makes it totally unreadable, so in this book, if we are talking about a binary number, it will always be separated every 4 digits for clarity.{1} All computers operate on binary data, so why do we use hex numbers? Take a test. Copy these two binary numbers: 1011 1000 0110 1010 1001 0101 0111 1010 0111 1100 0100 1100 0101 0110 1111 0011 Now copy these two hex numbers: B86A957A 7C4C56F3 As you can see, you recognize hex numbers faster and you make fewer mistakes in transcription with hex numbers. ADDITION AND SUBTRACTION The rules for binary addition are easy: 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 (carry 1 to the next digit left) similarly for binary subtraction: 0 - 0 = 0 0 - 1 = 1 (borrow 1 from the next digit left) 1 - 0 = 1 1 - 1 = 0 On the 8086, you can have a 16 bit (binary digit) number represent a number from 0 - 65535. 65535 + 1 = 0 (65536). For binary numbers, the boundary is 65535/0. You count up or down through that boundary. The 8086 is a mod 65536 machine. That means the things that are equivalent to 35631 mod 65536 are:{2} ____________________ 1. This will not be true of the actual assembler code, since the assembler demands an unseparated number. 2. 35631 + 29905 = 65536. -29905 = 35631 (mod 65536) Chapter 0.2 - Binary and Hex Numbers xvii ____________________________________ (3*65536 + 35631) (3*65536 - 29905) (2*65536 + 35631) (2*65536 - 29905) (1*65536 + 35631) (1*65536 - 29905) ( 0 + 35631) ( 0 - 29905) (-1*65536 + 35631) (-1*65536 - 29905) (-2*65536 + 35631) (-2*65536 - 29905) (-3*65536 + 35631) (-3*65536 - 29905) The unsigned number 35631 and the signed number -29905 look the same. They ARE the same. In all addition, they will operate in the same fashion. The unsigned number will use CF (the carry flag) and the signed number will use OF (the overflow flag). On all 16 bit computers, 0-32767 is positive and 32768 - 65535 is negative. Here's 32767 and 32768. 32767 0111 1111 1111 1111 32768 1000 0000 0000 0000 32768 and all numbers above it have the left bit 1. 32767 and all numbers below it have the left bit 0. This is how to tell the sign of a signed number. If the left bit is 0 it's positive and if the left bit is 1 it's negative. TWO'S COMPLEMENT In base 10 we had 10s complement to help us with negative numbers. In base 2, we have 2s complememt. 0 = 65536 = 65535 + 1 so we have: 1 0000 0000 0000 0000 = 1111 1111 1111 1111 + 1 To get the negative of a number, we subtract: -49 = 0 - 49 = 65536 - 49 = 65535 - 49 + 1 (65536) 1111 1111 1111 1111 + 1 (49) 0000 0000 0011 0001 result 1111 1111 1100 1110 + 1 -> 1111 1111 1100 1111 (-49) ; - - - - - -21874 (65536) 1111 1111 1111 1111 + 1 (21874) 0101 0101 0101 0111 result 1010 1010 1010 1000 + 1 -> 1010 1010 1010 1001 (-21847) ; - - - - - -11628 (65536) 1111 1111 1111 1111 + 1 (11628) 0010 1101 0110 1100 result 1101 0010 1001 0011 + 1 -> 1101 0010 1001 0100 (-11628) ; - - - - - The PC Assembler Tutor xviii ______________________ -1764 (65536) 1111 1111 1111 1111 + 1 (1764) 0000 0110 1110 0100 result 1111 1001 0001 1011 + 1 -> 1111 1001 0001 1100 (-1764) ; - - - - - Notice that since: 1 - 0 = 1 1 - 1 = 0 when you subtract from 1, you are simply switching the value of the subtrahend (that's the number that you subtract). 1 -> 0 0 -> 1 1 becomes 0 and 0 becomes 1. You don't even have to think about it. Just switch the 1s to 0s and switch the 0s to 1s, and then add 1 at the end. Well do one more: -348 (65536) 1111 1111 1111 1111 + 1 (348) 0000 0001 0101 1100 result 1111 1110 1010 0011 + 1 -> 1111 1110 1010 0100 (-348) Now two more, this time without the crutch of having the top number visible. Remember, even though you are subtracting, all you really need to do is switch 1s to 0s and switch 0s to 1s, and then add 1 at the end. -658 (658) 0000 0010 1001 0010 result 1111 1101 0110 1101 + 1 -> 1111 1101 0110 1110 (-658) ; - - - - - -31403 (34103) 0111 1010 0100 0111 result 1000 0101 1011 1000 + 1 -> 1000 0101 1011 1001 (-31403) SIGN EXTENSION If you want to use larger numbers, it is possible to use multiple words to represent them.{3} The arithmetic will be done 16 bits at a time, but by using the method described in Chapter 0.1, it is possible to add and subtract numbers of any length. One normal length is 32 bits. How do you convert a 16 bit to a 32 bit number? If it is unsigned, simply put 0s to the left: 0100 1100 1010 0111 -> 0000 0000 0000 0000 0100 1100 1010 0111 ____________________ 3. On the 8086, a word is 16 bits. Chapter 0.2 - Binary and Hex Numbers xix ____________________________________ What if it is a signed number? The first thing we need to know about signed numbers is what is positive and what is negative. Once again, for reasons of symmetry, we choose positive to be from 0000 0000 0000 0000 0000 0000 0000 0000 to 0111 1111 1111 1111 1111 1111 1111 1111 (hex 00000000 to 7FFFFFFF) and we choose negative to be from 1000 0000 0000 0000 0000 0000 0000 0000 to 1111 1111 1111 1111 1111 1111 1111 1111 (hex 10000000 to FFFFFFFF).{4} This longer number system cycles from 1111 1111 1111 1111 1111 1111 1111 1111 to 0000 0000 0000 0000 0000 0000 0000 0000 (hex FFFFFFFF to 00000000). Notice that by using binary numbers we are innundating ourselves with 1s and 0s. If it is a positive signed number, it is still no problem (recall that in our 16 bit system, a positive number is between 0000 0000 0000 0000 and 0111 1111 1111 1111, a negative signed number is between 1000 0000 0000 0000 and 1111 1111 1111 1111). Just put 0s to the left. Here are some positive signed numbers and their conversions: (1974) 0000 0111 1011 0110 -> 0000 0000 0000 0000 0000 0111 1011 0110 (1) 0000 0000 0000 0001 -> 0000 0000 0000 0000 0000 0000 0000 0001 (3909) 0000 1111 0100 0101 -> 0000 0000 0000 0000 0000 1111 0100 0101 If it is a negative number, where did its representation come from in our 16 bit system? -x -> 1111 1111 1111 1111 + 1 -x = 1111 1111 1111 1111 - x + 1. This time it won't be FFFFh + 1 but FFFFFFFFh + 1. Let's have some examples. (Here we have 8 bits to the group because there is not enough space on the line to accomodate 4 bits to the group). 16 BIT SYSTEM 32 BIT SYSTEM -1964 11111111 11111111 + 1 11111111 11111111 11111111 11111111 + 1 00000111 10101100 00000000 00000000 00000111 10101100 11111000 01010011 + 1 11111111 11111111 11111000 01010011 + 1 11111000 01010100 11111111 11111111 11111000 01010100 ____________________ 4. Once again, the sign will be decided by the left hand digit. If it is 0 it is a positive number; if it is 1 it is a negative number. The PC Assembler Tutor xx ______________________ -2867 11111111 11111111 + 1 11111111 11111111 11111111 11111111 + 1 00001011 00110011 00000000 00000000 00001011 00110011 11110100 11001100 + 1 11111111 11111111 11110100 11001100 + 1 11110100 11001101 11111111 11111111 11110100 11001101 -182 11111111 11111111 + 1 11111111 11111111 11111111 11111111 + 1 00000000 10110110 00000000 00000000 00000000 10110110 11111111 01001001 + 1 11111111 11111111 11111111 01001001 + 1 11111111 01001010 11111111 11111111 11111111 01001010 As you can see, all you need to do to sign extend a negative number is to put 1s to the left. Can't those 1s on the left become 0s when we add that 1 at the end? No. In order for that to happen, the right 16 bits must be 1111 1111 1111 1111. But that can only happen if the number to be negated is 0: 1111 1111 1111 1111 1111 1111 1111 1111 + 1 -0000 0000 0000 0000 1111 1111 1111 1111 1111 1111 1111 1111 + 1 -> 0000 0000 0000 0000 0000 0000 0000 0000 In all other cases, adding 1 does not carry anything out of the right 16 bits. It is impossible to truncate one of these 32 bit numbers to a 16 bit number without making the results unreliable. Here are two examples: +1,687,451 00000000 00011001 10111111 10011011 -> 10111111 10011011 (-16485) -3,524,830 11111111 11001010 00110111 00100010 -> 00110111 00100010 (+14114) Truncating has changed both the sign and the absolute value of the number. Chapter 0.3 - LOGIC =================== xxi Programs use numbers a lot. But they also ask questions that require a yes/no answer. Is there an 8087 chip in the computer? Is there a color monitor; how about a monochrome monitor? Is there keyboard input waiting to be processed? Are you going to get lucky on your date on Friday? Or, since you are a computer programmer, are you going to have a date this month? Did the file open correctly? Have we reached end of file? In order to combine these logical questions to our heart's content, we need a few operations: AND, OR, XOR (exclusive or), and NOT. AND If we have two sentences "A" and "B", then ("A" AND "B") is true if (and only if) both "A" and "B" are true. "It is raining and I am wet" is true only if both "It is raining" and "I am wet" are true. If one or both are false, then 'A and B' is false. A shortcut for writing this is to use a truth table. A truth table tells under what conditions an expression is true or false. All we need to know is whether each component expression is true or false. T stands for true, F for false. "A" "B" "A" AND "B" T T T T F F F T F F F F Notice that the truth table does NOT say anything about whether the expression makes sense. The sentence: "It's hot and I am sweating." is a reasonable expression which may or may not be true. It will be true if both "It is hot" and "I am sweating" are true. But the sentence: "The trees are green and Quito is the Capital of Ecuador." is pure garbage. It does not satisfy our idea of what a sensible expression should be, and should NEVER be evaluated by means of a truth table. The warranty on a truth table is, if the expression makes sense, then the truth table will tell you under what conditions it is true or false. If the expression does not make sense, then the truth table tells you nothing. Fortunately, this problem really belongs to philisophical logic. When you use logical operators in your program, there will be a ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor xxii ______________________ well defined reason for each use. If you start doing screwy things, your program probably won't run. OR There are two different types of OR alive and kicking in the English language - the exclusive OR (A or B but not both) and the inclusive OR (A or B or both). A mother tells her child "You can have a piece of cake or a piece of candy." Does this mean that he can have both if he wants? Of course not. He can have one or the other, but not both. This is XOR, the exclusive or. The truth table for this is: "A" "B" "A" XOR "B" T T F T F T F T T F F F 'A XOR B' is true if exactly one of them is true. If they both are true 'A XOR B' is false. If neither is true, 'A XOR B' is false. Examples of XOR are: 1) We will either go to Italy or to Japan on our vacation. 2) I'll either have a tuna salad or a chef's salad. 3) He'll either buy a Lamborghini or a BMW. Consider this sentence: "To go to Harvard, you need to have connections or to be very smart." Do we want this to mean that if you have connections but are very smart, you are automatically excluded from going to Harvard? No. We want this to mean one or the other or perhaps both. Sometimes you write this as 'and/or' if you want to be absolutely clear. This is the inclusive OR. The truth table for OR is: "A" "B" "A" OR "B" T T T T F T F T T F F F 'A OR B' is true if one or both of them are true. If both are false, then it is false. Examples of OR are: 1) They'll either go to Italy or to Austria on their vacation. 2) I'll have either steak or shrimp at The Sizzler. 3) He'll buy either a paisley tie or a rep tie. The three sentences for XOR and OR mimic each other on purpose. In the English language, you know which type of OR is being used by what is being talked about. You know intuitively which one Chapter 0.3 - Logic xxiii ___________________ applies. If someone buys two different ties you are not suprised. If someone buys two expensive cars at the same time you are quite surprised.{1} With very few exceptions, if you confuse the two you are doing it on purpose. If your father says "You can have the car on Friday night or on Saturday night." and you don't understand which OR applies, it's not his fault. NOT The final logical operation is NOT. The sentence: "It is not raining." is false if it is raining, and true otherwise. The truth table is: "A" NOT "A" T F F T This is self-explanatory. Amazingly enough, this is all you need to know about logic to be a quality programmer. Trying to make very complex combinations of these logical operations may be fun for philosophy, but it is death to a program. KISS is the operative word in programming.{2} ____________________ 1. Especially if his job is working the cash register at Sears. 2. Which, if you don't know, is Keep It Simple, Stupid. Chapter 0.4 - MEMORY ==================== xxiv The basic unit of memory on 8086 machines is the byte.{1} This means that in every memory cell we can store a number from 0 to 255. Each memory cell has an address. The first one in memory has address 0, then address 1, then 2, then 3, etc. The registers on the 8086 are one word (two bytes) long. This means that any register can store and operate on a number from 0 to 65,535. (It also has some registers which can operate on bytes and can store and operate on numbers from 0 to 255.) Memory is physically external to the 8086. Registers are physically internal to the 8086; they are actually on the chip. One of the ways that we can access memory on the 8086 is by putting the address of a memory cell in a register and then telling the 8086 that we want to use the data at that memory address. Since each byte has its own address, and since we can't have a number larger that 65,535 in any one register, it is impossible to address more than 65,535 bytes with a single register. Back in the dark ages, that might have been enough memory, but it sure isn't enough nowdays. Intel solved the problem by creating SEGMENTS. Each segment is 65,536 bytes long, going from address 0 to address 65,535.{2} You tell the 8086 where you want to go in memory by telling it which segment you are in and what the address is within that segment. Segments are numbered from 0 to 65,535. That is, there are 65,536 of them. As a design decision, Intel decided that a segment should start every 16 bytes. This decision was made in the late 70's and is cast in stone. On the 8086, a segment starts every 16 bytes. Here is a list of some segments, with the segment number and the segment starting address in both decimal and hex. SEGMENT NUMBER STARTING ADDRESS 0d 0h 0d 00h 1d 1h 16d 10h 2d 2h 32d 20h 3d 3h 48d 30h 4d 4h 64d 40h ____________________ 1. As it is on all computers. 2. The last segments are actually less that 65,535, as will be explained later. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 0.4 - Memory xxv ____________________ 200d C8h 3,200d C80h 21694d 54BEh 347,104d 54BE0h 51377d C8B1h 822,032d C8B10h One thing you should note is that in hex, the segment number is the same as the starting address, except that the starting address has an extra 0 digit on the right. These segments overlap. A new segment starts every 16 bytes, and they are all 65,535 bytes long. This means that with the exception of the first 16 bytes, any address in memory is in more that one segment. Here are some addresses. The word "offset" means the number of bytes from the beginning of the segment. (It is possible for a memory cell to have an offset 0). The offset is shown in both decimal (d) and hex (h): memory address 55 (37h) Seg # Offset Offset 0 55d 37h 1 39d 27h 2 23d 17h 3 7d 7h Thus the address 55 is in 4 different segments, and can be addressed relative to any one of them. memory address 17,946 (461Ah) Seg # Offset Offset 0 17,946d 461Ah 1 17,930d 460Ah 2 17,914d 45FAh ... ... 1120 26d 1Ah 1121 10d 0Ah The address 17,946 is in 1122 different segments, and can be addressed relative to any of them. Notice that as the segment number goes up one segment, the offset address goes down 16 bytes (10h). Above the address 65,519, every memory cell can be addressed by 4,096 different segments. Because of the way the addresses are generated on the 8086, the maximum memory address possible is 1,048,575 (FFFFF hex.) This means that the final segments are less than 65,536 bytes long. In this table "Address" is the starting address of the segment. All the following numbers are decimal: Segment # Address Max Offset 65,535d 1,048,560d 15d 65,534d 1,048,544d 31d 65,533d 1,048,528d 47d ... ... ... The PC Assembler Tutor xxvi ______________________ 64,000d 1,024,000d 24,575d Let's look at these same numbers in hex: Segment # Address Max Offset FFFFh FFFF0h Fh FFFEh FFFE0h 1Fh FFFDh FFFD0h 2Fh ... ... ... FA00h FA000h 5FFFh The maximum addressable number is FFFFFh, which is why these segments are cut short. There are 4,095 segments at the top which are less than 65,536 bytes long. Don't worry, though, because this top section of memory belongs to the operating system, and your programs will never be put there. It will never affect you. Back in the late 70s, a million bytes of memory seemed like a lot. In the 60s, large mainframe computers had only a half-million bytes of memory. In the 70s memory was still exhorbitantly expensive. Nowdays, however, you practically need a megabyte just to blow your nose. But this segmentation system is cast in stone, so there is no way to get more memory on the 8086.{3} In the beginning, when we make a program, we will use one segment for the machine code, one segment for permanant data, and one segment for temporary data. If we need it, this gives us 196,608 bytes of usable memory right off the bat. As you will see by the time we are finished, ALL memory is addressable - we just need to do more work to get to it all. All this talk about segments and offsets may have you concerned. If you have to keep track of all these offsets, programming is going to be very difficult.{4} Not to worry. It is the assembler's job to keep track of the offsets of every variable ____________________ 3. This megabyte rule is unalterable. EMS (extended memory) is actually a memory swapping program. On the 28086 and 80386 you can have more than one megabyte of memory but the program can only access one megabyte. The program reserves a section of its one megabyte for a transfer area. It then calls EMS which transfers the data from this extended memory to the transfer area. It is in effect a RAM disk. It is like using a hard disk but is much faster. If Intel had bitten the bullet with the 80286 and said that a segment would start every 256 bytes instead of every 16 bytes, we would have 16 megabytes of directly accessible memory instead of 1 megabyte. Hindsight is such a wonderful thing. 4. Remember, an offset is just how many bytes a memory cell is from the beginning of the segment. Chapter 0.4 - Memory xxvii ____________________ and every label in your program.{5} Which segments your program uses is decided by the operating system when it loads your program into memory. It puts some of this information into the 8086. At the start of the program, you put the rest of the information into the 8086, and then you can forget about segments. NUMBERS IN MEMORY The largest number you can store in a single byte is 255. If you are calculating the distance from the sun to Alpha Centauri in inches, obviously one byte is not enough. Unfortunately, the 8086 can't really handle large numbers like that.{6} It can handle numbers which are 16 bits (2 bytes) long. However, with subprograms supplied with all compilers, we can handle large numbers, though rather slowly if we don't use an 8087. All these different programs need a standard way to write numbers in memory, and this standard is supplied by Intel. The standard is : (1) integers can be 1, 2, or 4 bytes long. This corresponds to -128 to +127 , -32,768 to +32,767, and -2,147,483,648 to +2,147,483,647. (2) scientific floating point numbers which have an exponent and can be very large. They come as 4 byte and 8 byte numbers. We will not deal with them at all, but we need to know how they are stored. (3) Commercial or BCD numbers which occupy 1/2 byte per digit. Since some of the 8086 instructions are concerned with these we will cover them, but if you are not curious about them, you can skip that section. The standard is a 10 byte number. Let's look at a number. For the rest of this section, all numbers will be hex, and if a number is longer than one byte, we will display it with a blank space between each byte. If it is a one byte number - e.g. 3C, we know exactly where we are going to put it. But what if a number is 4 bytes long - e.g. 2D F5 33 0A - and we want to put it in memory starting at offset 264. We have two choices: 2D F5 33 0A Address Choice 1 Choice 2 267 2D 0A 266 F5 33 265 33 F5 ____________________ 5. As in other languages, a label is a name that marks a place in the code. In BASIC, labels are actually numbers (such as 500 in the instruction GOTO 500). Labels are frowned on in Pascal and C, but are the lifeblood of assembler language. 6. But fortunately, the 8087 can. The PC Assembler Tutor xxviii ______________________ 264 0A 2D Neither choice is better than the other. Choice 1 puts the right-most byte in low memory, choice 2 puts the right-most byte in high memory.{7} The right-most byte is called the LEAST SIGNIFICANT BYTE because it has the least effect on a number, while the left-most byte is called the MOST SIGNIFICANT BYTE because it has the most effect on a number. In fact, Intel picked choice #1 for the 8086 (which has the least significant byte in low memory), and Motorola picked choice #2 for the 68000 (which has the most significant byte in low memory). This is consistent for both the 8086 and the 8087: THE LEAST SIGNIFICANT BYTE IS ALWAYS IN LOW MEMORY: EACH NUMBER IN MEMORY STARTS WITH THE LEAST SIGNIFICANT BYTE. Remember that, and you'll save yourself some trouble. One problem you will run up against is that when we draw pictures of memory, we often draw from left to right, that is: ADDRESS 264 265 266 267 When we do that, things start looking wierd. For 2D F5 33 0A we have: ADDRESS 264 265 266 267 VALUE 0A 33 F5 2D This is exactly backwards. Remember, memory doesn't go from left to right, it goes UP from 0, and THE LEAST SIGNIFICANT BYTE IS ALWAYS IN LOW MEMORY. You will certainly make some mistakes till you get used to being consistent about this. The right hand digit of a number is always in low memory. If you think of memory as being VERTICAL: 1E A3 07 B5 Value Address 1E 4782 A3 4781 07 4780 B5 4779 rather than being LEFT TO RIGHT: Address 4779 4780 4781 4782 Value B5 07 A3 1E you will be much better off. ____________________ 7. Low memory always means the low addresses and high memory always means the high addresses. Chapter 0.5 - STYLE =================== xxix Finally, it is time to say a word about style. Assembler code is by its nature difficult to read. It divides any concept into a number of sequential steps. If you have the BASIC statement: MINUTES = DAYS * 1440 You get the idea because you can scan the line to see what is wanted. The assembler code for the above line is: {1} mov ax, days mov bx, 1440 mul bx mov minutes, ax In BASIC, the concept was imbedded in the expression. In assembler it was lost. This means two things. First, you must be religious about documenting every step. If you come back to something two or three months later and you haven't documented what you are doing, it may take you longer to figure out what you did than it would to completely rewrite what you did. Secondly, if you are a person who likes code like this: x = (y + k) / (z - 4) you are headed for big trouble. At the assembler level it is CRITICAL that you give every variable a name that signifies exactly what it is. If you are counting the number of correct answers, then the variable should be: correct_answers not 'K'. If you are looking at the remainder from a division, then the variable should be: remainder not 'R'. If you try to use short names like 'x', 'k' and 'y', I will guarantee that for every minute you save by not having to type in long variable names, you will lose 10 minutes by not being able to figure out what is going on when you reread the code. In this tutorial we will use multiple words connected by underscores: first_positive_prime median_age oldest_student ____________________ 1. Don't worry about what this code does. You will learn soon enough. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor xxx ______________________ The Microsoft assembler allows 31 significant characters in a name. Even though there are several other characters allowed, we will use only letters, the underscore, and numbers (where appropriate): approximation1 approximation2 approximation3 This should make your code similar to well written C or Pascal code, and greatly increase the readability of the code. Chapper 1 - Some Simple Programs ================================ 1 It is now time to start writing assembler code. One of the problems with writing in assembler is that there is no way to get input into the program or output from the program until you are very far along with learning assembler language. This is a Catch-22 situation. You can't learn assembler easily without access to input and output, and you can't write i/o routines till you know assembler.{1} Help is at hand. Included on this disk is a file called asmhelp.obj. It is actually a series of programs that will allow you to get input from the keyboard and print output to the screen. It has some other features which will be explained later. The second problem at the start is that every assembler program has a lot of overhead. These are standard instructions and formats that you need to get the program to work AT ALL. This disk contains templates that contain all the overhead, so to write a program you just make a copy of the template and enter the code and data at the appropriate place. By the end of this sequence of lessons, you will know how to make templates yourself and know the meaning of each word in the template. For now, you have to have faith that what is written is necessary, and that you will learn the meaning of everything later. Let's start. At the end of this chapter is the template we will use for now - temp1.asm. These templates are in the subdirectory \template. Let's call the first program prog1.asm (very original). All programs in assembler must have the file extension .asm so make a copy by giving the command: >copy \template\temp1.asm prog1.asm You are now ready to enter code. Open up prog1.asm with your editor, and take a look at it. It should look the same as temp1.asm. Where it says "put name of program here" - that is for your personal use so you can see the program name while in the editor. The assembler ignores everything after a semicolon. All the lines that start with a semicolon are there for visual separation or for comments. The lines with asterisks separate segments. Yes, the assembler is going to make this program into three segments. You should put all code between the line labeled "START CODE BELOW THIS LINE" and the line labeled "END CODE ABOVE THIS LINE". ____________________ 1 Just to give you an idea of how contradictory the situation is, asmhelp.obj was written in assembler language and consists of about 3500 lines (that's about 50 pages), yet you need to be using it from day one. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 2 ______________________ Later you will get more flexibility. Also notice the lines starting with the word EXTRN. Those lines tell the assembler that the subroutines (such as print_num) are in a different file and must be found when this program is linked. The assembler enters each EXTRN name in a list and records each place that an EXTRN subroutine is requested. It is possible that one of these subroutines is never requested. That's fine. However, every one of these subroutines must be present at link time or you will get a link error. This is true even if the subroutine is on the list but never requested. Since the overhead is so long, and I am so lazy, the template will never be included in the description of the program. What you will get is the name of the template, then any data you need, then the code. It will look like this: TEMP1.ASM ; - - - - - START DATA BELOW THIS LINE the data is written here ; - - - - - END DATA ABOVE THIS LINE ; - - - - - START CODE BELOW THIS LINE the code is written here ; - - - - - END CODE ABOVE THIS LINE If there is no data, no data section will be included. If there is data, it should be written in the segment named DATASTUFF between the lines "START DATA..." and "END DATA ...". The code should be written between the lines that say "START CODE ..." and "END CODE ...". For our first program, the description will look like this: TEMP1.ASM ; - - - - - START CODE BELOW THIS LINE first_label: call get_num call print_num jmp first_label ; - - - - - END CODE ABOVE THIS LINE That's all we need. If we needed to write all the overhead (starting with the line "main proc far") we would have: main proc far start: push ds sub ax, ax push ax mov ax, DATASTUFF mov ds, ax ; + + + + + + + + + + + Chapter 1 - Some Simple Programs 3 ________________________________ first_label: call get_num call print_num jmp first_label ; + + + + + + + + + + + ret mainendp etc. You can see that a simple four line program has blossomed into a monster. I'm assuming some intelligence on your part. Until further notice, the code goes between the "START CODE' and the "END CODE" lines and the data goes in the DATASTUFF segment between the "START DATA" and the "END DATA" lines. It's time to type in the program listed above. Be careful when you type. When you are done I'll explain it. In assembler we need a way to label different spots in the code. We use labels. A LABEL is a name (at the beginning of a line) which is immediately followed by a colon. A label doesn't generate any code. The assembler merely keeps track of where the label is for future use. The label we are using is named first_label. The CALL instruction tells the assembler to call the subroutine listed after the call.{2} We are calling two subroutines; first get_num which gets a number, then print_num, which prints a number in a variety of styles. Finally, JMP tells the assembler that you want to jump to the label listed after it. It is the same as GOTO in BASIC. If you look at the program, you will notice that we have an infinite loop. It was designed that way. It takes a fair amount of code to exit gracefully, so we will always exit ungracefully. When you are tired of the program, simply press Control-C. That should get you out. That way you can try out something an indefinite number of times, and when you have finished you can press CTRL-C to quit the program. One warning about machine language before we start. There is no safety net, so before you start a machine language program, make sure all files are closed (i.e. that you have no other programs in memory). We will NEVER open a file in one of our programs. ____________________ 2 I am using the words subroutine, routine, program and procedure (the technical word) interchangebly throughout the book. A program is actually a group of one or more procedures, but I'm not going to be too strict about it. Context should tell whether we are talking about a single procedure or a whole program. The PC Assembler Tutor 4 ______________________ Your programs are almost certain to wind up in zombie space from time to time. If that happens, your choices are (1) hit CTRL-C. If that doesn't work, then (2) hit CTRL-ALT-DEL. As a last resort, you can (3) hit a reset button or shut the machine down. For that reason, memorize this mantra: BACKUP YOUR PROGRAMS AND BACKUP YOUR BACKUPS. Double check that you have typed in the assembler code correctly. Now it's time to assemble it. I am assuming that you have the Microsoft assembler.{3} Type: >masm prog1 ; The first thing you will see is the copyright notice: Microsoft (R) Macro Assembler Version 5.10 Copyright (C) Microsoft Corp 1981, 1988. All rights reserved. The file extension .asm is unnecessary; it's understood. The semicolon is to speed things up. If you don't use it, the assembler will ask you if you want to change any of the default choices. If you type just masm: >masm then you need to give the name of the assembler text file on the first line: Source filename [.ASM]: prog1 Object filename [prog1.OBJ]: Source listing [NUL.LST]: Cross-reference [NUL.CRF]: but press ENTER for the other options. If you type masm prog1: >masm prog1 You don't want to change any of the default settings: Object filename [prog1.OBJ]: Source listing [NUL.LST]: Cross-reference [NUL.CRF]: When the assembler asks you about options, hit the ENTER key. If you have made any errors, the assembler will tell you which line they are on and give you a description of the problem. Make a hard copy of them on the printer, then use your editor and find the line. Unfortunately, at this stage of the game, it will be ____________________ 3 If you are using A86, then consult A86.APP. If you are using Turbo Assembler, then consult TASM.APP. They are both located in the \APPENDIX subdirectory on disk 3. Chapter 1 - Some Simple Programs 5 ________________________________ very difficult for you to figure out what the problem is. You will have to struggle through the first 4 or 5 programs before things start getting easier. All the programs on these disks have been compiled on a Microsoft v5.1 assembler. They have assembled. They have been run, and they work. Don't tamper with the template and copy the code exactly and everything should work. If you haven't made any errors, the assembler will say: 0 Warning Errors 0 Severe Errors LINKING The assembler has given you back another program named prog1.obj - the same name with the extension .obj. I am assuming that you have all used the linker with compiled programs. If you haven't, you may be getting in over your head by using machine language. All the extra subroutines are in a program called asmhelp.obj. Its pathname is \asmhelp\asmhelp.obj. You want to put it in the root directory of your current drive. In the whole book, we will assume that its pathname is: \asmhelp.obj If you put it somewhere else, you will have to modify the pathname whenever it appears. Link the two modules by writing: >link prog1+\asmhelp ; The copyright notice will appear: Microsoft (R) Overlay Linker Version 3.61 Copyright (C) Microsoft Corp 1983-1987. All rights reserved. This time the file extensions are understood to be .obj. The semicolon is to avoid having to make default choices. If you type: >link then you need to put the module names after the first prompt: Object Modules [.OBJ]: prog1+\asmhelp Run File [PROG1.EXE]: List File [NUL.MAP]: Libraries [.LIB]: but press ENTER for the other choices. If you type: link prog1+\asmhelp You need to do nothing extra: Run File [PROG1.EXE]: The PC Assembler Tutor 6 ______________________ List File [NUL.MAP]: Libraries [.LIB]: When the linker asks for choices, simply press the ENTER key. The linker gives the executable file the name of the first object file on the line, so you should always put your program first and asmhelp.obj second. If there are no errors, you are ready to go. If there are errors, once again, they will be very difficult to trace. Go back and check everything from the beginning. You are now ready to run the program. Type: >prog1 The program will start. The first thing you will see is a copyright notice. The PC Assembler Helper Version 1.0 Copyright (C) 1989 Chuck Nelson All rights reserved. It appears the first time you call a subprogram in the module asmhelp.obj. The program will request a number. Give it any legal signed or unsigned number. It should be no longer than 5 digits. Press ENTER, and it will display the possible ways that that number can be thought of by the computer. Enter any decimal number 4410 HEX SIGNED UNSIGNED CHAR BINARY 113AH +04410 04410 11 : ** 0001000100111010 Enter any decimal number 30486 HEX SIGNED UNSIGNED CHAR BINARY 7716H +30486 30486 w 16 ** 0111011100010110 If the signed or unsigned number doesn't look the same as what you entered, then the number you entered is too big for a 16 bit computer. For signed numbers, the limits are +32767 to -32768 and for unsigned numbers, the limits are 0 to 65535. Enter any decimal number -64661 HEX SIGNED UNSIGNED CHAR BINARY 036BH +00875 00875 03 k ** 0000001101101011 Enter any decimal number 94547 HEX SIGNED UNSIGNED CHAR BINARY 7153H +29011 29011 q S * 0111000101010011 Lets look at the numbers. Each type of output is labeled. After a hex number, there is an 'H' and after the characters, there is a '*'. This is always true. Every time you print a hex number, there will be an 'H', and every time you print a character, there Chapter 1 - Some Simple Programs 7 ________________________________ will be a '*'. This is so you will always know what is being printed. Also notice that a signed integer ALWAYS has a sign and an unsigned integer NEVER has a sign. Not all characters are visible. Ascii 0 - 32 are invisible (32 is a blank). On the PC, ascii 33-255 are visible, but ascii 127 and ascii 255 are problematic. Therefore, if the ascii code is 0-32, 127 or 255, that character will be printed as a hex number, not a character, and print_num will signal the event by printing a double asterisk '**' instead of a single one. This has happened in the first two examples. ( 11 : ** ) and ( w 16 ** ). The first one is the hex number 11 followed by the character ':' and the second one is the character 'w' followed by the hex number 16. Both are signalled by the double asterisk '**' instead of the single asterisk '*'. Do a few examples. When you are done looking at the numbers, press CTRL-C and you will exit the program. Enter any decimal number ^C PROGRAM 2 The second program is almost the same as the first one. The program takes input from the keyboard and displays it in a variety of styles. This time, however, it is going to ask for different inputs: ascii, hex, binary and decimal. If you make an error in the input, the subroutine will prompt you again for the input. Here's the program: TEMP1.ASM ;+ + + + + + + + + + START CODE BELOW THIS LINE first_label: call get_num ; 1 to 5 digit signed or unsigned call print_num call get_ascii ; 1 or 2 characters call print_num call get_binary ; a 1 to 16 bit binary number call print_num call get_hex ; a 1 to 4 digit hex number call print_num jmp first_label ;+ + + + + + + + + + END CODE ABOVE THIS LINE The things to the right of the semicolons are comments. You do not need to type them in if you don't want to. Once again, assemble the program. (There should be no warning or severe errors. If something is wrong, it is most likely a typing error.) Then link it with asmhelp.obj. Remember - your program should be The PC Assembler Tutor 8 ______________________ the first one listed.{4} If all is well, run the program. It will ask you for a number (that is a signed or unsigned number), ascii characters, a binary number, and a 4 digit hex number (0-9,A-F). Enter any decimal number 27959 HEX SIGNED UNSIGNED CHAR BINARY 6D37H +27959 27959 m 7 * 0110110100110111 Enter one or two ascii characters $% HEX SIGNED UNSIGNED CHAR BINARY 2425H +09253 09253 $ % * 0010010000100101 Enter a two byte binary number 0101111001100010 HEX SIGNED UNSIGNED CHAR BINARY 5E62H +24162 24162 ^ b * 0101111001100010 Enter a two byte hex number 784d HEX SIGNED UNSIGNED CHAR BINARY 784DH +30797 30797 x M * 0111100001001101 Once again, this is an infinite loop, so in order to quit, you need to hit CTRL-C. The purpose of these first two programs is to remind you that the computer doesn't care whether you think you are storing binary numbers, characters, hex numbers, signed numbers or unsigned numbers. They all wind up in the computer as a series of 1s and 0s, and you can use these 1s and 0s any way you like. It's up to you to keep track of them. If you feel comfortable with the way we are writing, assembling and linking programs, you are ready to start looking at the 8086 itself. ____________________ 4 For your convenience, there is a batch file on the disk called asmlink.bat. Its pathname is \asmhelp\asmlink.bat. It is one line long and looks like this: link %1+\asmhelp ; If you use this batch file, you will never have an order problem. If your file is named myfile.asm, then type: >asmlink myfile Chapter 1 - Some Simple Programs 9 ________________________________ ; TEMP1.ASM The first assembler template ; put name of program here ; - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - STACKSEG SEGMENT STACK 'STACK' dw 100 dup (?) STACKSEG ENDS ; - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - DATASTUFF SEGMENT PUBLIC 'DATA' ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE DATASTUFF ENDS ; - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - CODESTUFF SEGMENT PUBLIC 'CODE' EXTRN print_num:NEAR , get_num:NEAR EXTRN get_ascii:NEAR , get_hex:NEAR , get_binary:NEAR ASSUME cs:CODESTUFF, ds:DATASTUFF main proc far start: push ds ; set up for return sub ax,ax push ax mov ax, DATASTUFF mov ds,ax ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE ret main endp CODESTUFF ENDS ; - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - END start The PC Assembler Tutor 10 ______________________ SUMMARY CALL CALL calls a subroutine. call get_num JMP JMP jumps to the indicated label. JMP label47 LABEL A label is a name at the beginning of a line which is followed by a colon. It is used to mark a spot in the program. label47: There are three different things which will be mentioned from time to time, so it's best to define them now. ASSEMBLER INSTRUCTIONS (CODE) is the text that you type in and give to the assembler. MACHINE CODE is the code that the assembler generates. After some adjustment by the linker, it is readable by the 8086. It is the actual code that controls the program. MICROCODE is the code that is imbedded in the 8086 itself. Each instruction has its own set of mini instructions within the 8086. This is the MICROCODE. Chapter 2 - Data ================ 11 Before you start using data, you need to know what data looks like. It is not necessary for the data to have a name. For instance, the following definition is perfectly legal: db "Mary had a little lamb." Unfortunately, the assembler has no way to find it. The normal thing is to start the line with a name, and then give the definition of the data. The assembler processes the data line by line, so a definition on one line does not carry over to another line. We can have: poem db "Mary had a little lamb," Notice that names for data don't have colons after them. What if we wanted to continue the poem? It isn't going to fit all on one line. No problem. All we need to do is define the following lines without a name. poem db "Mary had a little lamb," db "It's fleas were white as snow," db "And everywhere that Mary went," db "She scratched and scratched and scratched." The assembler still can't find lines 2-4, but starting at the first byte of "poem", it can go all the way through the poem one byte after the other. By the way, there are no carriage returns in the poem right now. They will come later. So we have the name part, the db part, and the data part. What is that db anyway. It stands for Define Byte. Whenever you give the name "poem" to the assembler, it knows that you want to deal with the data one byte at a time. If you try working a word at a time, you will get an assembler error. The legal definitions are: DB define byte [ 1 byte ] DW define word [ 2 bytes ] DD define doubleword [ 2X2 bytes = 4 bytes ] DQ define quadword [ 4X2 bytes = 8 bytes ] DT define ten-byte [ 10 bytes ] DF define farword [ 6 bytes - used for 80386 only ] Every time you use one of these directives, the assembler allocates the number of bytes in brackets for EACH variable. For instance in: db "Mary had a little lamb," each character inside the quotes is a variable. That's 23 variables X 1 byte = 23 bytes. In: ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 12 ______________________ dq 0, 1, 2, 3, 4 each number is a variable. 5 variables X 8 bytes = 40 bytes. Notice from these examples that you can have more than one variable on a line but they all share the same defining type. What do you do if you have an uninitialized variable, i.e. you don't know its starting value? Easy as pie. Here's a four byte variable: some_data dd ? The question mark lets the assembler know that you didn't forget the number but rather you didn't know the number. The commas are separators. When you write a comma, the assembler expects another piece of data on the line. If it doesn't get the number, it is an error. That means there can be no commas inside a number. dw 32,421 is two variables: 32 and 421. What if you want to make an array? The assembler has a directive for that too: dw 150 dup ( 400 ) The 'dup' is for duplicate. This makes 150 two byte copies and puts the number 400 in each one. db 273 dup ( 'c' ) This makes 273 one byte copies and puts the letter 'c' in each one. dd 459 dup ( 1, 2, 3, 4, 5 ) This makes 459 copies of what is inside the parentheses. That means ( 5 variables X 4 bytes ) X 459 for a total of 9180 bytes. Starting from the beginning of the array, we will have the sequence: 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3,... dq 20000 dup ( 455 ) This makes 20000 eight byte copies and causes an assembler error because there is a limit of 65,536 bytes for the data and you have used 160,000 bytes (20,000 X 8). db 7 dup ( 'Mary had a little lamb,') This makes 7 copies of 'Mary had ..' which is 23 bytes, for a total of 161 bytes. dw 39 dup ( 28 dup ( 0 ) ) Chapter 2 - Data 13 ________________ The assembler even supports nesting, so you can make a multi-dimensional array. This is a 39 X 28 array initialized to zero. 39 copies of 28 two byte numbers is 2184 bytes. The standard form for arrays is (1) first define the data type, (2) then say how long the array is followed by the keyword "dup" and (3) put the initial value inside the parentheses. What if you don't know the initial value? Simple: dw 347 dup ( ? ) The question mark lets the assembler know that you don't know. DEFINING NUMBERS What kinds of data can you have? 1. A single character inside single or double quotes: 'a' , "&" , '|' 2. A string inside single or double quotes: "Mary had a little lamb," 'Mary had a little lamb,' Each character is stored as a byte, and the bytes are stored consecutively. If the array starts at address 2743, 2743 = 'M', 2744 = 'a', 2745 = 'r', 2746 = 'y', 2747 = ' ', etc. As usual in these instances, if you want a double quote inside a double quoted string or a single quote inside a single quoted string, you need to use a pair: "Mary asked her fleas ""Why don't you join the circus?""" 'Mary asked her fleas "Why don''t you join the circus?"' 3. A decimal number. Decimal is the default: 27, 44, 641, 89 4. A hex number. A hex number must start with a number, so if the highest digit is A - F, there must be a 0 in front.{1} b77h is illegal, 0b77h is legal. All hex numbers must be followed by an 'h': 0a162H , 0329H , 0DDDh , 7h 5. An octal (base 8) number. An octal is followed either by the ____________________ 1 When the assembler looks at something it needs to know whether it is a name or a number. Is 'A7' a name or a hex number? Is '3D' a name or a number? To solve this problem, all assemblers and all compilers insist that -> if the first character is a number, it's a number; if the first character is not a number, it is not a number. That is why you can't start a variable name with a number. The PC Assembler Tutor 14 ______________________ letter q or the letter o: 641q , 2345o , 1472o 6. A binary number. A binary number is followed by a b: 0100100b , 1b , 01001000111010b Any of these types can be mixed on a line. For instance: db "Mary had a little lamb," , 13 , 10 13 followed by 10 is CRLF, the PC signal for a carriage return. A string in the C language ends with the number 0. If we wanted a C string with CRLF, we would have: db "Mary had a little lamb," , 13 , 10 , 0 Another mixed example: dw 7 , 010010b , 0FFFFh , 037q is dopey but legal. You can also have an equation, as long it resolves to a number. This calculation is done by the assembler, so the values of variables are not allowed: dw ( ( 19 * 7 * 25 ) + 6 ) / ( 9 + 7 ) is legal, but: data1 dw 25 data2 dw 7 dw ( ( 19 * 7 * data1 ) + 6 ) / ( 9 + data2 ) is illegal. Everything must be a constant. Remember that when the assembler starts calculating it might truncate the partial answers, so don't get too fancy. Chapter 2 - Data 15 ________________ SUMMARY The assembler works one line at a time. Each line with data must start with a data type declaration (after an optional name.) DATA TYPES DB define byte ( 1 byte ) DW define word ( 2 bytes ) DD define doubleword ( 2X2 bytes = 4 bytes ) DQ define quadword ( 4X2 bytes = 8 bytes ) DT define ten-byte ( 10 bytes ) DF define farword ( 6 bytes - used for 80386 only ) COMMON INTEGER TYPES TYPE MAX SIGNED MAX UNSIGNED byte -128/+127 255 word -32768/+32767 65535 doubleword -2147483648/+2147483647 4294967295 Note that the max. negative integer is 1 larger than the max. positive integer. POSSIBLE BASES FOR CONSTANTS b binary data o,q octal data d decimal data (default) h hex data (must start with a number 0 - 9) ARRAY DEFINITIONS d* num1 dup ( data1 ) Using the d* data type (db, dw, dd, dq, etc.) make num1 copies of data1 (data1 may be either a single piece of data or a group of data.) MULTIPLE DATA ON ONE LINE Different data elements on the same line are separated by commas. All elements on the same line have the same data type. Chapter 3 - ASMHELP =================== 16 We are now going to introduce both the 8086 registers and a program for looking at them. You are going to get information flying at you at a rapid pace, so read both this and the next chapter carefully and slowly. REGISTERS The 8086 has a number of registers. Remember, registers are places for storing data that are internal to the 8086 chip. They are much faster, but there are very few of them. There are 6 registers that you can use for addition and subtrac- tion of word (2 byte) sized numbers, as well as logical opera- tions on word (2 byte) numbers or data. These registers are AX, BX, CX, DX, SI, DI. In addition, there is a register which works the same way, but has a special function in all high-level languages (Basic, Pascal, C, etc.). This is BP, the base pointer. There is one more register that performs the same operations as the above seven, but it is RESERVED for special use and should never be used for anything. It is called SP (the stack pointer). There are 4 registers that tell the 8086 which memory segments you are in. They just sit there and help the 8086 find things in memory. You will learn how they work later. They are CS, DS, SS, ES. (That is Code Segment, Data Segment, Stack Segment and Extra Segment respectively). There is the flags register which contains all the information the 8086 needs to evaluate its state. We will learn about this later. The flags register has no name, and there are machine instructions for manipulating individual flags in the register. Finally, there is IP, the instruction pointer, which points to the machine instructions. You have no direct access to this, which is good because you would screw it up for certain. The 8086 handles the IP automatically and correctly. One word (two bytes or 16 bits) is the largest piece of data that the 8086 can handle naturally. It is possible to handle larger things, but we do it through software (which is slower), not hardware (which is faster). Sometimes we want to handle things one byte at a time as when we work with characters. The 8086 gives us this possibility by letting us divide the AX, BX, CX, and DX registers into an upper half and a lower half. For any or all of these registers, we can replace one 2 byte register by two 1 byte registers. The data in the full register stays the same, but we can look at each half. The two parts of AX are called AH (for A high) and AL (for A low). |------AX------| 0000000000000000 |--AH--||--AL--| Chapter 3 - ASMHELP.OBJ 17 _______________________ This is the 16 bit binary number 0 in the AX register. Using AX allows us to manipulate all 16 bits. Using AH allows us to manipulate the upper 8 bits (without affecting the lower 8 bits), and using AL allows us to manipulate the lower 8 bits without affecting the upper 8 bits. Similarly, for BX we have BH, BL, for CX we have CH, CL, and for DX we have DH, DL. SHOW_REGS We have named all the registers, now let's take a look at them. Included in the module asmhelp.obj is a program called show_regs. It shows all the above registers on the top 10 lines of your screen and allows you to enter data normally underneath. When you call show_regs, it puts the current value of all the registers on the screen. Those values stay on the screen until the next call - i.e. the program does not change what is on the screen even though the registers may be changing value. You need to call show_regs every time that you want to see the current values of the registers. The first time you call show_regs, it clears the screen so you should call it right at the beginning of the program in order to initialize the screen. This time we want temp2.asm for a template; we will call this program prog3.asm, so make a copy of temp2.asm and give it the new name. Let's take a look at it. The only differences are (1) there are a lot more programs in the EXTRN statements and (2) in the data segment DATASTUFF there are these definitons: ax_byte db 2 bx_byte db 2 cx_byte db 2 etc. These will be used for show_regs later, but you need to learn a few assembler instructions first. Here's our next program: TEMP2.ASM + + + + + + + + START CODE BELOW THIS LINE call show_regs label_one: call get_hex call show_regs call get_num call show_regs jmp label_one + + + + + + + + END CODE ABOVE THIS LINE The PC Assembler Tutor 18 ______________________ That's all the program does. It asks for a number, then calls show_regs to show us what is in the registers. Note that one of the numbers is hex while the other number is decimal. Compile this, and link it with >link prog3+\asmhelp ; and we're ready to go. The program reads information in the computer to find out what kind of monitor you have and where the screen output goes. It then puts the register information on the top lines. If it doesn't appear there, we have a screwup somewhere. The text should appear in black and white, but if you have a color monitor you can make it a blue background with white letters.{1} *********************** SCREEN SHOT *************************** AX 19825 SI 00000 BX 00000 DI 00256 CX 00255 BP 17113 DX 02596 SP 00508 CS 0AA4H DS 0A54H ES 0A24H SS 0A34H IP 0018H OF DF IEF TF SF ZF AF PF CF + x + x E COUNT 00003 ----------------------------------------------------------------- The PC Assembler Helper Version 1.0 Copyright (C) 1989 Chuck Nelson All rights reserved. Enter a two byte hex number 4df9 Enter any decimal number +19825 Enter a two byte hex number ***************************************************************** This is how the screen looks after entering first a hex number, then a decimal number. The numbers in the registers will probably be different for you. Note that AX contains the last number that was entered. On the left side you will see the AX, BX, CX and DX ____________________ 1 To make a blue background and white letters, insert the code "call set_blue" before the FIRST "call show_regs". i.e.: call set_blue call show_regs label_one: etc. This only works if it is allowed by the video board. Chapter 3 - ASMHELP.OBJ 19 _______________________ registers. For the time being, these registers will display unsigned numbers. On the right are the SI, DI, BP and SP registers. They are also unsigned numbers for the moment. Below that are the segment registers CS, DS, ES, and SS and the instruction pointer (IP). These are hex numbers and will always be hex numbers. The bottom line has OF, DF, IEF, etc. These are the flags, and the marking underneath them (either a blank or some character) tells how they are set. Finally we have COUNT. This has nothing to do with the 8086. It is a counter that is incremented each time you call show_regs. Keep entering numbers and watch the registers. You will notice that three things are changing - AX, IP and COUNT. AX has the last number you entered and IP keeps changing. Write down the value of IP each time it changes. It goes back and forth between two numbers. That is because you call show_regs in two different places in the loop, {2} and those are two different places in memory where the 8086 is reading the machine code. Why is AX changing? You may have wondered in prog1.asm and prog2.asm how that information was going back and forth between your program and asmhelp.obj. The answer is that in all the programs in asmhelp.obj, if you need to pass information, it is passed via register AX. This is not the normal way to pass information. The normal way is more elegant but more complicated. We will cover that much later. The counter, of course, increases by 1 each time you call show_regs. Try entering a few more numbers and then it's time to go on to the next program. MOVING DATA Obviously, we want to move data from memory to the 8086, from the 8086 to memory, and between registers. We have the following possibilities: (1) move from a register to memory (2) move from memory to a register (3) move a constant to memory (4) move a constant to a register (5) move from one register to another register That's it. There is no 8086 instruction that moves a single word or a byte from one place in memory to another. The move mnemonic for the 8086 is "mov".{3} We need some sample data: ____________________ 2 IP actually has three different values, since you call show_regs once before you enter the loop. 3 A mnemonic is a name for a machine instruction, which sounds like what the instruction is supposed to do - MOV for move, SUB for subtract, IMUL for integer multiplication, etc. The PC Assembler Tutor 20 ______________________ EXAMPLE DATA this_year dw 1989 total dw ? average dw ? time dw 7 age db ? ; I hope you aren't older than 255 poem db "In 1492 Columbus played a mean kazoo." secret_code db 3Bh character db ? Here is some sample code. To move from register to memory, we have: mov total, ax mov time, si mov age, cl mov character, bh The first thing that strikes the eye is that the destination is on the left and the source is on the right.{4} This is standard for the 8086 instruction set, and it's going to take some getting used to. DESTINATION ON LEFT, SOURCE ON RIGHT. You are going to blow this one from time to time, so always double check to make sure that they are in the right order. Also note that the 1 byte registers are matched up with 1 byte variables, and 2 byte registers are matched up with 2 byte data. If the sizes don't match, the assembler will complain.{5} Thus: mov age, ax is an illegal instruction. For examples of memory to register moves, we have: mov ch, secret_code mov di, this_year mov dl, poem ; moves 'I' to dl mov bx, time ____________________ 4 The computer community likes the words "destination" and "source". "Source" means FROM, and "destination" means TO. The 8086 instruction set is designed: MOV TO, FROM which is exactly opposite to the way you would say it in an English sentence. For the 8086, it is always destination on the left, source on the right. 5 Half register and full register operations have different machine codes, and the assembler needs to know which code to use, so the two things must be the same number of bytes. Chapter 3 - ASMHELP.OBJ 21 _______________________ Once again: (1) DESTINATION ON LEFT, SOURCE ON RIGHT, and (2) the sizes of the two objects must match. For constant to memory we have: mov total, 2986 mov age, 36h mov secret_code, 0110100b mov average, (27/5) + 3 The arithmetic is done by the assembler and anything that is made up totally of constants is legal. Thus: mov average, (((64+27)*51 )/(196-82)) is legal but: mov average, this_year/time is illegal. The assembler makes either a one byte or a two byte constant to match the size of the destination. The constants for "total" and "average" are two byte constants while those for "age" and "secret_code" are one byte constants. The constants must be within range, that is -129 is too negative for a byte, 256 is too positive for a byte, -32769 is too negative for a word, 65536 is too positive for a word. The assembler will give an error if the constant is out of range. You can also move a constant to a register: mov al, 'c' mov ax, 'c' mov di, 46280 mov bl, 99 The same rules apply. The constant must be within range and the assembler will make a constant the same size as the destination register (one or two bytes). These constants are actually imbedded in the machine code by the assembler, and are unchangable. Lastly, you can move data from one register to another: mov ax, cx ; from cx to ax mov ah, bl ; from bl to ah mov dl, dh ; from dh to dl mov di, dx ; from dx to di All this is summarized at the end of the chapter. It's time for program #4. All this program is going to do is get input and the move it to different registers. We are still using temp2.asm. Here's the program: TEMP2.ASM The PC Assembler Tutor 22 ______________________ ;+ + + + + + + + + + START DATA BELOW THIS LINE byte_data db ? word_data dw ? ;+ + + + + + + + + + END DATA ABOVE THIS LINE ;+ + + + + + + + + + START CODE BELOW THIS LINE call show_regs This_is_a_very_long_label_name: call get_hex ; (1) call show_regs_and_wait mov dx, ax ; (2) call show_regs_and_wait mov byte_data, al ; (3) mov ch, byte_data call show_regs_and_wait mov word_data, ax ; (4) mov di, word_data call show_regs jmp This_is_a_very_long_label_name ;+ + + + + + + + + + END CODE ABOVE THIS LINE There is a data section in this one, so copy those variables into your data section. Here is what the program does. (1) it gets a hex number from the keyboard, (2) it moves the number in ax to dx, (3) it moves one byte from al to ch via the variable byte_data, and (4) it moves two bytes from ax to di via word_data. There are two different subprograms - show_regs and show_regs_and_wait. They do the same thing except that show_regs_and_wait waits for you to hit the ENTER key before continuing. The computer works so fast that we wouldn't be able to see the changes in the screen if we didn't have a way of pausing. You can use COUNT on the screen to keep track of exactly where you are in the loop. Assemble program 4, link it to asmhelp.obj, and watch it work. There are two things to notice here. First, we are entering a hex number, but AX is displaying an unsigned number. It is not self-evident that the unsigned number in AX is the same as the hex number that you are entering. Secondly, though CH is changing, there doesn't seem to be any relationship between the number in AX and the number in CX. We will solve both problems in the next chapter. Chapter 3 - ASMHELP.OBJ 23 _______________________ SUMMARY MOVING DATA You can: (1) move from a register to memory (2) move from memory to a register (3) move a constant to memory (4) move a constant to a register (5) move from one register to another register The constants are actual constants which are imbedded in the machine code. REGISTERS The normal registers are AX, BX, CX, DX, SI, DI AND BP. AX, BX, CX, and DX can be divided into AH-AL, BH-BL, CH-CL AND DH-DL. The 'H' is for high and the 'L' is for low. SP is committed and may not be used. The segment registers are CS, DS, ES, SS. The instruction pointer (IP) is not available to you. The register that holds the flags is manipulated by special machine instructions. ASMHELP.OBJ Call show_regs to see what is in the 8086 registers. Count increments by one every time you call show_regs. show_regs_and_wait is the same as show_regs except that it waits for you to hit the ENTER key to allow you time to look at the screen. Call set_blue at the outset if you have a color card and a color monitor and you want to have a blue background. get_num gets a signed or unsigned number from the keyboard. (1-5 digits). It does no range checking to see whether the number is too big. All other input routines check to see if a number is too large for its data size. get_hex gets a hex number from the keyboard. (1-4 digits) get_ascii gets characters from the keyboard. (1 or 2 characters) get_binary gets a binary number from the keyboard (1 - 16 digits) print_num prints a number as hex, signed, unsigned, char, and binary. All data transfer to or from ASMHELP.OBJ is via the AX register. Chapter 4 - SHOW_REGS ===================== 24 We got started using the program show_regs in the last chapter, but we have already run into problems. The hex number doesn't look the same once we put it in the register - that's because what we are seeing in the arithmetic registers is an unsigned number. Also, when we moved a byte from AL to CH, it was clear that something had moved, but it wasn't clear what the number was. There are two problems here: (1) We want to use data in hex, binary, ascii, unsigned and signed format depending on what we are doing in the program. (2) Some of the registers can be used as half registers, so we want a whole register when we need it and a half register when we need it. Nooooooo problem. There are eight registers whose formats we want to be able to change: AX, BX, CX, DX, SI, DI, BP and SP. We need to give each one a code to tell it what to display. The code is the following: signed number 1 unsigned number 2 binary number 3 hex number 4 ascii 5 Also, we need to know whether AX, BX, CX and DX are half or full registers. The code for that is: half register 128 (80 hex) full register 0 We will need to do two things - set up the codes, then tell show_regs about the code. We'll begin by setting up the codes. First let's start with SI, DI, BP and SP. They must be full registers, so the half register information is irrelevant.{1} In the data section is a set of data starting ax_byte, bx_byte ... sp_byte. That is where you need to put the code. Don't change the order of these variables. Just put the correct formatting code in the appropriate byte. mov si_byte, 3 will display SI as a binary number. ____________________ 1 show_regs is very forgiving. It only recognizes half registers where appropriate, and if you screw up on the format code, it just makes it an unsigned number. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 4 - SHOW_REGS 25 _____________________ mov bp_byte, 4 will display BP as a hex number. mov di_byte, 1 will display DI as a signed number. That's pretty easy. If you are using AX, BX, CX, or DX as a full register, they are exactly the same. mov dx_byte, 5 ; full register, character format mov ax_byte, 2 ; full register, unsigned format mov bx_byte, 3 ; full register, binary If we use half registers we need to pass more information. show_regs needs to know that it is half registers, not full registers; it also needs to know what the left format is and what the right format is. This is easier than it sounds but will take a little getting used to. The right nibble (half byte) gets the right format and the left nibble (half byte) gets the left format. Then you add 128 to the total. This works easily in hex. 13h means that the left half register is (1 = signed) and the right register is (3 = binary). Then we add (128 = 80h) for a total of 93h. Here are some more examples. Remember, 8 + 1 = 9, 8 + 2 = A, 8 + 3 = B, 8 + 4 = C, 8 + 5 = D C4h ; 80h + 44h left and right are hex A5h ; 80h + 25h left is unsigned, right is ascii B1h ; 80h + 31h left is binary, right is signed D2h ; 80h + 52h left is ascii, right is unsigned 94h ; 80h + 14h left is signed, right is hex There is a summary at the end of the chapter giving all the commands and codes for show_regs. It is important to take some time here and learn to make the registers look the way we want, because later on we have machine instructions for signed numbers, for ascii characters, for binary numbers, and you need to see what the registers look like in the appropriate formats. Spending a little time right now will save you a lot of time later on. If you don't like using hex numbers you can use decimal numbers: code = type_of_register + left code + right code where full register = 0, half register = 128d and: NUMBER FORMAT LEFT CODE RIGHT CODE signed 16d 1d unsigned 32d 2d binary 48d 3d hex 64d 4d ascii 80d 5d Therefore, left binary, right hex = 128 + 48 + 4 = 180d. Some The PC Assembler Tutor 26 ______________________ more examples: left ascii, right signed = 128 + 80 + 1 = 209d left signed, right binary = 128 + 16 + 3 = 147d left hex, right unsigned = 128 + 64 + 2 = 192d We can put in the code the same way as before. mov ax_byte, 192d mov dx_byte, 147d mov cx_byte, 0D2h mov bx_byte, 94h We now have the codes in out program, but show_regs doesn't know about them. In order to give the information to show_regs, we call set_reg_style. set_reg_style makes a copy of your information for use by show_regs. The next time that you call show_regs, it will use the new register formats. There is a small problem, however. The information we have is eight bytes long, and ax is only two bytes long. How do we pass the information? The answer is: we don't pass the information. Instead, we pass the address of the information. If you look in the data segment of temp2.asm, you will see that ax_byte is the first byte of this data. We pass the address of ax_byte (the first byte) and set_reg_style knows that that address plus the following 7 addresses have the information that it needs. There is a special machine instruction for putting an address in a register - it is LEA or load effective address. It looks like this: lea ax, ax_byte This instruction says: put the address of ax_byte in the AX register. Combined with the call, we have: lea ax, ax_byte call set_reg_style Before we start writing programs with set_reg_style, we will run a pre-existing program called SETREGS.EXE. Its pathname is \ASMHELP\SETREGS.EXE. It puts the same (pseudo) random number in all arithmetic registers except SP, then requests a formatting code for each register. After cycling through all the registers, it asks you to press ENTER. It then puts a new random number in the registers and starts the cycle again. The hex codes are displayed on the screen before each request. As usual, use Control-C to exit the program. Here is what the screen might look like after the first cycle. The prseudo random number 2571 will be the same, but your formatting might be different: Chapter 4 - SHOW_REGS 27 _____________________ *********************** SCREEN SHOT *************************** AX +02571 SI +02571 BH 00001010 BL 0B ** DI 02571 CX 0A 0B ** BP 0000101000001011 DH 0A ** DL 0B ** SP 00C4H CS 0AA4H DS 0A42H ES 0A25H SS 0A35H IP 0115H OF DF IEF TF SF ZF AF PF CF x + x + O x COUNT 00009 ---------------------------------------------------------------- hex = C0h or 4h; ascii = D0h or 5h Enter a code for sp. Enter a one byte hex number 4 Press ENTER to continue ***************************************************************** The formats I have used are: AX full register (signed) BX half registers (binary, ascii) CX full register (ascii) DX half registers (ascii, ascii) SI full register (signed) DI full register (unsigned) BP full register (binary) SP full register (hex) Cycle through the registers a couple of times. If you make them binary, they get longer, if you make them hex or ascii they get shorter; a sign appears if they are signed, and you can change from full to half registers for AX, BX, CX and DX. You will always be able to tell what kind of number show_regs is printing because (1) a signed number always has a + or - in front of it, (2) a hex number always has an h after it, (3) a binary number is 8 digits long for a half register or 16 digits long for a full register, and (4) an ascii has an asterisk after it. Just as with print_num, if the ascii character has one of the values 0-32, 127 or 255, it will print a hex number and show a double asterisk '**' to signal the event. (5) If none of the above is true, then it is an unsigned decimal number. If you have a feel for what's happening, it is time to take a mini-test. This is an untimed test, so just make sure that it is correct. I'll give you a particular style, and you figure out the code for that style. The answers are at the bottom of the page. You don't have to memorize the codes. You should be using the summary at the end of the chapter for this quiz. 1. full register, binary 2. half register, left ascii, right hex 3. half register, left signed, right unsigned 4. full register, ascii 5. half register, left binary, right ascii 6. half register, left hex, right signed 7. half register, left unsigned, right binary The PC Assembler Tutor 28 ______________________ 8. full register, hex 9. full register, signed 10. half register, left signed, right binary. If you feel comfortable with what's going on and are able to do set the registers with the help of the summary, we are ready to move on. Here are the answers.{2} ____________________ 2 Here are the answers, both in hex and decimal. PROBLEM HEX DECIMAL 1. 3h 3d 2. D4h 212d 3. 92h 146d 4. 5h 5d 5. B5h 181d 6. C1h 193d 7. A3h 163d 8. 4h 4d 9. 1h 1d 10. 93h 147d These things are slow to calculate. It took me about a minute per problem to do both the hex and binary. Chapter 4 - SHOW_REGS 29 _____________________ SUMMARY The registers may be displayed in signed, unsigned, binary, hex, and ascii formats. The basic codes for this are: signed 1 unsigned 2 binary 3 hex 4 ascii 5 In addition you need to add the register type. They are: full register 0 half register 128d or 80h For the left half register, we have: FORMAT LEFT HEX LEFT DECIMAL signed 10h 16d unsigned 20h 32d binary 30h 48d hex 40h 64d ascii 50h 80d Since the left code is of interest only when the half register type is being used, we simply add 80h and come up with: FORMAT LEFT CODE RIGHT CODE signed 90h 1h unsigned A0h 2h binary B0h 3h hex C0h 4h ascii D0h 5h Or we add 128d and have: FORMAT LEFT CODE RIGHT CODE signed 144d 1d unsigned 160d 2d binary 176d 3d hex 192d 4d ascii 208d 5d SETTING THE FORMATS Formats are set by calling set_reg_style. The address of ax_byte must be in AX. The standard assembler instructions for this are: The PC Assembler Tutor 30 ______________________ lea ax, ax_byte call set_reg_style set_reg_style makes a copy of your format data. It changes nothing on the screen. The next time that you call show_regs, it will use the new formatting data. The correct order for the data in the data segment is: ax_byte, bx_byte, cx_byte, dx_byte, si_byte, di_byte, bp_byte, sp_byte. They are, of course, all byte sized data. Chapter 5 - Addition and Subtraction ==================================== 31 The first arithmetic operations we will look at are addition and subtraction, but before we do that, we need to look at one instruction that controls program flow. LOOP We already have JMP which sends you to a label: jmp label3 sends the program to label3, wherever that is in the code. Sometimes we want to repeat a section of code a specific number of times and then go on. For this, we have LOOP. LOOP decrements the CX register by 1. If CX is not zero after being decremented, LOOP jumps to the label indicated. If CX is zero after being decremented, LOOP falls through. The 8086 does not have general purpose registers. A general purpose register is a register that can be used for ALL instructions. There are a number of instructions on the 8086 which must be done with specific registers, and LOOP is the first one we meet. LOOP always looks at the CX register. This first program lets you enter a number and then loops that many times so you can watch the CX register. As usual, you exit the program by hitting Control-C. We use temp2.asm. temp2.asm ; - - - - START CODE BELOW THIS LINE call show_regs ; initialize outer_loop: call get_unsigned mov cx, ax ; number to cx inner_loop: call show_regs_and_wait loop inner_loop jmp outer_loop ; - - - - END CODE ABOVE THIS LINE A very simple program. As always, link it with asmhelp.obj. Get_unsigned gets a two byte number (less than 65536) and puts it in AX. We put that number in CX, and then watch the program loop. Make sure you use show_regs_and_wait, or everything will happen too fast for you to see. Try entering 0. On the first pass, loop will decrement CX from 0 to 65535. If CX is 0 when you enter, you ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 32 ______________________ have to repeat the loop 65536 times before you exit the loop. Hit Control-C now to exit. Throughout the book, I will use label names that end in '_loop' to indicate that they are the destination of a jump or loop instruction. The single word "loop" is a reserved word and may not be used as a label - it can only be used as an instruction. The addition program will have 4 sections and LOOP will give us the ability to do each section a limited number of times before going on to the next section. ADDITION If you read the introductory section on numbers carefully, you know that it is the same instruction for both signed and unsigned addition. The 8086 sets the flags correctly for both signed and unsigned addition. For signed addition, the following flags are set: OF the overflow flag is set (1) if the result is too negative or too positive, that is, if the result in the register does not show the correct result of the addition. It is cleared (0) otherwise. ZF the zero flag is 1 if the result is zero, and is cleared (0) if the result is non-zero. SF the sign flag is set (1) if the result is NEGATIVE and is cleared (0) if the result is POSITIVE. Zero is considered a positive number. For unsigned addition, the following flags are set: CF the carry flag is set (1) if the result is too large (over 255 for byte and over 65535 for word operations). It is cleared (0) otherwise. ZF the zero flag is the same as above. In addition, there are two more flags (PF the parity flag and AF the auxillary flag) which will be set or cleared; we will learn about them later. Show_regs shows all the flags. The setting for each flag is underneath its name. For the flags OF, ZF and CF, there is an 'X' if the flag is set and a blank if the flag is cleared. SF, the sign flag, is '-' if the flag is set and '+' if the flag is cleared. The addition program is fairly long because there are four things to look at - unsigned word addition, unsigned byte addition, signed word addition and signed byte addition. For that reason, it has already been typed in for you. It is called ADD1.ASM and its pathname is \XTRAFILE\ADD1.ASM. Print out a copy of it. Chapter 5 - Addition and Subtraction 33 ____________________________________ There are four blocks of code which are almost identical except the calls are a little different and two blocks refer to whole registers while the other two refer to half registers. At the head of each block is code to set the appropriate register styles for show_regs. SI, DI, and BP are not used and are set to 0 to make the screen easier to read. Here is the first block of code, which is typical. ; - - - CODE ; UNSIGNED WORD ADDITION mov ax_byte, 2 ; ax, bx, dx unsigned mov bx_byte, 2 mov dx_byte, 2 lea ax, ax_byte ; call set_reg_style call set_reg_style mov cx, 3 ; 3 iterations unsigned_loop: mov ax, 0 ; clear the registers for visibility mov bx, 0 mov dx, 0 call show_regs call get_unsigned ; first number to ax call show_regs push ax ; temporarily save ax call get_unsigned ; second number to bx mov bx, ax pop ax ; get ax back mov dx, ax ; copy of ax to dx add dx, bx ; dx (=ax) + bx call show_regs_and_wait loop unsigned_loop ; - - - CODE First, we set AX, BX, and DX for the appropriate register style. Here it is unsigned full register. We then put 3 in CX so we can have 3 iterations with loop. Upon entering the loop, AX, BX, and DX are cleared for reasons of visibility. We don't want the screen cluttered up with numbers. Get_unsigned gets a two byte unsigned number and returns it in AX. We want the first number to be visually on the top (which is AX), but there is a problem here. In order to get the second number we need to call get_unsigned again, and it is going to put another number in AX. We need to temporarily store the first number while we bring in the second number and transfer it to bx. There is a special 8086 instruction to do this, it is called PUSH. Push temporarily stores a word. The word can be either a full register or a word (two bytes) in memory. You can have either: variable1 dw 10000 push ax push variable1 The PC Assembler Tutor 34 ______________________ These are stored in a special place called the stack which we will talk about much later. When you want it back, you use the instruction POP. POP gets back the LAST thing that you pushed onto the stack. Things come off the stack in REVERSE order of how they were put on. push variable1 push variable2 push variable3 push variable4 pop variable4 pop variable3 pop variable2 pop variable1 is the correct order. This is used for temporary storage only, and the only thing which is accessable is the last thing which you PUSHed on the stack. We push AX to store it temporarily, call get_unsigned again and transfer the number to BX. We then pop AX to get the number back. The situation now is: the first number is in AX, the second number is in BX. For the actual addition, we transfer AX to DX and then add DX and BX. AX and BX contain the two numbers, and DX contains the result. Then you must press ENTER to continue. LOOP will jump to 'unsigned_loop' two times. The third time it will fall through to the next section of code. This program illustrates a hallmark of assembler code. It normally takes scads of code just to do something simple. Assemble add1.asm and link it with asmhelp.obj. Run it: ******************** SCREEN SHOT ****************************** AX 17428 SI 00000 BX 19755 DI 00000 CX 00003 BP 00000 DX 37183 SP 00508 CS 0AA5H DS 0A55H ES 0A25H SS 0A35H IP 004DH OF DF IEF TF SF ZF AF PF CF x + x - E COUNT 00004 ---------------------------------------------------------------- The PC Assembler Helper Version 1.0 Copyright (C) 1989 Chuck Nelson All rights reserved. Enter a number from 0 to 65535 17428 Enter a number from 0 to 65535 19755 Press ENTER to continue ***************************************************************** This is the screen after the first addition. I have added 17428 (AX) and 19755 (BX). The result 37183 is in DX. CX is still 3 because it hasn't LOOPed yet. Chapter 5 - Addition and Subtraction 35 ____________________________________ Notice that even though it is the same assembler instruction: add in all four blocks of code, it is doing both signed and unsigned addition correctly. When you are doing signed addition, you want to look at OF, the overflow flag, SF, the sign flag, and ZF, the zero flag after each addition to see how they are set. When you do unsigned addition, you want to look at CF, the carry flag, and ZF, the zero flag to see how they are set. Play around with this for a while, and then it is time for the next program. As in all 8086 instructions, the order is: add destination, source We add both numbers, and put the result in the destination, the thing on the left. There are five different types of addition you can do, (just as there are five different types of moves). They are: 1. add two registers 2. add a register to a variable (memory) 3. add a variable (memory) to a register 4. add a constant to a variable (memory) 5. add a constant to a register Here's a program that does all 5 things. Use template.asm to make this program. template.asm is almost the same as the other ones we have used. It has a few changes. First, it now lists ALL the subroutines you can call in asmhelp.obj.{1} Appendix 1 (\APPENDIX\APP1.DOC) contains a description of all the subroutines, what they do, and how they are called. Second, the size of STACKSEG is larger. We don't need this large of a stack now; it is for later. Finally, there is a section: ; + + + + + + + + + + PUT SUBROUTINES BELOW THIS LINE ; + + + + + + + + + + PUT SUBROUTINES ABOVE THIS LINE for subroutines. Ignore this. This is for later. From now on, we will always use template.asm unless it is explicitly stated that something else is being used. Here's the program: template.asm ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE ____________________ 1 This does not change the size of the .EXE file by even one byte, but it adds a lot of information to the .OBJ file, so they are much larger. The PC Assembler Tutor 36 ______________________ variable1 dw ? variable2 dw ? ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE call show_regs outer_loop: call get_unsigned ; first number to ax push ax ; store ax call get_unsigned ; second number to bx mov bx, ax pop ax ; restore ax mov variable1, ax ; first number to variable1 mov variable2, bx ; second number to variable2 ; add 2 registers mov cx, ax ; cx + bx add cx, bx ; add register to memory add variable1, bx mov dx, variable1 ; put in dx for display ; add memory to register mov si, ax add si, variable2 ; add a constant to memory add variable2, 25 mov di, variable2 ; put in di for display ; add a constant to a register mov bp, bx add bp, 25 call show_regs jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE The program puts the first number in AX and the second number in BX. It then proceeds to do the same addition (first number plus second number) three times. These are: 1. CX = add two registers (CX + BX) 2. DX = add a register to memory (variable1 + BX) 3. SI = add memory to a register (SI + variable2) Finally it adds a constant (second number + 25). These are: 4. DI = add a constant to memory (variable2 + 25) 5. BP = add a constant to a register (BP + 25) On the 8086, it is not possible to add two things in memory. That is: add variable1, variable2 is an illegal instruction. Instead, you need to write: mov ax, variable2 Chapter 5 - Addition and Subtraction 37 ____________________________________ add variable1, ax SUBTRACTION It is now time to do some subtraction. The instruction is SUB: sub destination, source subtracts source from destination and stores it in destination, the thing on the left. sub ax, cx ; (ax - cx) -> ax In order to do subtraction we are going to modify add1.asm, so make a copy and call it sub1.asm: >copy add1.asm sub1.asm How many instructions do we need to change to modify the program? Four. add dx, bx -> sub dx, bx add dl, bl -> sub dl, bl Each of these is changed twice, and we are ready to roll. Assemble it, link it, and run it. Once again we want to look at the flags at the end of each subtraction. For unsigned subtraction, look at ZF, the zero flag, and CF, the carry flag. This time, CF will be set if the result is below zero. For signed subtraction, look at OF, the overflow flag, SF, the sign flag, and ZF, the zero flag. As with addition, subtraction changes PF, the parity flag and AF the auxillary flag. They don't concern us. As with addition, there are five possibilities for subtraction. They are: 1. subtract one register from another 2. subtract a register from a variable (memory) 3. subtract a variable (memory) from a register 4. subtract a constant from a variable (memory) 5. subtract a constant from a register the code for these is: sub cx, bx ; (cx - bx) -> cx sub variable1, bx ; (variable1 - bx) -> variable1 sub si, variable2 ; (si - variable2) -> si sub variable2, 25 ; (variable2 - 25) -> variable2 sub bp, 25 ; (bp - 25) -> bp You can copy add2.asm to sub2.asm if you want and change the five ADD instructions to SUB instructions. It will then do those five types of subtraction. The PC Assembler Tutor 38 ______________________ SIGNED AND UNSIGNED NUMBERS What should you do if you are doing unsigned addition or subtraction and the carry flag gets set? It depends. Sometimes it makes a difference, sometimes it doesn't. If you have an error handling routine, then you can call it with the following code: add ax, bx jnc go_on call error_handler go_on: JC and JNC are conditional jump instructions. JC (jump on carry) jumps if the carry flag is set (1) and JNC (jump on not carry) jumps if the carry flag is not set (0). Using reverse logic here, we skip the error handler if everything is ok. For signed numbers, it is certainly an error if there is overflow. You are making mathematical calculations and you now have invalid data. One possibility is to do the same as above but with the overflow flag. add ax, bx jno go_on call error_handler go_on: JO and JNO are two more conditional jump instructions. JO (jump on overflow) jumps if the overflow flag is set (1) and JNO (jump on not overflow) jumps if the overflow flag is not set (0). We use the same logic here. However, there is one special instruction for signed numbers, and that is INTO (interrupt on overflow). It is possible to have an error handler external to your program. It sits permanantly in memory. When you make a signed arithmetic error, INTO interrupts your program and goes to the external error handler. The code looks like this: add ax, bx into You probably don't have an error handler in your computer right now. In that case, INTO simply goes looking for it and returns when it can't find it. Let's find out if you have an error handler installed. Once again, use template.asm ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE mov ax_byte, 1 ; signed register style mov bx_byte, 1 mov cx_byte, 1 lea ax, ax_byte call set_reg_style call show_regs Chapter 5 - Addition and Subtraction 39 ____________________________________ outer_loop: call get_signed push ax call get_signed mov bx, ax pop ax mov cx, ax add cx, bx into call show_regs jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE This is basically the same thing as before, but using AX, BX, and CX. They are set for signed style, and then we get two signed numbers and add them. The result is in CX. Right after the addition instruction is INTO. If the result is too positive or too negative, OF will be set and INTO will look for the error handler. Assemble this program and link it with asmhelp.obj. Try both numbers that do not set the overflow flag and numbers that do set the overflow flag. Did anything different happen when the overflow flag was set? If nothing different happened, you don't have an error handler for INTO. Included on the disks is an error handler. It is called INTO.COM, and it's pathname is \XTRAFILE\INTO.COM. When you run it: >into it will install itself and then return to the command prompt: > INTO.COM will stay in memory until you reboot or shut off the machine. INTO.COM provides the type of sophisticated error handling that you might want to use in a real program. Install (run) INTO.COM, and then try the previous program again, both with numbers that cause an overflow and numbers that don't cause an overflow. The PC Assembler Tutor 40 ______________________ SUMMARY ADD performs both signed and unsigned addition. It can: 1. add two registers 2. add a register to a variable (memory) 3. add a variable (memory) to a register 4. add a constant to a variable (memory) 5. add a constant to a register SUB performs both signed and unsigned subtraction. It can: 1. subtract one register from another 2. subtract a register from a variable (memory) 3. subtract a variable (memory) from a register 4. subtract a constant from a variable (memory) 5. subtract a constant from a register The flags affected by both ADD and SUB are: CF the carry flag (for unsigned). Set if the 0/65535 (0/255) border was crossed. ZF the zero flag (for signed and unsigned). Set if the result is 0. SF the sign flag (for signed). Set if the result is negative. OF the overflow flag (for signed). Set if the result was too negative or too positive. PF the parity flag and AF, the auxillary flag The following jump instructions are conditional on the setting of the flags: JC jump on carry, JNC, jump on not carry JO jump on overflow. JNO, jump on not overflow LOOP LOOP decrements cx by 1. If cx is then not zero, it jumps to the named label. If cx is zero, it falls through to the next instruction. INTO If the overflow flag is set, INTO (interrupt on overflow) interrupts the program and goes to an external error handler if one exists. It returns immediately if one doesn't exist. PUSH and POP PUSH stores either a register or a word (in memory) in a temporary storage area. POP retrieves the last word PUSHed. Chapter 6 - Multiplication and Division ======================================= 41 Unlike addition and subtraction, where the result can be in either memory or in one of the registers, the multiplication and division instructions have a rigid format. MULTIPLICATION You can multiply a one byte number by a one byte number and get a two byte result, or you can multiply a one word number by a one word number and get a two word result. The first number MUST be in AL for the byte operation or in AX for the word operation. The second number may be a register or a memory location (but not a constant). The result is in AH:AL for the byte operation and DX:AX for the word operation. Our possibilities are: AL X (one byte register or memory) -> AH:AL AX X (one word register or memory) -> DX:AX Is there a difference between signed and unsigned numbers? Yes, a very big difference. For the byte operation FFh = -1 signed but FFh = 255 unsigned. -1 X -1 = 1 = 0001h. 255 X 255 = 65025 = FE01h. These are two completely different answers. You need to tell the 8086 whether you want signed multiplication or unsigned multiplication. The 8086 does the rest. Let's look at both signed and unsigned multiplication. We'll do byte multiplication for unsigned numbers and word multiplication for signed numbers. The instruction for unsigned multiplication is MUL. The instruction for signed multiplication is IMUL. AX or AL is understood to be the register, so it is not in the code. The instructions are: variable1 db ? variable2 dw ? mul bx ; unsigned word from a register mul variable1 ; unsigned byte from memory imul ch ; signed byte from a register imul variable2 ; signed word from memory No AX or AL. It's understood. Here's our program: template.asm ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE answer1 dw ? answer2 dw ? ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ; - - - - - START CODE BELOW THIS LINE mov cx, 0 ; clear cx for visual effect ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 42 ______________________ outer_loop: ; unsigned byte multiplication mov ax_byte, 0A2h ; half regs, unsigned mov bx_byte, 0A2h ; half regs, unsigned lea ax, ax_byte call set_reg_style mov ax, 0 ; clear regs mov bx, 0 mov dx, 0 call show_regs call get_unsigned_byte ; get two unsigned bytes call show_regs push ax ; save the first number call get_unsigned_byte mov bl, al pop ax ; restore the first number call show_regs_and_wait mul bl ; unsigned multiplication call print_unsigned ; display the result (ax) call show_regs_and_wait ; signed word multiplication mov ax_byte, 01h ; full reg, signed mov bx_byte, 01h mov dx_byte, 01h lea ax, ax_byte call set_reg_style mov ax, 0 ; clear regs mov bx, 0 call show_regs call get_signed ; get two numbers call show_regs push ax ; save the first number call get_signed mov bx, ax pop ax ; restore the first number call show_regs_and_wait imul bx ; signed multiplication push ax ; save result mov answer1, ax ; display 4 byte result mov answer2, dx lea ax, answer1 call print_signed_4byte pop ax ; restore result call show_regs_and_wait jmp outer_loop ; - - - - - END CODE ABOVE THIS LINE If the answer for the unsigned byte multiplication is greater than 255, it will be difficult to read the answer from the half Chapter 6 - Multiplication and Division 43 _______________________________________ registers, so we print out the whole AX register. If the answer for the signed word multiplication is greater than +32767 or is less than -32768, the answer will be unreadable in the DX:AX registers. We move the answer to memory, and then call print_signed_4byte. As with set_reg_style, the data is too long to be put in AX, so we pass the address of the first byte of data with: lea ax, answer1 and then call print_signed_4byte. Everything from PUSH AX to POP AX is designed to do that. Do MUL and IMUL set any flags? Yes. For byte multiplication, if AL contains the total answer, the 8086 clears the OF and CF flags. If part of the answer is in AH, then the 8086 sets both the OF and CF flags. For word multiplication, if AX contains the total answer, the 8086 clears the OF and CF flags. If part of the answer is in DX, then the 8086 sets both the OF and CF flags. What do we mean by the total answer? This is simple for unsigned multiplication. If AH (or DX for word) is 0, then the total answer is in AL (or AX for word). It is more complicated for signed multiplication. Consider word multiplication. +30000 X +2 = +60000. But that's less than 65536, so it is completely contained in AX, right? WRONG. The leftmost bit of AX contains the sign. If the signed result is out of the range -32768 to +32767, information about the absolute value of the number is corrupting information about the sign of the number. AX will have the wrong number and the wrong sign. Only by combining AX with DX will you get the correct answer. Similarly for byte multiplication with AL, if the result is not in the range -128 to +127, The leftmost (sign) bit will be corrupted, and only by looking at AH:AL will you be able to get the correct result. If CF and OF are set, you need to look at both registers to evaluate the number. You might want to do error handling, so once again, you can have: mul bx jnc go_on call error_handler go_on: using the same reverse logic as before (if nothing is wrong, skip the error handler). We can also use: mul bx into if there is an INTO error handler. DIVISION Division operates in the same way as multiplication. Word The PC Assembler Tutor 44 ______________________ division operates on the DX:AX pair and byte division operates on the AH:AL pair. There are two instructions, DIV for unsigned division and IDIV for signed division. After the division: byte AL = quotient, AH = remainder word AX = quotient, DX = remainder Both DIV and IDIV operate on BOTH registers. For bytes, they consider AH:AL a single number. This means that AH must be set correctly before the division or you will get an incorrect answer. For words, they consider DX:AX a single number. This means that DX must be set correctly before the division, or the result will be incorrect. Why did Intel include AH and DX in the division? Wouldn't it have been easier to use just AH (or AX for word division) and put the quotient and remainder in the same place? These instructions are actually designed for dividing a long number (4 or 8 bytes). How it works is pretty slick; you'll find out about it later in the book. How do you set AH and DX correctly? For unsigned numbers, that's easy. Make them 0: mov al, variable mov ah, 0 div cl ; unsigned byte division For signed division, set AH or DX to 0 (0000h) if it is a positive number and set them to -1 (FFFFh) if the number is negative. This is just standard sign extension that was covered in the chapter on numbers. Fortunately for us, Intel has provided instructions which do the sign extension for us. CBW (convert byte to word) correctly extends the signed number in AL through AH:AL. CWD (convert word to double) correctly extends the signed number in AX through DX:AX. The code is mov ax, variable5 cwd idiv bx ; signed word division Of course with these two instructions you can convert a byte to a double word. mov al, variable6 cbw cwd idiv bx ; signed word division first converting to a word, then to a double word. For the division program, we are going to use the multiplication program and make some small changes. Make a copy of your multiplication program: >copy mult.asm div.asm and then make the following changes: Chapter 6 - Multiplication and Division 45 _______________________________________ MULTIPLICATION DIVISION ; unsigned byte ; unsigned byte pop ax pop ax mov ah, 0 call show_regs_and_wait call show_regs_and_wait mul bl div bl ; signed word ; signed word pop ax pop ax cwd call show_regs_and_wait call show_regs_and_wait imul bx idiv bx The calls to print_unsigned and print_signed_4byte are irrelevant, so you may either delete them or ignore the output. All we did was change the multiplication instruction to division and prepare the upper register correctly (AH for byte, DX for word). That's all. Assemble, link, and run it. Try out both positive and negative numbers and see what the remainder looks like. Also notice the sign extension just before the division. Remember, for division, the results are in the following places: byte AL = quotient, AH = remainder word AX = quotient, DX = remainder Now divide by 0. Ka-pow! You should have exited the program and gotten an error message. Unlike the other arithmetical errors where you have the option of ignoring them or making an error handler for them, the 8086 considers division by 0 a major no-no. When the 8086 detects division by zero,{1} it interrupts the program and goes to the zero-divide handler (which is external to the program). Normally, this just exits the program since the data is now worthless. ____________________ 1 What it actually detects is that the quotient is too large to fit in the lower register (AL for byte or AX for word). As long as the upper register is correctly sign extended, the only time this can happen is when you divide by 0. If the upper register is NOT sign extended correctly, you can have zero divide errors all over the place, even though you aren't dividing by 0. As an example, if AH:AL contain 3275 and bl contains 10, then: div bl will give a quotient of 327 ( > 255) and will generate a zero divide error. The PC Assembler Tutor 46 ______________________ SUMMARY MUL and IMUL MUL performs unsigned multiplication and IMUL performs signed multiplication. For bytes, the multiplicand is in AL and the result is in the AH:AL pair. For words, the multiplicand is in AX and the result is in the DX:AX pair. If the total result is contained in the lower register, CF and OF are cleared (0). If part of the result is in the upper register, CF and OF are set (1). The multiplier may be either a register or a variable in memory. variable1 db ? variable2 dw ? mul variable1 ; unsigned byte mul cx ; unsigned word imul bl ; signed byte imul variable2 ; signed word DIV and IDIV DIV performs unsigned division. IDIV performs signed division. For bytes, the dividend is the AH:AL pair. For words, the dividend is the DX:AX pair. In byte division, AH must be correctly prepared before the division. For word division, DX must be correctly prepared before the division. The divisor may be either a register or a variable in memory. variable1 db ? variable2 dw ? div variable1 ; unsigned byte div cx ; unsigned word idiv bl ; signed byte idiv variable2 ; signed word The quotient and remainder are as follows: byte AL = quotient, AH = remainder word AX = quotient, DX = remainder No flags are affected. If the quotient is too large for the lower register, or if you divide by zero, a zero divide program interrupt occurs. CORRECT SIGN EXTENSION To prepare for division, you must correctly sign extend the lower register into the upper register. For unsigned division, zero the upper register (AH = 0 or DX = 0). For signed division, use CBW and CWD. CBW (convert byte to word) extends a signed number in AL through AH:AL. CWD (convert word to double) extends a signed number in AX through DX:AX Chapter 7 - LOGIC ================= 47 There are a number of operations which work on individual bits of a byte or word. Before we start working on them, it is necessary for you to learn the Intel method of numbering bits. Intel starts with the low order bit, which is #0, and numbers to the left. If you look at a byte: 7 6 5 4 3 2 1 0 that will be the ordering. If you look at a word: 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 that is the ordering. The overwhelming advantage of this is that if you extend a number, the numbering system stays the same. That means that if you take the number 45 : 7 6 5 4 3 2 1 0 0 0 1 0 1 1 0 1 (45d) and sign extend it: 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 each of the bits keeps its previous numbering. The same is true for negative numbers. Here's -73: 7 6 5 4 3 2 1 0 1 0 1 1 0 1 1 1 (-73d) 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 1 (-73d) In addition, the bit-position number denotes the power of 2 that it represents. Bit 7 = 2 ** 7 = 128, bit 5 = 2 ** 5 = 32, bit 0 = 2 ** 0 = 1. {1}. Whenever a bit is mentioned by number, e.g. bit 5, this is what is being talked about. AND We will use AND as the prototype. There are five different ways you can AND two numbers: ____________________ 1 I'm using the Fortran convention for showing exponents. That is, 2 ** 7 is 2 to the 7th, 3 ** 19 is 3 to the 19th. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 48 ______________________ 1. AND two register 2. AND a register with a variable 3 AND a variable with a register 4. AND a register with a constant 5. AND a variable with a constant That is: variable1 db ? variable2 dw ? and cl, dh and al, variable1 and variable2, si and dl, 0C2h and variable1, 01001011b You will notice that this time the constants are expressed in hex and binary. These are the only two reasonable alternatives. These instructions work bit by bit, and hex and binary are the only two ways of displaying a number bitwise (bit by bit). Of course, with hex you must still convert a hex digit into four binary digits. The table of bitwise actions for AND is: 1 1 -> 1 1 0 -> 0 0 1 -> 0 0 0 -> 0 That is, a bit in the result will be set if and only if that bit is set in both the source and the destination. What is this used for? Several things. First, if you AND a register with itself, you can check for zero. and cx, cx If any bit is set, then there will be a bit set in the result and the zero flag will be cleared. If no bit is set, there will be no bit set in the result, and the zero flag will be set. No bit will be altered, and CX will be unchanged. This is the standard way of checking for zero. You can't AND a variable that way: and variable1, variable1 is an illegal instruction. But you can AND it with a constant with all the bits set: and variable1, 11111111b If the bit is set in variable1, then it will be set in the result. If it is not set in variable1, then it won't be set in the result. This also sets the zero flag without changing the variable. AND is also used in masks, which will be covered at the end of the chapter. Chapter 7 - Logic 49 _________________ Finally, there is a variant of AND called TEST. TEST does exactly the same thing as AND but throws away the results when it is done. It does not change the destination. This means that it can check for specific things without altering the data. It has the same possibilities as AND: variable1 db ? variable2 dw ? test cl, dh test al, variable1 test variable2, si test dl, 0C2h test variable1, 01001011b will set the flags exactly the same as the similar AND instructions but will not change the destination. We need a concrete example, and for that we'll turn to your video card. In text mode, your screen is 80 X 25. That is 2000 cells. Each cell has a character byte and an attribute byte. The character byte has the actual ascii number of the character. The attribute byte says what color the character is, what color the background is, whether the character is high or low intensity and whether it blinks. An attribute byte looks like this: 7 6 5 4 3 2 1 0 X R G B I R G B Bits 0,1 and 2 are the foreground (character) color. 0 is blue, 1 is green, and 2 is red. Bits 4, 5, and 6 are the background color. 4 is blue, 5 is green, and 6 is red. Bit 3 is high intensity, and bit 7 is blinking. If the bit is set (1) that particular component is activated, if the bit is cleared (0), that component is deactivated. The first thing to notice is how much memory we have saved by putting all this information together. It would have been possible to use a byte for each one of these characteristics, but that would have required 8 X 2000 bytes = 16000 bytes. If you add the 2000 bytes for the characters themselves, that would be 18000 bytes. As it is, we get away with 4000 bytes, a savings of over 75%. Since there are four different screens (pages) on a color card, that is 18000 X 4 = 72000 bytes compared to 4000 X 4 = 16000. That is a huge savings. We don't have the tools to access these bytes yet, but let's pretend that we have moved an attribute byte into dl. We can find out if any particular bit is set. TEST dl with a specific bit pattern. If the zero flag is cleared, the result is not zero so the bit was on. If the zero flag is set, the result is zero so that bit was off test dl, 10000000b ; is it blinking? test dl, 00010000b ; is there blue in the background? test dl, 00000100b ; is there red in the foreground? The PC Assembler Tutor 50 ______________________ If we look at the zero flag, this will tell us if that component is on. It won't tell us if the background is blue, because maybe the green or the red is on too. Remember, test alters neither the source nor the destination. Its purpose is to set the flags, and the results go into the Great Bit Bucket in the Sky. OR The table for OR is: 1 1 -> 1 1 0 -> 1 0 1 -> 1 0 0 -> 0 If either the source or the destination bit is set, then the result bit is set. If both are zero then the result is zero. OR is used to turn on a specific bit. or dl, 10000000b ; turn on blinking or dl, 00000001b ; turn on blue foreground After this operation, those bits will be on whether or not they were on before. It changes none of the bits where there is a 0. They stay the same as before. XOR The table for XOR is: 1 1 -> 0 1 0 -> 1 0 1 -> 1 0 0 -> 0 That is, if both are on or if both are off, then the result is zero. If only one bit is on, then the result is 1. This is used to toggle a bit off and on. xor dl, 10000000b ; toggle blinking xor dl, 00000001b ; toggle blue foreground Where there is a 1, it will reverse the setting. Where there is a 0, the setting will stay the same. This leads to one of the favorite pieces of code for programmers. xor ax, ax zeros the ax register. There are three ways to zero the ax register: mov ax, 0 sub ax, ax xor ax, ax Chapter 7 - Logic 51 _________________ The first one is very clear, but slightly slower. For the second one, if you subtract a number from itself, you always get zero. This is slightly faster and fairly clear.{2} For the third one, any bit that is 1 will become 0, and and bit that is 0 will stay 0. It zeros the register as a side effect of the XOR instruction. You'll never guess which one many programmers prefer. That's right, XOR. Many programmers prefer the third because it helps make the code more obsure and unreadable. That gives a certain aura of technical complexity to the code. NEG and NOT NOT is a logical operation and NEG is an arithmetical operation. We'll do both here so you can see the difference. NOT toggles the value of each individual bit: 1 -> 0 0 -> 1 NEG negates the value of the register or variable (a signed operation). NEG performs (0 - number) so: neg ax neg variable1 are equivalent to (0 - AX) and (0 - variable1) respectively. NEG sets the flags in the same way as (0 - number). ; negvsnot.asm ; compares the operations NEG and NOT ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE mov ax_byte, 1 ; signed mov bx_byte, 1 ; signed mov cx_byte, 1 ; signed mov si_byte, 3 ; binary mov di_byte, 3 ; binary mov bp_byte, 3 ; binary mov dx, 0 ; not used, so clear lea ax, ax_byte call set_reg_style call show_regs outer_loop: call get_signed ; get number mov bx, ax ; move it to all registers mov cx, ax mov si, ax mov di, ax mov bp, ax not bx ; NOT the second row down ____________________ 2 This is one of the first instructions in the template files. The PC Assembler Tutor 52 ______________________ not di neg cx ; NEG the third row down neg bp call show_regs jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE This is set up so the left registers are signed and the right registers are binary. The top registers retain the original value, the second registers down will do NOT and the third registers down will do NEG. Put a number in. On the right side, you will see that NOT (in DI) reverses the bit pattern, while on the left, NEG (in CX) negates the number. Do a few more. This is always true. But they seem to be related. In fact, BX will always be 1 too negative. Why? Remember that in the introduction on numbers, when we changed signs, we had: negative of number = one's complement + 1 but the one's complement is exactly NEG. It switches the value of each bit. To get the values in the third row (CX and BP), simply add 1 to the values in the second row (BX and DI). Remember, NOT is a logical operation, NEG is an arithmetic operation. MASKS To explain masks, we'll need some data, and we'll use the attribute byte for the monitor. Here it is again: 7 6 5 4 3 2 1 0 X R G B I R G B Bits 0,1 and 2 are the foreground (character) color. 0 is blue, 1 is green, and 2 is red. Bits 4, 5, and 6 are the background color. 4 is blue, 5 is green, and 6 is red. Bit 3 is high intensity, and bit 7 is blinking. What we want to do is turn certain bits on and off without affecting other bits. What if we want to make the background black without changing anything else? We use and AND mask. and video_byte, 10001111b Bits 0, 1, 2, 3 and 7 will remain unchanged, while bits 4, 5 and 6 will be zeroed. This will make the background black. What if we wanted to make the background blue? This is a two step process. First we make the background black, then set the blue background bit. This involves first the AND mask, then an OR mask. and video_byte, 10001111b or video_byte, 00010000b Chapter 7 - Logic 53 _________________ The first instruction shuts off certain bits without changing others. The second turns on certain bits without effecting others. The binary constant that we are using is called a mask. You may write this constant as a binary or a hex number. You should never write it as a signed or unsigned number (unless you are one of those people who just adores making code unreadable). If you want to turn off certain bits in a piece of data, use an AND mask. The bits that you want left alone should be set to 1, the bits that you want zeroed should be set to 0. Then AND the mask with the data. If you want to turn on certain bits in a piece of data, use an OR mask. The bits that you want left alone should be set to 0. The bits that you want turned on should be set to 1. Then OR the mask with the data. Go back to AND and OR to make sure you believe that this is what will happen. The PC Assembler Tutor 54 ______________________ SUMMARY For AND, TEST, OR, and XOR you can have the following combinations: 1. two register 2. a register with a variable 3 a variable with a register 4. a register with a constant 5. a variable with a constant AND is a bitwise logical operation. 1 1 -> 1 1 0 -> 0 0 1 -> 0 0 0 -> 0 TEST does the same thing as AND except that the result is discarded. It is used for setting the flags without altering the data. OR is a bitwise logical operation. 1 1 -> 1 1 0 -> 1 0 1 -> 1 0 0 -> 0 XOR is a bitwise logical operation. 1 1 -> 0 1 0 -> 1 0 1 -> 1 0 0 -> 0 You can use NOT and NEG with either a register or a variable in memory. NOT is a bitwise logical operation 0 -> 1 1 -> 0 NEG is an arithmetic operation. NUMBER -> - NUMBER. It gives the negative of a signed number. MASKS If you want to turn off certain bits of a piece of data, use an AND mask. The bits that you want left alone should be set to 1, the bits that you want zeroed should be set to 0. Then Chapter 7 - Logic 55 _________________ AND the mask with the data. If you want to turn on certain bits of a piece of data, use an OR mask. The bits that you want left alone should be set to 0. The bits that you want turned on should be set to 1. Then OR the mask with the data. Chapter 8 - Shift and Rotate ============================ 56 There are seven instructions that move the individual bits of a byte or word either left or right. Each instruction works slightly differently. We'll make a standard program and then substitute each instruction into that program. SAL - SHL The instructions SHL (shift logical left) and SAL (shift arithmetic left) are exactly the same. They have the same machine code. They shift each bit to the left. How far? That depends. There are two (and only two) forms of this instruction. All other shift and rotate instructions have these two (and only these two) forms as well. The first form is: shl al, 1 Which shifts each bit to the left one bit. The number MUST be 1. No other number is possible. The other form is: shl al, cl shifts the bits in AL to the left by the number in CL. If CL = 3, it shifts left by 3. If CL = 7, it shifts left by 7. The count register MUST be CL (not CX). The bits on the left are shifted out of the register into the bit bucket, and zeros are inserted on the right. The easy way to understand this is to fire up the standard program. Remember, from now on we always use template.asm. ;sal.asm ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE mov ax_byte, 0A3h ; half reg, low reg binary mov bx_byte, 0A4h ; half reg, low reg hex mov cx_byte, 0A1h ; half reg, low reg signed mov dx_byte, 0A2h ; half reg, low reg unsigned lea ax, ax_byte call set_reg_style mov ax, 0 ; clear registers mov bx, 0 mov cx, 0 mov dx, 0 mov di, 0 mov bp, 0 call show_regs outer_loop: call get_hex_byte ; get number and put in registers mov bl, al mov cl, al ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 8 - Shift and Rotate 57 ____________________________ mov dl, al mov si, 8 ; 8 iterations of the loop and al, al ; set the flags call show_regs_and_wait shift_loop: sal al, 1 sal bl, 1 sal cl, 1 sal dl, 1 call show_regs_and_wait dec si jnz shift_loop jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE This standard program is with bytes, not words. This is because if we had used words we would have performed 16 individual shifts and that would have been time consuming and boring. First we set the style to half registers. Notice that one is binary, one is hex, one is signed and one is unsigned. That covers all bases. All the registers are then cleared. It would be nice to use the loop instruction, but CX is committed, so we make our own loop instruction. We move 8 into SI. The loop instructions are: dec si jnz shift_loop DEC decrements a register or a variable by 1. Its counterpart INC increments a register or variable by 1. JNZ (jump if not zero) jumps to 'shift_loop' if SI is not zero. We get a hex byte in AL and put the same byte in BL, CL, and DL. This way we will be able to see what is happening in binary, hex, signed and unsigned. Before starting, we have: and al, al This is there to set the flags correctly before starting. All four are shifted left one bit each time, and then we look at the result. Assemble, link and run it. Enter the number 7. In binary, that is (0000 0111). Take a look at the flags before starting. It is a positive number so SF shows '+'. ZF is not set. PF shows 'O'. O stands for odd. Every time you perform an arithmetic or logical operation, the 8086 checks parity. Parity is whether the number contains an even or odd number of 1 bits. This contains 3 1 bits, so the parity is odd. The possible settings are 'E' for even and 'O' for odd.{1} SAL checks for parity (though some of the other instructions don't). Now press ENTER. It will shift left 1 and you will have (0000 1110). What does the unsigned number say now? 14. Press ENTER again. (0001 1100) What does the unsigned number say? 28. Again (0011 1000) 56. Again (0111 0000) 112. Notice that ____________________ 1 This is for use by communications programs. The PC Assembler Tutor 58 ______________________ the signed number reads +112. Look at the CF and OF. They are both cleared. Things are going to change now. Press ENTER again. (1110 0000). SF is now '-'. OF, the overflow flag is set because you changed the number from positive to negative (from +112 to -32). What is the unsigned number now? 224. CF is cleared. PF is '0'. Shift again. (1100 0000) OF is cleared because you didn't change signs. (Remember, the leftmost bit is the sign bit for a signed number). PF is now 'E' because you have two 1 bits, and two is even. CF is set because you shifted a 1 bit off the left end. Keep pressing ENTER and watch SF, OF, CF, and PF. Let's look at the unsigned numbers we had until we started shifting 1 bits off the left end. We started with 7, then had 14, 28, 56, 112, 224. This instruction is multiplying by 2. That's right, and it is MUCH faster than multiplication (about 50 times faster). Far and away the fastest way to multiply a register by 2, 4 or 8 is to use sal. ; by 2 ;by 4 ; by 8 sal di,1 sal di, 1 sal di, 1 sal di, 1 sal di, 1 sal di, 1 For a register, it is faster to use a series of 1 shifts than to load cl. For a variable in memory, anything over 1 shift is faster if you load cl. Do a few more numbers to see what is happening both with the number and the flags. CF always signals when a 1 bit has been shifted off the end. SAR and SHR Unlike the left shift instruction, there are two completely different right shift instructions. SHR (shift logical right) shifts the bits to the right, setting CF if a 1 bit is pushed off the right end. It puts 0s in the leftmost bit. Make a copy of SAL.ASM and replace the four instructions: sal al, 1 sal bl, 1 sal cl, 1 sal dl, 1 with SHR. We'll call the new program SHR.ASM. Run this one too. Instead of 7, use E0h (1110 0000) which is 224d. The first time you shift (0111 0000) the OF flag will be set because the sign changed. Keep shifting, noting the flags and the unsigned number. This time we have 224, 112, 56, 28, 14, 7, 3, 1. It is dividing by two and is once again MUCH faster than division. For a single shift, the remainder is in CF. For a shift of more than one bit, you lose the remainder, but there is a way around this which we will discuss in a moment. Do some more numbers till you are comfortable with the flags and the operation. If you want to divide by 16, you will shift right four times, so Chapter 8 - Shift and Rotate 59 ____________________________ you'll lose those 4 bits. But those bits are exactly the value of the remainder. All we need to do is: mov dx, ax ; copy of number to dx and dx, 0000000000001111b ; remainder in dx mov cl, 4 ; shift right 4 bits shr ax, cl ; quotient in ax Using a mask, we keep only the right four bits, which is the remainder. SAR SAR (shift arithmetic right) is different. It shifts right like SHR, but the leftmost bit always stays the same. This will make more sense when you run the program. Make another copy, call it SAR.ASM, and change the four instructions to SAR. The flags operate the same as for SHR and SHL. The overflow flag will never change since the left bit will always stay the same. First enter 74h (+116). We will be looking at the signed numbers only. Copy down the signed numbers as you go along. They should be: 116, 58, 29, 14, 7, 3, 1, 0, 0. Now try 8Ch (-116). The numbers you should get are: -116, -58, -29, -15, -8, -4, -2, -1, -1. They started out the same, then they got off by one. The negative numbers are one too negative. Try 39h (+57). The numbers here are: 57, 28, 14, 7, 3, 1, 0, 0, 0. Just as it should be for division by 2. Now try C7 (-57). Here the numbers are: -57, -29, -15, -8, -4, -2, -1, -1, -1. This time it went screwy right off the bat. Once again, the negative numbers are one too negative. SAR is an instruction for doing signed division by 2 (sort of). It is, however, an incomplete instruction. The rule for SAR is: SAR gives the correct answer if the number is positive. It gives the correct answer if the number is negative and the remainder is zero. If the number is negative but there is a remainder, then the answer is one too negative. The reason for this is a little complex, but we need to add some code if we want to do signed division.{2} For SHR, the remainder part was optional. Here it is not. We need to know whether the remainder is zero or not. For this example we will do a word shift left by 6. That's dividing by 64. remainder_mask dw 002Fh ; 63 call get_signed ; number in ax mov bx, ax ; copy in bx and bx, remainder_mask ; the remainder mov cl,6 ; shift right 6 bits sar ax, cl jns continue ; is it positive? ____________________ 2 Both the code and the reasons will be explained (but not proved) in the summary. The PC Assembler Tutor 60 ______________________ and bx, bx ; is the remainder zero? jz continue inc ax continue: We get the remainder, then shift right 6 bits. Upon finishing SAR, the sign flag will be set correctly. Here is yet another jump. This one is JNS (jump on not sign) jumps if the sign flag is NOT set, that is if the number is positive. If it is positive, then everything is ok so we skip ahead. If the number is negative, then we check to see if there was a remainder. If there wasn't, everything is ok, so we go ahead. If there was a remainder, then we INC (add 1) ax. Is the remainder correct? If the number was positive, the remainder is correct, but if the number was negative, then we need to do one more thing. After INC, but before 'continue' we have a SUB instruction: inc ax sub bx, 64 ; correct the remainder continue: Why that is the correct number will be explained in the summary. What a lot of work when we could simply write: mov cx, 64 call get_signed cwd ; sign extend idiv cx ; signed division Is there any advantage to this instruction? Not really. Remember that the more you shift, the longer it takes. If you shift 2, then it's about 1/3 faster than division. If you shift 14, then it is only 15% faster than division. Considering that even a slow PC can do 25000 divisions a second, you must be in serious need of speed to use this. In any case, you will never or almost never use SAR for signed division, while you will find lots of opportunity to use SHR and SHL for unsigned multiplication and division. ROR and ROL ROR (rotate right) and ROL (rotate left) rotate the bits around the register. We will just do one program since they operate the same way, only in opposite directions. Make another copy of SAL.ASM and put in ROR in the appropriate spots. Enter a number. This time you will notice that the bits, rather than dissapearing off the end, reappear on the other side. They rotate around the register. The only flags that are defined are OF and CF. OF is set if the high bit changes, and CF is set if a 1 bit moves off the end of the register to the other side. Do a few more, and we'll go on to the last two instructions. Chapter 8 - Shift and Rotate 61 ____________________________ RCR and RCL RCR (rotate through carry right) and RCL (rotate through carry left) rotate the same as the above instructions except that the carry flag is involved. Rotating right, the low bit moves to CF, the carry flag and CF moves to the high bit. Rotating left, the high bit moves to CF and CF moves to the low bit. There are 9 bits (or 17 bits for a word) involved in the rotation. Make yet another copy of the program, and change those 4 instructions to RCR. Also, since we have 9 bits instead of 8, change the loop count to 9 from 8: mov si, 9 Enter a number and watch it move. Before you start moving, look at CF and see if there is anything in it. There are only two flags defined, OF and CF. Obviously, CF is set if there is something in it. OF is wierd. In RCL (the opposite instruction to the one we are using), OF operates normally, signalling a change in the top (sign) bit. In RCR, OF signals a change in CF. Why? I don't have the slightest idea. You really have no need for the OF flag anyways, so this is unimportant. Well, those are the seven instructions, but what can you do with them besides multiply and divide? First, you can work with multiple bit data. The 8087 has a word length register called the status register. Looking at the upper byte: 15 14 13 12 11 10 9 8 X X X bits 11, 12 and 13 contain a number from 0 to 7. The data in this register is not directly accessable. You need to move the register into memory, then into an 8086 register. If you want to find what this number is, what do you do? mov bx, status_register_data mov cl, 3 ror bx, cl and bh, 00000111b we rotate right 3 and then mask off everything else. The number is now in BH. We could have used SHR if we wanted. Another 8087 register is the control register. In the upper byte it has: 15 14 13 12 11 10 9 8 X X a number from 0 to 3 in bits 10 and 11. If we want the information, we do the same thing: mov bx, control_register_data mov cl, 2 ror bx, cl The PC Assembler Tutor 62 ______________________ and bh, 00000011b and the number is in BH. You are now going to write a program that inputs an unsigned number and prints out its hex representation. Here it is: ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE mov ax_byte, 0A5h ; half regs, right ascii mov bx_byte, 4 ; hex mov dx_byte, 4 ; hex lea ax, ax_byte call set_reg_style call show_regs outer_loop: call get_unsigned mov bx, ax mov dx, ax mov cx, 4 inner_loop: push cx ; save cx mov cl, 4 rol bx, cl ; rotate left 1/2 byte mov al, bl ; copy to al and al, 0Fh ; mask off upper 1/2 byte cmp al, 10 ; < 10, 0 - 9 ; > 9 A - F jae use_letters add al, '0' ; change to ascii jmp print_it use_letters: add al, 'A' - 10 ; 10 = 'A' print_it: call print_ascii_byte call show_regs_and_wait pop cx loop inner_loop jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE AL will be shown in ascii while BX and DX will be in hex. We save the original number in DX. Since the first thing we want to print is the left hex character, we rotate left, not right. We move the low byte to AL, mask off everything but the low hex number and then convert to an ascii character. If it is 0 - 9, we add '0' (the character, not the number). If it is > 9, we add "'A' - 10" and get a letter (if the number is 10, we get 'A'). JAE means jump if above or equal, and is an unsigned comparison.{3} ____________________ 3 You are getting innundated with conditional jump instructions. Don't worry. As long as you understand each one when you run across it, you don't have to remember it. All jump instructions will be covered soon. Chapter 8 - Shift and Rotate 63 ____________________________ Finally, we print the ascii character that is in AL.{4} Another thing to notice is that just inside the loop we push CX. That is because we use CL for the ROL instruction. It is then POPped just before the loop instruction. This is typical. CX is the only register that can be used for counting in indexed instructions. It is common for indexing instructions to be nested, so you temporarily store the old value of CX while you are using CX for something different. push cx ; typical code for a shift mov cl, 7 shr si, cl pop cx Finally, let's multiply large numbers by 2. Here's the code: ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE byte1 db ? byte2 db ? byte3 db ? byte4 db ? error_message db "Result is too large.", 0 ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE outer_loop: lea ax, byte1 ; get 4 byte number call get_unsigned_4byte shl byte1, 1 rcl byte2, 1 rcl byte3, 1 rcl byte4, 1 jnc go_on lea ax, error_message call print_string go_on: lea ax, byte1 call print_unsigned_4byte jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE This will require some explaination. Get_unsigned_4byte gets a number from 1 to four billion. We put it in memory. Normally, the following instructions would be done word by word. We are doing them byte by byte so you can see the mechanics of the situation. The low byte is shifted left 1 bit. This doubles it, but may shift a 1 bit from the high bit into CF. If it does, then it will be present when we rotate byte2. That moves CF into the low bit and moves the high bit into CF. We do it again. And again. If there is an unsigned overflow, it will be signalled by CF being ____________________ 4 Any subroutine in ASMHELP.OBJ that involves a one byte input or output has the data in AL. The PC Assembler Tutor 64 ______________________ set after: rcl byte4, 1 JNC (jump on not carry) will skip the error message if everything is ok. Print_string prints a zero terminated string, that is a C string which is terminated by the number (not the character) 0. Finally, we print the number. A word about large numbers in ASMHELP.OBJ. It is assumed that you would like to use commas if you could. Any data type over 1 word long allows commas. The following are considered the same by ASMHELP.OBJ in its input routines: 23546787 2,3,5,4,6,7,8,7 23,,5,46,,78,7 23,546787 23,546,787 It always prints commas correctly in the print routines. Chapter 8 - Shift and Rotate 65 ____________________________ SUMMARY All shift and rotate instructions operate on either a register or on memory. They can be either 1 bit shifts: sal cx, 1 ror variable1, 1 shr bl, 1 or shifts indexed by CL (it must be CL): rcl variable2, cl sar si, cl rol ah, cl SHL and SAL SHL (shift logical left) and SAL (shift arithmetic left) are exactly the same instruction. They move bits left. 0s are placed in the low bit. Bits are shoved off the register (or memory data) on the left side, and CF indicates whether the last bit shoved was a 1 or a 0. It is used for multiplying an unsigned number by powers of 2. SHR SHR (shift logical right) does the same thing as SHL but in the opposite direction. Bits are shifted right. 0s are placed in the high bit. Bits are shoved off the register (or memory data) on the right side and CF indicates whether the last bit shoved off was a 0 or a 1. It is used for dividing an unsigned number by powers of 2. SAR SAR (shift arithmetic right) shifts bits right. The high (sign) bit stays the same throughout the operation. Bits are shoved off the register (or memory data) on the right side. CF indicates whether the last bit shoved off was a 1 or a 0. It is used (with difficulty) for dividing a signed number by powers of 2. ROR and ROL ROR (rotate right) and ROL (rotate left) rotate the bits of a register (or memory data) right and left respectively. The bit which is shoved off one end is moved to the other end. CF indicates whether the last bit moved from one end to the other was a 1 or a 0. RCR and RCL The PC Assembler Tutor 66 ______________________ RCR (rotate through carry right) and RCL (rotate through carry left) rotate the bits of a register (or of memory data) right and left respectively. The bit which is shoved off the register (or data) is placed in CF and the old CF is placed on the other side of the register (or data). INC INC increments a register or a variable by 1. inc ax inc variable1 DEC DEC decrements a register or a variable by 1. dec ax dec variable1 The following is fairly technical. It is only for those willing to wade their way through a turgid explaination. If you don't understand it, forget it. CODE FOR SHL If you are shifting an UNSIGNED number right by 'X' bits, it is the same as dividing by (2 ** X) 1 bit = (2**1 = 2), 2 bits = (2**2 = 4), 7 bits = (2**7 = 128). This is the same as dividing by a number which is all 0s except the Xth bit which is 1 (for 0 we have 0000 0001, for 1 we have 0000 0010, for 3 we have 0000 1000, for 7 we have 1000 0000). The remainder mask will be this number minus 1 (for 0 we have 0000 0000, for 1 we have 0000 0001, for 3 we have 0000 0111, for 7 we have 0111 1111). CODE FOR SAR The order of numbers is important for SAR. If you start with 0 and add 1 each time, the actual sequence of signed numbers that you get (from the bottom up) is: -1 -2 . . -32767 -32768 +32767 +32766 . . 3 2 1 0 Chapter 8 - Shift and Rotate 67 ____________________________ The positive numbers are increasing in absolute value while the negative numbers are decreasing in absolute value. If you divide by shifting and there is no remainder, then the quotient is exact. If there is a remainder, the quotient will truncate towards 0 IN THE ABOVE DIAGRAM. This means that positive numbers will truncate down, while the negative numbers will truncate towards -32768, and will be one too negative. If the number was positive, the remainder will be positive and will be exactly the same as for SHR. If the number was negative, then things are more complicated. We'll take division by 32 as an example. If we divide by 32 (0010 0000) the remainder mask will be 31 (0001 1111). If the number is negative, then what we get when we AND the mask: and ax, 00011111b is not the remainder but (remainder + 32). In order to get the actual negative remainder, we need to subtract 32. This gives us (remainder + 32 - 32). remainder mask = divisor - 1 negative remainder correction = NEG divisor. Chapter 9 - Jumps ================= 68 So far we have done almost exclusively sequential programming - the order of execution has been from one instruction to the next until the very end, where the jump instruction has brought us back to the top. This is not the normal way programs work. In order to have things like DO loops, FOR loops, WHILE loops, CASE constructions and IF-THEN-ELSE constructs, we need to make decisions and to be able to go to different blocks of code depending on our decisions. Intel has provided a wealth of conditional jumps to answer all our needs. All of them are listed in a summary at the end of this chapter. The thing we will do most often is compare the size of two numbers. In BASIC code: IF A < B THEN we need to see if A < B . If that is true, we do one thing, if it is false, we do something else. One thing that we need to watch out for is whether A and B are signed or unsigned numbers. Let A = F523h (signed -2781; unsigned 62755) and B = 59E0h (signed +23008; unsigned 23008). If A and B are signed numbers, then A < B. However, if they are unsigned numbers, then A > B. In C and PASCAL, the compiler takes care of whether you want signed or unsigned numbers (BASIC assumes that it is always signed numbers). You are now on the machine level, and must take care of this yourself. To compare two numbers, you subtract one from the other, and then evaluate the result (less than 0, 0, or more than 0). To compare A and B, you do A minus B. To compare AX and BX, you can: sub ax, bx and then evaluate it. Unfortunately, if you do that, you will destroy the information in AX. It will have been changed in the process. Intel has solved this problem with the CMP (compare) instruction. cmp ax, bx subtracts BX from AX, sets the flags, and throws the answer away. CMP is exactly the same as SUB except that AX remains unchanged. We probably don't want to save the result anyway. If we do, we can always use: sub ax, bx We have subtracted BX from AX. There are now three possibilities. (1) AX > BX so the answer is positive, (2) AX = BX so the answer is zero, or (3) AX < BX so the answer is negative. But are these ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 9 - Jumps 69 _________________ signed or unsigned numbers? The 8086 uses the same subtract (or CMP) instruction for both signed or unsigned numbers. You have to tell the 8086 in the following instruction whether you were talking about signed or unsigned numbers. cmp ax, bx ja some_label asks the machine to compare AX and BX. It then says that AX and BX were UNSIGNED numbers and the machine should jump to "some_label" if AX was above BX. cmp ax, bx jg some_label asks the machine to compare AX and BX. It then says that AX and BX were SIGNED numbers and the machine should jump to "some_label" if AX was greater than BX. The 8086 makes the decision by looking at the flags. They give it complete information about the result. (The decision making rules are in the summary). In our example on the previous page, we had A = -2781 (unsigned 62755) and B = +23008 (unsigned 23008). If we put A in AX and B in BX, then the instruction JA will execute the jump, while the instruction JG will not execute the jump. What happens if the 8086 doesn't execute the jump? It goes on to the next instruction. As I said before, the 8086 has LOTS of conditional jumps. I am going to give you a list of them now, but be forewarned that the mnemonics for these suckers are confusing. Rather than doing something sensible like JUG for 'jump unsigned greater' and JSG for 'jump signed greater' - things that would be easy to distinguish, they have JA for 'jump above' (unsigned) and JG for 'jump greater' (signed).{1} Therefore, use the summary at the end of the chapter. With enough use, you will remember which is which. The arithmetic jump instructions have two forms, a positive one and a negative one. For instance: ja ; jump above jnbe ; jump not (below or equal) are actually the same machine instruction. You use whichever one makes the program clearer. We will have examples later to illustrate both positive and negative mnemonics. Here are the signed and unsigned instructions and their meaning. The ____________________ 1 This wierd use of the words above and greater, below and less, is so confusing that my copy of the Microsoft assembler manual v5.1 has it reversed on one page. It calls the signed jumps unsigned and the unsigned jumps signed. And that's the one place where it SHOULD be right. The PC Assembler Tutor 70 ______________________ equivalent mnemonics will appear in pairs. THESE ARE THE SIGNED JUMP INSTRUCTIONS jg ; jump if greater jnle ; jump if not (less or equal){2} jl ; jump if less jnge ; jump if not (greater or equal) je ; jump if equal jz ; jump if zero jge ; jump if greater or equal jnl ; jump if not less jle ; jump if less or equal jng ; jump if not greater jne ; jump if not equal jnz ; jump if not zero These are self-explainatory, keeping in mind that these apply only to signed numbers. THESE ARE THE UNSIGNED JUMP INSTRUCTIONS ja ; jump if above jnbe ; jump if not (below or equal) jb ; jump if below jnae ; jump if not (above or equal) je ; jump if equal jz ; jump if zero jae ; jump if above or equal jnb ; jump if not below jbe ; jump if below or equal jna ; jump if not above jne ; jump if not equal jnz ; jump if not zero These apply to unsigned numbers, and should be clear. JZ, JNZ, JE and JNE work for both signed and unsigned numbers. After all, zero is zero. ____________________ 2 I was trying to decide whether or not to put in the parentheses. If there are two things after the "not" the "not" applies to both of them. By the rules of logic, not (less or equal) == not less AND not equal. If you don't understand this, try to find someone who can explain it to you. Chapter 9 - Jumps 71 _________________ Before we get going, there is one more thing you need to know. The unconditional jump: jmp some_label can jump anywhere in the current code segment. XXXX is the current number in CS, then jmp can go from XXXX offset 0 to XXXX offset 65535. Conditional jumps are something else entirely. ALL conditional jumps (including the loop instruction) are limited range jumps. They are from -128 to +127. That is, they can go backward up to 128 bytes and they can go forward up to 127 bytes.{3} You will find that you will get assembler errors because the conditional jumps are too far away, but don't worry because we can ALWAYS fix them. You will find out later how to deal with them. As in the other arithmetic instructions, there are five forms of the CMP instruction. 1. compare two registers 2. compare a register with a variable 3. compare a variable with a register 4. compare a register with a constant 5. compare a variable with a constant These look like: cmp ah, dl cmp si, memory_data cmp memory_data, ax cmp ch, 127 cmp memory_data, 14938 Here are some decisions we might have to make in programs. (1) We are writing a program to print hex numbers on the screen. We have one routine for 0 - 9 and a different one for A - F. Sound familiar? cmp al, 10 jb decimal_routine ; here we are past the jump, so we can start the A - F ; routine. (2) We want to fire everyone over the age of 55 (this is an unsigned number since there are no negative ages): ____________________ 3 But they don't jump from the beginning of the machine instruction, they jump from the END of the machine instruction (which is two bytes long). This means that they have an effective range of from -126 bytes to +129 bytes from the BEGINNING of the instruction. The PC Assembler Tutor 72 ______________________ cmp employee_age, 55 ja find_a_reason_for_termination ; start of routine for younger employees here. (3) You want to know if you need to go to a loanshark: mov di, bank_balance cmp unpaid_bills, di jg gotta_go_see_Vinnie ; continue normal routine here (4) Notice that the last one could have also been written: mov di, bank_balance cmp di, unpaid_bills jl gotta_go_see_Vinnie ; continue normal routine though to my eye the first one seems clearer. (5) You have the results from two calculations, the first one in DI and the second one in CX. You need to go to different routines depending on which is larger. If they are the same, you exit: cmp di, cx jg routine_one ; di is greater jl routine_two ; cx is greater jmp exit_label ; they are equal We had two conditional jumps in a row and both of them were able to look at the results of the CMP instruction because the first jump (JG) did not change any of the flags. This is true of all jumps - conditional jumps, the unconditional jump, and LOOP. They never change any of the flags, so you can have two jumps in a row and be certain that the flags will not be altered. (6) Your dad has promised to buy you a Corvette if you don't get suspended from school more than three times this semester. Here's his decision code: cmp number_of_suspensions, 3 jng buy_him_the_corvette ; better luck next time JUMP OUT OF RANGE If the code you are jumping to is more than -128 or +127 bytes from the end of a conditional jump, you will get an assembler error that the jump is out of range. This can always be fixed. Take this code: Chapter 9 - Jumps 73 _________________ cmp ax, bx jg destination_label further_code: ; continue the program If 'destination_label' is out of range, what you need to do is jump to 'further_code' with a conditional jump (you are within range of 'further_code') and use JMP (which can go anywhere in the segment) to go to 'destination_label'. To switch the jump, you simply negate the jump condition: jg -> jng je -> jne jne -> je jbe -> jnbe We use reverse logic. Originally, if the condition was met we jumped. If the condition was not met we continued. Now, if the condition is NOT met, we jump, and if the condition is NOT not met (yes, there are two NOTs) which means it was met, we continue, and this sends us to the JMP instruction. Make sure you believe this works correctly before going on. The code then looks like this: cmp ax, bx jng further_code jmp destination_label further_code: ; continue the program This is the standard way of handling the situation. The PC Assembler Tutor 74 ______________________ SUMMARY CMP CMP performs the same operation as subtraction, but it does not change the registers or variables, it only sets the flags. It is the cousin of TEST. As usual, there are five possibilities: 1. compare two registers 2. compare a register with a variable 3. compare a variable with a register 4. compare a register with a constant 5. compare a variable with a constant THESE ARE THE SIGNED JUMP INSTRUCTIONS jg ; jump if greater jnle ; jump if not (less or equal) jl ; jump if less jnge ; jump if not (greater or equal) je ; jump if equal jz ; jump if zero jge ; jump if greater or equal jnl ; jump if not less jle ; jump if less or equal jng ; jump if not greater jne ; jump if not equal jnz ; jump if not zero THESE ARE THE UNSIGNED JUMP INSTRUCTIONS ja ; jump if above jnbe ; jump if not (below or equal) jb ; jump if below jnae ; jump if not (above or equal) je ; jump if equal jz ; jump if zero jae ; jump if above or equal jnb ; jump if not below jbe ; jump if below or equal jna ; jump if not above jne ; jump if not equal jnz ; jump if not zero Chapter 9 - Jumps 75 _________________ THESE JUMPS CHECK A SINGLE FLAG These come in opposite pairs jc ; jump if the carry flag is set jnc ; jump if the carry flag is not set jo ; jump if the overflow flag is set jno ; jump if the overflow flag is not set jp or jpe ; jump if parity flag is set (parity is even) jnp or jpo ;jump if parity flag is not set (parity is odd) js ; jump if the sign flag is set (negative ) jns ; jump if the sign flag is not set (positive or 0) THIS CHECKS THE CX REGISTER jcxz ; jump if cx is zero Why do we have this instruction? Remember, the loop instruction decrements CX and then checks for 0. If you enter a loop with CX set to zero, the loop will repeat 65536 times. Therefore, if you don't know what the value of CX will be when you enter the loop, you use this instruction just before the loop to skip the loop if CX is zero: jcxz after_the_loop loop_start: . . . . loop loop_start after_the_loop: INFORMATION ABOUT JUMPS The unconditional jump (JMP) can go anywhere in the code segment. All other jumps, which are conditional, can only go from -128 to +127 bytes from the END of the jump instruction (that's from -126 to +129 bytes from the beginning of the instruction). Jumps have no effect on the 8086 flags. How does the 8086 decide whether something is more, less, or the same? The rules for unsigned numbers are easy. If you subtract a larger number from a smaller, the subtraction will pass through zero and will set the carry flag (CF). If they are the same, the result will be zero and it will set the zero flag (ZF). If the first number is larger, the machine will clear the carry flag and the zero flag will be cleared (ZF = 0). Therefore, for unsigned numbers, we have: The PC Assembler Tutor 76 ______________________ First number is: above CF = 0 ZF = 0 equal CF = 0 ZF = 1 not equal CF = ? ZF = 0 below CF = 1 ZF = 0 All other unsigned compares are combinations of those three things: jae = ja OR je (CF = 0 and ZF = 0) or ZF = 1 jbe = jb OR je CF = 1 or ZF = 1 When you have a negative condition, it is much easier to look at its equivalent positive condition to figure out what is going on: jnae is the same as jb CF = 1 jnbe is the same as ja CF = 0 ZF = 0 SIGNED NUMBERS This section is not for the fainthearted. It is not necessary, so if you find yourself getting confused, just remember that if you see documentation talking about a jump where the overflow flag equals the sign flag or the overflow flag doesn't equal the sign flag, it is talking about SIGNED numbers. Zero is zero, so we won't concern ourselves with it here. It is exactly the same. If A and B are two signed word sized numbers and we have: cmp A, B then we can have four different cases: 1) If A is just a little greater than B [(A - B) <= +32767], then the result will be a small positive number, and there will be no overflow. SF = 0, OF = 0. 2) If A is much greater than B [+32767 < (A - B)], then the result will be too positive and it will overflow from positive to negative. This will set both the sign flag (it is now negative) and the overflow flag. SF = 1, OF = 1. 3) If A is a little less than B [-32768 <= (A - B)], that is if it is only a little negative, then the result is a small negative number, and there is no overflow. SF = 1, OF = 0. 4) If A is much less than B [(A - B) < -32768], then the result is a large negative number. It is too negative and overflows into a positive number. SF = 0, OF = 1. Recapping, for a positive result: 1) SF = 0, OF = 0 Chapter 9 - Jumps 77 _________________ 2) SF = 1, OF = 1 and for a negative result: 3) SF = 1, OF = 0 4) SF = 0, OF = 1 For positive results (and zero), SF = OF. For negative results, SF is not equal to OF. This, in fact, is how the 8086 decides a signed jump. If SF = OF, it's positive, if SF is not equal to OF, it's negative. If ZF = 1, then obviously they are equal. Here is the list: greater SF = OF ZF = 0 equal SF = OF ZF = 1 not equal ZF = 0 less SF is not equal to OF As with the unsigned numbers, if you have a negative condition, it is easier to change it into its equivalent positive condition and then figure out the requirements. For instance: jnge same as jl SF is not equal to OF jnl same as jge ( SF = OF and ZF = 0 ) or ( ZF = 1) If you think about it, this OF = SF stuff does make sense. We are subtracting two numbers. If the first one is greater, then the answer will be positive. It can either be a little positive as in (cmp 0, -1 ) = 1 or it can be very positive, as in (cmp 32767, -32768) = 65,535 (same as -1). If it is just a little positive, there is no overflow and it has a positive sign (SF = 0, OF = 0). If the difference gets large, then the number overflows from + to -. At that point OF = 1, but it now has a negative sign, so OF = SF. The flags MUST match. In the opposite case where the second number is greater, The answer is negative. It can either be a little negative as in (cmp 12, 13)= -1, or it can be very negative (cmp -32768, 32767) = -65535 =1. If it is a small difference, the sign is negative, but there is no overflow (SF = 1, OF = 0). As the difference gets larger, the number overflows from negative to positive so the sign flag is now positive, but the overflow flag is set (SF = 0, OF = 1). Those flags CAN'T match. Chapter 10 - Templates ====================== 78 Do you remember when you were younger and you needed to look up a word in the dictionary? It would define the word in terms of a second word which you didn't know so you would look that up too. Most likely that second word was either defined in terms of a third word you didn't know or it referred you back to the first word. This chapter is something like that. The items in the template file are interdependent. If you're lucky, everything will be clear by the time you have finished the chapter. If not, you'll have to reread it. There are four different things which operate on the assembler instructions which you write - the ASSEMBLER, the LINKER, the LOADER and the 8086. 1) The ASSEMBLER takes your text and turns it into the machine code that is used by the 8086. It is complete except that the addresses of data and subroutines might change during linking and loading. The assembler generates information called HEADER files which give the LINKER and LOADER the information they need to update these addresses in the machine code. This means that you can move the code anywhere in memory. 2) If your program is made up of more than one file, the LINKER links them together. It then makes it ready for running. If there is only one file, the linker makes it ready for running. It does this by updating the addresses of anything it has moved. It still leaves the HEADER files which contain the segment addresses. 3) At run time, the LOADER, which is part of the operating system, decides where to put your program in memory. It loads the program, and adjusts any segment addresses in the program to reflect where the program actually is in memory. It then gives control to the program. 4) The code is fixed at the time the 8086 takes over. Any addresses are constants and are unchangable. Keep this in mind as we work through the template file. THE .LST FILE The first thing we need to look at is segments. Let's look at a slightly modified version of the template file called segs.asm. Here it is. ;*********************************** ; segs.asm ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 10 - Templates 79 ______________________ ; - - - - - - - - - - - - - STACKSEG SEGMENT STACK 'STACK' variable4 dw 4444h dw 100h dup (?) STACKSEG ENDS ; - - - - - - - - - - - - - MORESTUFF SEGMENT PUBLIC 'HOHUM' variable2 dw 2222h MORESTUFF ENDS ; - - - - - - - - - - - - - DATASTUFF SEGMENT PUBLIC 'DATA' variable1 dw 1111h DATASTUFF ENDS ; - - - - - - - - - - - - - CODESTUFF SEGMENT PUBLIC 'CODE' EXTRN print_num:NEAR , get_num:NEAR ASSUME cs:CODESTUFF,ds:DATASTUFF ASSUME es:MORESTUFF,ss:STACKSEG variable3 dw 3333h main proc far start: push ds sub ax,ax push ax mov ax, DATASTUFF mov ds,ax mov ax, MORESTUFF mov es,ax mov cx, variable1 mov variable1, cx ret main endp CODESTUFF ENDS ; - - - - - - - - - - - - END start ;*************************** There is an extra segment put in that has the definition MORESTUFF SEGMENT PUBLIC 'HOHUM' The PC Assembler Tutor 80 ______________________ There is a variable defined in each segment including the stack segment. These variables all have numbers in them, and the numbers are in hex so they will be easy to read. There are only two external subroutines (neither of which is called). It is time to take a look at an assembler listing. ----- THIS IS FROM THE SCREEN ----- C>masm segs Microsoft (R) Macro Assembler Version 5.10 Copyright (C) Microsoft Corp 1981, 1988. All rights reserved. Object filename [segs.OBJ]: Source listing [NUL.LST]: segs Cross-reference [NUL.CRF]: ----------- If you don't put a semicolon after the filename with masm, you get some prompts. The first asks you if you want the object file name to be different from the asm file name. You may change either the name or the name and the extension. If you don't want to change either, just press ENTER. The second asks if you want a listing. Normally you don't, so you just press ENTER. This time we do, so we give it the same name as the assembler file. The assembler will generate a file SEGS.LST. Finally, it asks if you want the information needed to create a cross-reference file. We won't cover that. Once again, press ENTER. The assembler generates an object file and a listing. Here's the complete listing. ********************** Microsoft (R) Macro Assembler Version 5.10 9/2/89 09:50:54 Page 1-1 ; segs.asm ; - - - - - - - - - - - - - 0000 STACKSEG SEGMENT STACK 'STACK' 0000 4444 variable4 dw 4444h 0002 0100[ dw 100h dup (?) ???? ] 0202 STACKSEG ENDS ; - - - - - - - - - - - - - 0000 MORESTUFF SEGMENT PUBLIC 'HOHUM' 0000 2222 variable2 dw 2222h Chapter 10 - Templates 81 ______________________ 0002 MORESTUFF ENDS ; - - - - - - - - - - - - - 0000 DATASTUFF SEGMENT PUBLIC 'DATA' 0000 1111 variable1 dw 1111h 0002 DATASTUFF ENDS ; - - - - - - - - - - - - - 0000 CODESTUFF SEGMENT PUBLIC 'CODE' EXTRN print_num:NEAR , get_num:NEAR ASSUME cs:CODESTUFF,ds:DATASTUFF ASSUME es:MORESTUFF,ss:STACKSEG 0000 3333 variable3 dw 3333h 0002 main proc far 0002 1E start: push ds 0003 2B C0 sub ax,ax 0005 50 push ax 0006 B8 ---- R mov ax, DATASTUFF 0009 8E D8 mov ds,ax 000B B8 ---- R mov ax, MORESTUFF 000E 8E C0 mov es,ax 0010 8B 0E 0000 R mov cx, variable1 0014 89 0E 0000 R mov variable1, cx 0018 CB ret 0019 main endp 0019 CODESTUFF ENDS Microsoft (R) Macro Assembler Version 5.10 9/2/89 09:50:54 Page 1-2 ; - - - - - - - - - - - - END start Microsoft (R) Macro Assembler Version 5.10 9/2/89 09:50:54 Symbols-1 Segments and Groups: The PC Assembler Tutor 82 ______________________ N a m e Length Align Combine Class CODESTUFF . . . . . . . . . . . 0019 PARA PUBLIC 'CODE' DATASTUFF . . . . . . . . . . . 0002 PARA PUBLIC 'DATA' MORESTUFF . . . . . . . . . . . 0002 PARA PUBLIC 'HOHUM' STACKSEG . . . . . . . . . . . . 0202 PARA STACK 'STACK' Symbols: N a m e Type Value Attr GET_NUM . . . . . L NEAR 0000 CODESTUFF External MAIN . . . . . . . . F PROC 0002 CODESTUFF Length = 0017 PRINT_NUM . . . . L NEAR 0000 CODESTUFF External START . . . . . . . . L NEAR 0002 CODESTUFF VARIABLE1 . . . . . . L WORD 0000 DATASTUFF VARIABLE2 . . . . . . L WORD 0000 MORESTUFF VARIABLE3 . . . . . . L WORD 0000 CODESTUFF VARIABLE4 . . . . . . L WORD 0000 STACKSEG @CPU . . . . . . . . . . . . . . TEXT 0101h @FILENAME . . . . . . . . . . . TEXT segs @VERSION . . . . . . . . . . . . TEXT 510 54 Source Lines 54 Total Lines 21 Symbols 48006 + 428261 Bytes symbol space free 0 Warning Errors 0 Severe Errors ********************** As you can see, the listing, even for a short program, is very long. Let's take it apart section by section. The first large section is a copy of the text file except that there is information on the left. The number on the far left tells the offset address (in hex) from the beginning of the segment for each label, variable or instruction. In this section: 0000 3333 variable3 dw 3333h 0002 main proc far 0002 1E start: push ds 0003 2B C0 sub ax,ax 0005 50 push ax 0006 B8 ---- R mov ax, DATASTUFF 0009 8E D8 mov ds,ax Chapter 10 - Templates 83 ______________________ 000B B8 ---- R mov ax, MORESTUFF 000E 8E C0 mov es,ax 0010 8B 0E 0000 R mov cx, variable1 0014 89 0E 0000 R mov variable1, cx 0018 CB ret 0019 main endp "start" is at 0002h ,"mov cx, variable1" is at 0010h and "ret" is at 18h. The second set of numbers is the actual machine instructions in hex. These are the what the 8086 operates on. "push ds" is 1E, "mov ds, ax" is 8E D8, and "ret" is CB. The instructions can be from 1 - 6 bytes long. Notice the "R" after some of the instructions. The "R" stands for relocatable. This means that it is an address that might be changed by either the linker or the loader. We'll talk about that later. In any case, the object file keeps track of these so they can be changed if necessary. Also, go back to the complete listing and look at the four variables; you will see that the values have been put in the object code; that is, 1111h, 2222h, 3333h and 4444h. If we had had an error, the assembler would have placed an error message at the spot of the error in this part of the file. The next part of the .LST file is the segment listing. It tells how the segments are defined. N a m e Length Align Combine Class CODESTUFF . . . . . . . . 0019 PARA PUBLIC 'CODE' DATASTUFF . . . . . . . . 0002 PARA PUBLIC 'DATA' MORESTUFF . . . . . . . . 0002 PARA PUBLIC 'HOHUM' STACKSEG . . . . . . . . . 0202 PARA STACK 'STACK' We have the segment name, length, and some other information we'll talk about later. Notice that 'HOHUM' which is an artificial class, is dutifully listed with no complaints. Then comes the listing of all labels, variables, and procedure names. Symbols: N a m e Type Value Attr GET_NUM . . . . L NEAR 0000 CODESTUFF External MAIN . . . . . F PROC 0002 CODESTUFF Length = 0017 PRINT_NUM . . L NEAR 0000 CODESTUFF External START . . . . . L NEAR 0002 CODESTUFF VARIABLE1 . . . L WORD 0000 DATASTUFF The PC Assembler Tutor 84 ______________________ VARIABLE2 . . . L WORD 0000 MORESTUFF VARIABLE3 . . . L WORD 0000 CODESTUFF VARIABLE4 . . . L WORD 0000 STACKSEG It shows the segment and offset, whether they are bytes, words, processes etc. The "L" stands for label. The variables and procedures which are in an external file are so marked. Neither print_num nor get_num was called, but the assembler maintains a listing for them. Finally, some internal info for the assembler. @CPU . . . . . . . . . . . . . . TEXT 0101h @FILENAME . . . . . . . . . . . TEXT segs @VERSION . . . . . . . . . . . . TEXT 510 We will come back to parts of the .LST file, so make yourself comfortable with it. SEGMENTS It is now time for the nitty-gritty. We need to know what all those statements in the template file mean. Remember that there are four players in the game - (1) MASM, the Microsoft assembler, (2) LINK, the Microsoft linker, (3) the program loader and (4) the 8086 chip itself. Who does what to whom is the subject of this chapter. You will notice that there are three segments in all the template files, one for data, one for code, and one for the stack. How many segments can a program have? An unlimited number for code, an unlimited number for data, and one for the stack.{1} Although you can have an unlimited number of segments, you can use only four at any one time - two for regular data (referenced by the DS and ES registers), one for code (referenced by the CS register), and one for temporary data (referenced by the SS register). You don't have direct control over CS. You should NEVER change the value in SS. This means that you can only change which segments that ES and DS refer to. How do you do that? The 8086 does not allow you to move a constant into a segment register. Therefore it is a two step process. Put the constant into an arithmetic register (AX, BX, CX, DX, SI, DI or BP) and from there to the segment register. Suppose we have 327 different data segments in our file (named SEG1, SEG2, SEG3 ... SEG327) and we wanted to reference data in SEG27. The code would be: mov ax, SEG27 mov ds, ax ____________________ 1 Although if you REALLY need more space for a stack it is possible, if a little arcane. Chapter 10 - Templates 85 ______________________ This is the standard way to do it, and this is the same as the fourth and fifth instructions in the code segment of the template files where we are putting the address of DATASTUFF in ds. What is that SEG27 in the instruction (mov ax, SEG27)? It is a constant. When the assembler assembles the program, it makes note of the fact that you want to have the starting address of SEG27 in that instruction (you saw the "R" in the listing for the instruction 'mov ax, DATASTUFF'). Later the linker makes sure there is a SEG27 segment in the complete program, gives it a temporary segment address, and puts this temporary address in every place that references that segment address. This address is guaranteed to be adjusted. You will see why when we look at the linker .MAP file. Finally, the loader (which is the program that puts your program into memory) puts the segment where it wants and updates all references to the segment address to reflect where it now is. Thus, the program is complete only when this information is put in at run time. Each time you run the program SEG27 might be in a different place, but the loader will always update the references correctly. We named the segments SEG1, SEG2, etc. Does SEG have to be part of the segment name? Not on your life. Here are three perfectly acceptable segment definitions: CURLY SEGMENT LARRY SEGMENT MOE SEGMENT It is good practice to have 'SEG' as part of the segment name to remind you that these are segments, not variables, but this is a practice only, it is not a law. Any name you could use for a variable or a label you could use as a segment name. The reserved word SEGMENT after the name tells the assembler that this is the beginning of a segment with that name. You tell the assembler that you are starting a segment with 'SEGMENT' CURLY SEGMENT and you tell the assembler that you are finished with that segment with the reserved word ENDS (END [of] Segment): CURLY ENDS You need to put the name of the segment before the ENDS directive. In the template file, the data segment definition reads: DATASTUFF SEGMENT PUBLIC 'DATA' DATASTUFF is the segment name, but what are PUBLIC and 'DATA' there for? To understand this, we need to look at the linker. First, let's assemble temp1.asm (our first template file) just the way it is. The PC Assembler Tutor 86 ______________________ ---------- FROM THE SCREEN ---------- C>masm temp1.asm Microsoft (R) Macro Assembler Version 5.10 Copyright (C) Microsoft Corp 1981, 1988. All rights reserved. Object filename [temp1.OBJ]: Source listing [NUL.LST]: temp1 Cross-reference [NUL.CRF]: ---------- We have made the listing file so let's look at the segment information. N a m e Length Align Combine Class CODESTUFF . . . . . . . . . . . 000A PARA PUBLIC 'CODE' DATASTUFF . . . . . . . . . . . 0000 PARA PUBLIC 'DATA' STACKSEG . . . . . . . . . . . . 00C8 PARA STACK 'STACK' You will see that CODESTUFF is Ah (10d) bytes long, DATASTUFF has no data and is 0 bytes long, and STACKSEG is C8h (200d) bytes long. Now let's link temp1.obj and asmhelp.obj. ---------- FROM THE SCREEN ----- C>link temp1+asmhelp Microsoft (R) Overlay Linker Version 3.61 Copyright (C) Microsoft Corp 1983-1987. All rights reserved. Run File [TEMP1.EXE]: List File [NUL.MAP]: temp Libraries [.LIB]: ---------- This time we have made a listing file for the link process. It is called TEMP.MAP. Let's look at it. Start Stop Length Name Class 00000H 000C7H 000C8H STACKSEG STACK 000D0H 00540H 00471H DATASTUFF DATA 00550H 01944H 013F5H CODESTUFF CODE Program entry point at 0055:0000 This is what the map file looks like. There are still only three segments in the final executable file, STACKSEG, DATASTUFF and CODESTUFF. You will notice that the class name is still there, but the PUBLIC is missing. It's job is finished. "Start" says where the segment starts in the executable file, "Stop" says Chapter 10 - Templates 87 ______________________ where the segment ends in the executable file, and "Length" says the length in bytes of the segment. These numbers are 5 digit hex numbers instead of 4. That means that they are showing the total address. The segment number is the left 4 digits of 'Start'. STACKSEG is C8h (200d) bytes long like before. Although DATASTUFF had no data, it is now 471h (1137d) bytes long, and CODESTUFF was Ah (10d) bytes long before but now it is a whopping 13F5h (5109d) bytes long. What happened? The linker did its work. One of the things the linker does is combine things that we want to be in the same segment. It took the DATASTUFF segment from temp1.obj and appended the DATASTUFF segment from asmhelp.obj, combining them into one larger segment.{2} It took the CODESTUFF segment from temp1.obj and appended the CODESTUFF segment from asmhelp.obj, making them one large segment. Why did it do that? Because we put the word "PUBLIC" in the segment definition. When the assembler sees "PUBLIC" in the segment definition, it passes that information along to the linker in a header file.{3} When the linker has a segment which is "PUBLIC", it will append any other segment which (1) is "PUBLIC", (2) has the same name (i.e. CODESTUFF or DATASTUFF or CURLY etc.), and (3) has the same class name{4}. All three things must be true for the linker to combine them. We will actually check this out a little later to make sure you believe it. One other thing to notice is that the linker is allocating only as much space as is needed. It could allocate 65536 bytes for each segment defined, but it uses only as much as the program needs and then starts the next segment at the next segment starting address. This is efficient management of memory. What is the advantage of combining the smaller segments into one larger segment? For code, there is no big advantage. But for data, remember that every time we want to access data, we need to have the starting address of that particular segment in register ds. We do this by using: mov ax, DATASTUFF mov ds, ax If we have a number of data segments, every time we access data ____________________ 2 The linker always works from left to right. For each different type of segment, it starts with the first one it finds and then appends each succeeding one it finds. 3 A header is information for the linker or loader which is put in front of the machine code in an object file or an executable file. There are typically a number of headers in front of the machine code. 4 Remember that class names are somewhat arbitrary. I use 'CODE', 'DATA' and 'STACK' for clarity and because they are the standard Microsoft class names, but if you are not linking with anyone else's programs, you can use any class name you want. The PC Assembler Tutor 88 ______________________ we need to (1) make sure that ds contains the address of the correct data segment, and (2) if not, we need to write the code to change ds. This entails using a lot of code, can be confusing and is certainly error prone. With one data segment, you simply load ds with the correct address at the beginning of the program and then forget about it. This should be a rule for you. Unless you have truly humongous amounts of data (over 65535 bytes), ALWAYS put all your data in the same segment. Do you remember those dashes '----' in the assembler listing? That was because the assembler didn't have a segment address to put there. 0004 B8 ---- R mov ax, DATASTUFF 0007 8E D8 mov ds,ax 0009 8B 0E 0000 R mov cx, variable1 000D 89 0E 0000 R mov variable1, cx The linker now has a temporary address for the start of DATASTUFF (000D0h) so it will put the segment address (the left four hex bytes) in this spot. This is temporary, but will be updated by the loader. If variable1 has been moved, it will update that too. Why am I sure that these temporary segments will be moved? The segment address of STACKSEG is 0000h. The segment address of DATASTUFF is 000Dh (13h) and the segment address of CODESTUFF is 0055h (85d). But the operating sysyem owns the first several THOUSAND segments. The loader will load your program in much higher memory. They must move. So the linker combines all the segments we want to combine, and then it looks at the machine code and modifies every reference to the segments and to the variables which have been moved. That is a lot of work. For instance, when the linker appends asmhelp.obj, there are a hundred or so variables which it moves and a thousand or so references to those variables which it modifies. The linker does that every time you link a file with ASMHELP.OBJ. That's not too shabby. Do all three conditions need to be met for the linker to combine segments into one segment? 1) They have the same name 2) They have the same class name 3) They are both defined PUBLIC Joe Bob says check it out. Here are two .ASM files which contain a number of segments. Here's the first file: ;file1.asm ;- - - - - - - - - - - - - - - - - - - - STACKSEG SEGMENT STACK 'STACK' dw 100 dup (?) STACKSEG ENDS ;- - - - - - - - - - - - - - - - - - - - MORESTUFFA SEGMENT PUBLIC variable21 dw ? MORESTUFFA ENDS ;- - - - - - - - - - - - - - - - - - - - DATASTUFF SEGMENT PUBLIC 'DATA' variable1 dw ? DATASTUFF ENDS ;- - - - - - - - - - - - - - - - - - - - MORESTUFF SEGMENT PUBLIC 'DATA' variable2 dw ? MORESTUFF ENDS ;- - - - - - - - - - - - - - - - - - - - EVENMORESTUFF SEGMENT PUBLIC 'DATA' variable3 dw ? EVENMORESTUFF ENDS ;- - - - - - - - - - - - - - - - - - - - CODESTUFF SEGMENT PUBLIC 'CODE' ASSUME cs:CODESTUFF, ds:DATASTUFF ASSUME ds:MORESTUFF, es:MORESTUFFA, ds:EVENMORESTUFF main proc far start: push ds sub ax,ax push ax ret main endp CODESTUFF ENDS ;- - - - - - - - - - - - - - - - - - - - END start Here's the other file: ;file2.asm ; - - - - - - - - - - - - - - - - - - - - STACKSEG SEGMENT STACK 'STACK' dw 100 dup (?) STACKSEG ENDS ; - - - - - - - - - - - - - - - - - - - - NOTDATASTUFF SEGMENT PUBLIC 'DATA' variable4 dw ? NOTDATASTUFF ENDS ; - - - - - - - - - - - - - - - - - - - - The PC Assembler Tutor 90 ______________________ DATASTUFF SEGMENT PUBLIC 'DATA' variable5 dw ? DATASTUFF ENDS ; - - - - - - - - - - - - - - - - - - - - MORESTUFFA SEGMENT PUBLIC variable61 dw ? MORESTUFFA ENDS ; - - - - - - - - - - - - - - - - - - - - MORESTUFF SEGMENT PUBLIC 'CLASSOF68' variable6 dw ? MORESTUFF ENDS ; - - - - - - - - - - - - - - - - - - - - EVENMORESTUFF SEGMENT 'DATA' variable7 dw ? EVENMORESTUFF ENDS ; - - - - - - - - - - - - - - - - - - - - CODESTUFF SEGMENT PUBLIC 'CODE' ASSUME cs:CODESTUFF, ds:DATASTUFF, ds:NOTDATASTUFF ASSUME ds:MORESTUFF,ds:MORESTUFFA, ds:EVENMORESTUFF subroutine proc far ret subroutine endp CODESTUFF ENDS ; - - - - - - - - - - - - - - - - - - - - END You will notice that the two CODESTUFFs, the two DATASTUFFs, the two MORESTUFFAs and the two STACKSEGs each have the same definitions, but that (1) NOTDATASTUFF has a different name than DATASTUFF, (2) one MORESTUFF has a different class name from the other, (3) one EVENMORESTUFF is PUBLIC and the other is not, and (4) the two MORESTUFFAs have NO class name. Here's the segment information from file1.lst N a m e Length Align Combine Class CODESTUFF . . . . . . . . . . 0005 PARA PUBLIC 'CODE' DATASTUFF . . . . . . . . . . 0002 PARA PUBLIC 'DATA' EVENMORESTUFF . . . . . . . . 0002 PARA PUBLIC 'DATA' MORESTUFF . . . . . . . . . . 0002 PARA PUBLIC 'DATA' MORESTUFFA . . . . . . . . . . 0002 PARA PUBLIC STACKSEG . . . . . . . . . . . 00C8 PARA STACK 'STACK' and from file2.lst N a m e Length Align Combine Class CODESTUFF . . . . . . . 0001 PARA PUBLIC 'CODE' Chapter 10 - Templates 91 ______________________ DATASTUFF . . . . . . . 0002 PARA PUBLIC 'DATA' EVENMORESTUFF . . . . . 0002 PARA NONE 'DATA' MORESTUFF . . . . . . . 0002 PARA PUBLIC 'CLASSOF68' MORESTUFFA . . . . . . . 0002 PARA PUBLIC NOTDATASTUFF . . . . . . 0002 PARA PUBLIC 'DATA' STACKSEG . . . . . . . . 00C8 PARA STACK 'STACK' These are in alphabetical order. Before we link them together, let's think about what should happen if all three conditions must be met. Both CODESTUFF segments are PUBLIC with the same class name, so they should merge. Both DATASTUFF segments are PUBLIC with the same class name so they should merge. EVENMORESTUFF is PUBLIC in one file but not public in the other, so they should not merge. MORESTUFF is PUBLIC in both files, but they have different class names, so they should not merge. What about STACKSEG? The STACK combine type is similar to PUBLIC{1}, and they have the same class name, so they should merge. Finally, there are the MORESTUFFAs. They have the same name and are PUBLIC, but they have no class name. Will they combine? Let's see what happens. Here is the .MAP file from the command C> link file1+file2 Start Stop Length Name Class 00000H 0018FH 00190H STACKSEG STACK 00190H 001A1H 00012H MORESTUFFA 001B0H 001C1H 00012H DATASTUFF DATA 001D0H 001D1H 00002H MORESTUFF DATA 001E0H 001E1H 00002H EVENMORESTUFF DATA 001F0H 001F1H 00002H NOTDATASTUFF DATA 00200H 00201H 00002H EVENMORESTUFF DATA 00210H 00220H 00011H CODESTUFF CODE 00230H 00231H 00002H MORESTUFF CLASSOF68 Program entry point at 0021:0000 STACKSEG, DATASTUFF and CODESTUFF combined. MORESTUFFA combined. The others are separate. Doesn't this confuse the linker if it has more than one segment with the same name? No. The linker knows which variables are in which segments, and the names of the segments are not relevant. If you look at the class information from the linker listing, you will notice that all things in the same class are grouped together. The linker works from left to right on the command line, so for the above, it read file1.obj first and then read ____________________ 1 STACK tells the linker to combine any other segments which have STACK and the class type 'STACK' and it tells the loader to set the SS register to that segment, and set the SP register to point to the end of that segment. The PC Assembler Tutor 92 ______________________ file2.obj. It orders things (1) first by class (in the order encountered, and then (2) by segment (in the order encountered). For the linker ordering, a segment is like a subclass. Look through the assembler files to check that if you link in the order file1+file2, the order of encountering classes is 'STACK', empty, 'DATA', 'CODE', and 'CLASSOF68'. check the segment ordering also. What if we link the opposite way? > link file2+file1 Here's the listing: Start Stop Length Name Class 00000H 0018FH 00190H STACKSEG STACK 00190H 00191H 00002H NOTDATASTUFF DATA 001A0H 001B1H 00012H DATASTUFF DATA 001C0H 001C1H 00002H EVENMORESTUFF DATA 001D0H 001D1H 00002H MORESTUFF DATA 001E0H 001E1H 00002H EVENMORESTUFF DATA 001F0H 00201H 00012H MORESTUFFA 00210H 00211H 00002H MORESTUFF CLASSOF68 00220H 00234H 00015H CODESTUFF CODE Program entry point at 0022:0010 Assure yourself that this is the order the classes are encountered for file2+file1. Before we go on, let's summarize what we have so far. 1) In an .asm file, each segment starts with a name followed by the word SEGMENT. 2) Each segment ends with the name followed by the word ENDS (end of segment). This is the minimal segment definition: ; - - - - - SEG_A SEGMENT SEG_A ENDS ; - - - - - In addition, if you want to combine a segment with segments from other files in order to make one large segment, then all the segments to be combined must: 1) have the same name. 2) have the same class name (type) 3) be declared PUBLIC Chapter 10 - Templates 93 ______________________ ASSUME The next thing from the template file is the word ASSUME. Who is assuming what? ASSUME cs:CODESTUFF, ds:DATASTUFF This is for the assembler. It says that whenever you are working in the CODESTUFF segment, CS will be set to the segment address of the CODESTUFF segment. Whenever you are working in the DATASTUFF segment, DS will be set to the segment address of the DATASTUFF segment. The CS register takes care of itself, but it is your responsibility to make sure that DS actually points to the proper segment. If you just move a word from memory to a register: mov cx, variable1 the 8086 automatically thinks that it is in the DS segment. But it doesn't have to be that way. The 8086 has something called segment overrides. Here is the list: SEGMENT HEX VALUE CS 2E DS 3E ES 26 SS 36 An override is a 1 byte machine instruction that tells the 8086 that for the next instruction, the memory location will not reference the natural segment register; what it will reference is the segment register named in the override - CS if it is 2Eh, DS if it is 3Eh, ES if it is 26h, and SS if it is 36h. We could plug these in ourselves, but that is a lot of work. Fortunately, the assembler takes care of this for us. Let's look at the code from the very beginning of the chapter. ;*********************************** ; segs.asm ; - - - - - - - - - - - - - STACKSEG SEGMENT STACK 'STACK' variable4dw 4444h dw 100h dup (?) STACKSEG ENDS ; - - - - - - - - - - - - - MORESTUFF SEGMENT PUBLIC 'HOKUM' variable2 dw 2222h MORESTUFF ENDS ; - - - - - - - - - - - - - The PC Assembler Tutor 94 ______________________ DATASTUFF SEGMENT PUBLIC 'DATA' variable1 dw 1111h DATASTUFF ENDS ; - - - - - - - - - - - - - CODESTUFF SEGMENT PUBLIC 'CODE' EXTRN print_num:NEAR , get_num:NEAR ASSUME cs:CODESTUFF,ds:DATASTUFF ASSUME es:MORESTUFF,ss:STACKSEG variable3 dw 3333h main proc far start: push ds sub ax,ax push ax mov ax, DATASTUFF mov ds,ax mov ax, MORESTUFF mov es, ax mov cx, variable1 mov variable1, cx ret main endp CODESTUFF ENDS ; - - - - - - - - - - - - END start ;*************************** For the ASSUME statement we have: ASSUME cs:CODESTUFF,ds:DATASTUFF ASSUME es:MORESTUFF,ss:STACKSEG What we want to look at is this section of code: mov cx, variable1 mov variable1, cx Here is the listing of the offset address and machine code: 000E 8E C0 mov es,ax 0010 8B 0E 0000 R mov cx, variable1 0014 89 0E 0000 R mov variable1, cx Chapter 10 - Templates 95 ______________________ 0018 CB ret Variable1 is in DATASTUFF (ASSUME ds:DATASTUFF), and DS is the natural segment for variables. Now let's change the code to: mov cx, variable2 mov variable2, cx This is the ONLY change in the file. Variable2 is in MORESTUFF and we have - ASSUME es:MORESTUFF. Here's the listing when we assemble the modified file. 000E 8E C0 mov es,ax 0010 26: 8B 0E 0000 R mov cx, variable2 0015 26: 89 0E 0000 R mov variable2, cx 001A CB ret The assembler has put 26h as a segment override. When the 8086 looks at the machine code, it knows that those two instructions reference the es, not the ds, segment register. Also note that the code is now two bytes longer - one byte for each segment override. The "ret" instruction is at 1Ah (26d) instead of 18h (24d). Let's try it with: mov cx, variable3 mov variable3, cx Variable3 is in CODESTUFF and we have - ASSUME cs:CODESTUFF. Here's the listing: 000E 8E C0 mov es,ax 0010 2E: 8B 0E 0000 R mov cx, variable3 0015 2E: 89 0E 0000 R mov variable3, cx 001A CB ret The assembler put in the CS segment override. Now the 8086 knows that variable3 is in the CS segment. Finally: mov cx, varaible4 mov variable4, cx Variable4 is in STACKSEG and we have - ASSUME ss:STACKSEG. Here's the listing: 000E 8E C0 mov es,ax 0010 36: 8B 0E 0000 R mov cx, variable4 0015 36: 89 0E 0000 R mov variable4, cx 001A CB ret The PC Assembler Tutor 96 ______________________ Once again, the assembler put in a segment override. This time it was the SS override. That's nifty. We simply tell the assembler which segment register we will use for each segment and it does all the work. We will do more with segment overrides in the chapter on addressing modes. Remember, though, that it is your responsibility to see that at the time this code is used, the segment register actually contains the appropriate segment address. Is this ASSUME definition unique? That is, must there be a one to one correspondence between segments and registers, with each segment having its own register? No, not at all. Here are a two ASSUME statements, both of which are legal: ASSUME cs:COMSEG, ds:COMSEG, es:COMSEG, SS:COMSEG All four registers contain the address of the same segment. In fact, we will meet this statement when we talk about COM files. This is the only appropriate statement for a .COM file ASSUME ds:SEG_A, ds:SEG_B, es:SEG_C, es:SEG_D, es:SEG_A Four different segments, two of which are referenced by DS and three of which are referenced by ES. Remember, ASSUME tells the assembler that whenever you access something in that segment, the named register will be set to the starting segment address. What exactly does this mean to the assembler? Let's rearrange this a little: SEG_A ds, es SEG_B ds SEG_C es SEG_D es This is the list from the assembler's viewpoint. Suppose it has a variable that is in SEG_C. Does it need an override? Yes, it needs an ES override. Suppose it has a variable in SEG_A. Does it need an override? No, because DS is set to that segment. SUBROUTINES In assembler parlance, subroutines are called procedures. Why? You got me. In any case, whenever I say subroutine, process, subprogram, or anything like that, I mean a procedure. A procedure can have any name you want. You start a procedure by giving the name, using the reserved word 'proc' and then defining it as either near or far. my_procedure proc near is a near procedure with the name my_procedure. You end a procedure by giving the name and following it with the reserved word 'endp' (for end of procedure). Chapter 10 - Templates 97 ______________________ my_procedure endp What is a near procedure? It is one which is ALWAYS in the same segment as the calling program. When you call a near procedure, the value in CS stays the same, but IP (the instruction pointer) changes to the offset of the first byte of the procedure. The next instruction executed will be the first byte of the procedure. If a procedure is called even once from a different segment, then it MUST be a far procedure. my_procedure proc far my_procedure endp When you call a far procedure, the CS register is changed to the segment of the called procedure and IP (the instruction pointer) is set to the first byte of the procedure. This will be covered in the chapter on subroutines. How does the loader know where to start the program? The assembler tells the linker which tells the loader. How does the assembler know? You tell it. The last line of the file is the single word 'END'. That tells the assembler that you are done with the assembler code. If there is a word after the word 'END' (on the same line), then the assembler assumes that this word is the name of the label where the program starts. The first instruction executed will be whatever immediately follows that label. In the template files we have: END start so the label 'start:' indicates where the first instruction is. For an .EXE file, this can be anywhere at all, but we have it at the beginning. The label 'start:' is used for clarity, but we could just as easily have had: END zzyx4 The assembler would then look for the label 'zzyx4:' as the place to start the program. If you look at the link .MAP file from our file1+file2 example you will see: Program entry point at 0021:0000 That says that the starting address is CS = 0021h, IP = 0000h. Note that both CS and IP are different for the file2+file1 example: Program entry point at 0022:0010 where CS = 0022h and IP = 0010h. The initial offset was given to the linker by the assembler. The linker did any adjustment to the offset if it moved the code, and then it calculated the segment address itself. The PC Assembler Tutor 98 ______________________ RET When the loader loads the program, it puts the segment of the starting address in CS and the offset of the starting address in IP. This gives control to your program. When your program is done, how does it get back to the operating system? Good question. When the loader loads the program, it creates something called the PSP (program segment prefix). This is a 100h (256d) byte block of information and code. The first byte (offset 0000) of this block is an 8086 instruction for an orderly exit from a program. What we need to do is set CS to the PSP segment and set IP to 0000. Then the next instruction executed will be the orderly exit code. In talking about procedures, I said that when you call a far procedure, the 8086 puts the procedure's segment in CS and the procedure's offset in IP. But before that, it does two things: push CS ; these are the old CS and IP push IP ; this is not a real 8086 instruction {2} When you have a RET (return) instruction in a far procedure, the 8086 does the following: pop IP ; this is not a real 8086 instruction pop CS ; put back the old CS and IP so RET resets CS and IP to go back where it came from. That is its job. What has been pushed on the stack before starting your program? NOTHING. That's right. That means that if you execute ret at the end of your program, the 8086 will pop two pieces of garbage into IP and CS. Fortunately, when setting up a program, the loader ALWAYS puts the segment address of the PSP in DS., the data segment. All we need to do is PUSH DS (the PSP) and then PUSH 0 (offset 0000) and we have the address of our orderly exit code. If we then execute RET, it will POP these two items into IP and CS, sending us to our orderly exit code. That is what is at the beginning of the code section of the template file. We cannot PUSH a constant, so we manufacture a 0 with 'sub ax, ax'. The code is: push ds ; PSP segment sub ax, ax ; manufacture a 0 ____________________ 2 This is not actual 8086 code. You have no direct access to IP. This is, however, what the 8086 effectively does. Chapter 10 - Templates 99 ______________________ push ax ; offset = 0000 and the program is set up for the return. That's a lot of things together, so let's review. To exit a procedure we use RET, but for the starting procedure we need to return to the operating system. The PSP has the code for an orderly return at offset 0000. At load time, the loader puts the segment address of the PSP in DS. We push the PSP segment address and offset 0000 for later use by the RET instruction. We do this with: push ds ; PSP segment sub ax, ax ; manufacture a 0 push ax ; offset = 0000 These should be the first instructions in the program. Now that you have stored the PSP, DS is free for other use. You can now use DS to hold the segment address of your data. DS is used because that is the segment register that the 8086 expects unless told otherwise. You can't move a constant to a segment register, so this is a two step process: mov ax, DATASTUFF mov ds, ax EXTRN Finally, an EXTRN statement tells the assembler that the procedure or data is in another file and you did not forget it. For a procedure, you need to say whether it is NEAR (push old IP and put in new IP) or far (push old CS and IP; put in new CS and IP). Here is the assembler listing for five calls: E8 09CA R call near_routine 9A 15EE ---- R call far_routine E8 0000 E call near_external_routine 9A 0000 ---- E call far_external_routine E8 0000 E call get_unsigned The first two are in the same file, the next two are in an external file, and we have our friend 'get_unsigned'. 'R' means that the data may be changed, 'E' means that it is external, and will be done by the linker. The first four are labelled whether they are near or far. 'get_unsigned' is a near procedure. Notice that E8 is the near call while 9A is the far call. Also notice that the assembler reserves one word for the new IP in the near calls. If the call is in the same file, the assembler fills in this number, but if it is external the assembler sets it to 0. In the far calls the assembler reserves two words instead of one. The first word is again the new IP, which is either filled in or set to zero. The second word is for the segment address, and will be set by the linker. The PC Assembler Tutor 100 ______________________ Whew!!! It sure took a long time to go through all that and you still probably are unsure about some of this. Read the summary, and if you don't feel good about it, leave it for a day or two and reread it then. At the end of the book I will show you how you can simplify a lot of these things by using standardized segment names and some other standardized instructions. For now, you need to get used to what the structure of programs is, and we will continue using the same type of templates.{3} ____________________ 3 Just think of me as the computer equivalent of a woodshop teacher who forces you to use hand tools to make a coffee table rather than allowing you to use what you really want to be using - a chainsaw. Chapter 10 - Templates 101 ______________________ SUMMARY SEGMENTS Segments are defined by giving a name followed by the word SEGMENT. The end of a segment is signalled by the segment name, followed by the word ENDS (end of segment). ; - - - - - SOME_NAME SEGMENT SOME_NAME ENDS ; - - - - - (As always, anything after a comma is a comment and is ignored by the assembler). In addition, if you want to combine a segment with other segments, then all the segments to be combined must: 1) have the same name. 2) have the same type (class) 3) be declared PUBLIC THE STACK SEGMENT The stack segment may have any name you want, but should be declared " SEGMENT STACK 'STACK' ". This forces the loader to do certain initialization for you. If you don't declare it this way, you have to do the initialization yourself. ANY_NAME SEGMENT STACK 'STACK' EXTRN For procedures, an EXTRN statement tells the assembler that the procedure that you want to call is in a different file, that you didn't forget it. Procedures which are EXTRN must be declared either NEAR or FAR. The grammar is name colon NEAR or name colon FAR. EXTRN procedure1:NEAR, procedure2:FAR You may declare as many things on one line as will fit, but you need to separate them with commas. There can be no comma at the end. ASSUME An ASSUME statement tells the assembler that when a statement references that particular segment, the corresponding segment register will be set to that segment address. ASSUME es:MORESTUFF The PC Assembler Tutor 102 ______________________ tells the assembler that no matter what you do in other parts of the program, every time a variable in MORESTUFF is referenced, es will have the segment address of MORESTUFF. This is for the purpose of correct coding of segment overrides. SEGMENT OVERRIDES Normally, when the 8086 accesses a variable in memory, it does so via the DS segment register. This can be changed with a segment override. The assembler puts the correct segment override code in front of the instruction and the 8086 will use that segment register to access the data in memory. The override codes are: SEGMENT HEX VALUE CS 2E DS 3E ES 26 SS 36 CS CS is the code segment. When the 8086 processes machine code, it ALWAYS uses CS. There is no override. IP IP, the instruction pointer, gives the offset in CS of the next instruction to be processed. When the 8086 processes an instruction, it looks at IP, gets the next instruction and updates IP. This is totally automatic and internal to the 8086. You have no direct access to IP. PROCEDURES A procedure is declared by giving a name followed by the word 'proc' followed by either NEAR or FAR. A procedure is ended by giving the name, followed by 'endp' (end of procedure). ; - - - - - square_root proc far square_root endp ; - - - - - The words NEAR and FAR are for the assembler and the linker so they know whether to change just IP or both IP and CS in RET, the return statement as well as in CALL, the subroutine call. RET The assembler codes a near or a far return depending on whether you have declared a near or a far procedure. A NEAR return POPs Chapter 10 - Templates 103 ______________________ IP off of the stack while a FAR return POPs IP then POPs CS. Thus, a NEAR return stays in the same segment but a FAR return gets a new segment address in CS.{4} END The word END signals to the assembler that you are done with code. The assembler will ignore all following lines, whether they are blank or contain code. If the line with END has a name after the word END, then the assembler assumes that this is the name of a label where execution will begin at run time. That means that the instruction at 'label:' will be the first instruction executed in the program. SETUP In order to setup the program in the beginning you need to (1) PUSH the segment address of the PSP (which is in DS), then push 0 (the offset of the orderly return code). Following this you need to put the segment address of the data segment in DS. The code for all of this is: push ds ; PSP seg address is in ds sub, ax, ax ; 0 push ax ; push 0000 offset mov ax, DATA_SEG ; data segment address to ds mov ds, ax ____________________ 4 Of course, it is possible for CS to keep the same value if the calling procedure is is the same segment. Chapter 11 - Addressing Modes and Pointers ========================================== 104 In this chapter we are going to cover all possible ways of getting data to and from memory with the different addressing modes. Read this carefully, since it is likely this is the only time you will ever see ALL addressing possibilities covered. The easiest way to move data is if the data has a name and the data is one or two bytes long. Take the following data: ; ----- variable1 dw 2000 variable2 db -26 variable3 dw -589 ; ----- We can write: mov variable1, ax mov cl, variable2 mov si, variable3 and the assembler will write the appropriate machine code for moving the data. What can we do if the data is more than two bytes long? Here is some more data: ; ----- variable4 db "This is a string of ascii data." variable5 dd -291578 variable6 dw 600 dup (-11000) ; ----- Variable4 is the address of the first byte of a string of ascii data. Variable5 is a single piece of data, but it won't fit into an 8086 register since it is 4 bytes long. Variable6 is a 600 element long array, with each element having the value -11000. In order to deal with these, we need pointers. Some of you will be flummoxed at this point, while those who are used to the C language will feel right at home. A pointer is simply the address of a variable. We use one of the 8086 registers to hold the address of a variable, and then tell the 8086 that the register contains the address of the variable, not the variable itself. It "points" to a place in memory to send the data to or retrieve the data from. If this seems a little confusing, don't worry; you'll get the hang of it quickly. As I have said before, the 8086 does not have general purpose registers. Many instructions (such as LOOP, MUL, IDIV, ROL) work only with specific registers. The same is true of pointers. You may use only BX, SI, DI, and BP as pointers. The assembler will give you an error if you try using a different register as a pointer. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 11 - Addressing Modes 105 _____________________________ There are two ways to put an address in a pointer. For variable4, we could write either: lea si, variable4 or: mov si, offset variable4 Both instructions will put the offset address of variable4 in SI.{1} SI now 'points' to the first byte (the letter 'T') of variable4. If we wanted to move the third byte of that array (the letter 'i') to CL, how would we do it? First, we need to have SI point to the third byte, not the first. That's easy: add si, 2 But if we now write: mov cl, si we will generate an assembler error because the assembler will think that we want to move the data in SI (a two byte number) to CL (one byte). How do we tell the assembler that we are using SI as a pointer? By enclosing SI in square brackets: mov cl, [si] since CL is one byte, the assembler assumes you want to move one byte. If you write: mov cx, [si] then the assembler assumes that you want to move a word (two bytes). The whole thing now is: lea si, variable4 add si, 2 mov cl, [si] This puts the third byte of the string in CL. Remember, if a register is in square brackets, then it is holding the ADDRESS of a variable, and the 8086 will use the register to calculate where the data is in memory. What if we want to put 0s in all the elements of variable6? ____________________ 1 LEA stands for load effective address. Note that with LEA, we use only the name of the variable, while with: mov si, offset variable4 we need to use the word 'offset'. The exact difference between the two will be explained later. The PC Assembler Tutor 106 ______________________ Here's the code: mov bx, offset variable6 mov ax, 0 mov cx, 600 zero_loop: mov [bx], ax add bx, 2 loop zero_loop We add 2 to BX each time since each element of variable6 is a word (two bytes) long. There is another way of writing this: mov bx, offset variable6 mov cx, 600 zero_loop: mov [bx], 0 add bx, 2 loop zero_loop Unfortunately, this will generate an assembler error. Why? If the assembler sees: mov [bx], ax it knows that you want to move what is in AX to the address in BX, and AX is one word (two bytes) long so it generates the machine code for a word move. If the assembler sees: mov [bx], al it knows that you want to move what is in AL to the address in BX, and AL is one byte long, so it generates the machine code for a byte move. If the assembler sees: mov [bx], 0 it doesn't know whether you want a byte move or a word move. The 8086 assembler has implicit sizing. It is the assembler's job to look at each instruction and decide whether you want to operate on a byte or a word. Other microprocessors do things differently. On the Motorola 68000, the assembler uses explicit sizing. Each instruction must explicitly state whether it is a byte or a word.{2} On the 68000 you have: move.b #213, (A1) move.w #213, (A1) The first instruction says to move a byte (the number 213) to the address in register A1 while the second instruction says to move ____________________ 2 Any of you who use the 68000 assembler know that this is fudging the facts a little bit. Chapter 11 - Addressing Modes 107 _____________________________ a word (the number 213) to the address in register A1.{3} Back to the 8086. If the 8086 assembler looks at an instruction and it can't tell whether you want to move a byte or a word, it generates an error. When you use pointers with constants, you should explicitly state whether you want a byte or a word. The proper way to do this is to use the reserved words BYTE PTR or WORD PTR. mov [bx], BYTE PTR 213 mov [bx], WORD PTR 213 These stand for byte pointer and word pointer respectively. I find this terminology exceptionally clumsy, but that's life. Whenever you are moving a constant with a pointer, you should specify either BYTE PTR or WORD PTR. The Microsoft assembler makes some assumptions about the size of a constant. If the number is 256 or below (either positive or negative), you MUST explicitly state whether it is a byte or a word operation. If the number is 257 or above (either positive or negative), the assembler assumes that you want a word operation. Here's the previous code rewritten correctly: mov bx, offset variable6 mov cx, 600 zero_loop: mov [bx], WORD PTR 0 add bx, 2 loop zero_loop Let's add 435 to every element in the variable6 array: mov bx, offset variable6 mov cx, 600 add_loop: add [bx], WORD PTR 435 add bx, 2 loop add_loop How about multiplying every element in the array by 12? mov di, offset variable6 mov cx, 600 mov si, 12 mult_loop: mov ax, [di] imul si mov [di], ax add di, 2 loop mult_loop ____________________ 3 A1 is a 68000 register. The PC Assembler Tutor 108 ______________________ None of these examples did any error checking, so if the result was too large, the overflow was ignored. This time we used DI for a change of pace. Remember, we may use BX, SI, DI or BP, but no others. You will notice that in all these examples, we started at the beginning of the array and went step by step through the array. That's fine, and that's what we normally would do, but what if we wanted to look at individual elements? Here's a sample program: ; + + + + + START DATA BELOW THIS LINE ; poem_array db "She walks in Beauty, like the night" db "Of cloudless climes and starry skies;" db "And all that's best of dark and bright" db "Meet in the aspect ratio of 1 to 3.14159" character_count db 149 ; + + + + + END DATA ABOVE THIS LINE ; + + + + + START CODE BELOW THIS LINE mov bx, offset poem_array mov dl, character_count character_loop: sub ax, ax ; clear ax call get_unsigned_byte dec al ; character #1 = array[0] cmp al, dl ; out of range? ja character_loop ; then try again mov si, ax ; move char # to pointer register mov al, [bx+si] ; character to al call print_ascii_byte jmp character_loop ; + + + + + END CODE ABOVE THIS LINE You enter a number and the program prints the corresponding character. Before starting, we put the array address in BX and the maximum character count in DL. After getting the number from get_unsigned_byte, we decrement AL since the first character is actually poem_array[0]. The character count has been reduced by 1 to reflect this fact. It also makes 0 an illegal entry. Notice that the program checks to make sure you don't go past the end of the poem. This time we use BX to mark the beginning of the array and SI to count the number of the character. Once again, there are only specific combinations of pointers that can be used. They are: BX with either SI or DI (but not both) BP with either SI or DI (but not both) My version of the Microsoft assembler (v5.1) recognizes the forms [bx+si], [si+bx], [bx][si], [si][bx], [si]+[bx] and [bx]+[si] as the same thing and produces the same machine code for all six. Chapter 11 - Addressing Modes 109 _____________________________ We can get even more complicated, but to show that, we need structures. In databases they are called records. In C they are called structures; in any case they are the same thing - a group of different types of data in some standard order. After the group is defined, we usually make an array with the identical structure for each element of the array.{4} Let's make a structure for an address book. last_name db 15 dup (?) first_name db 15 dup (?) age db ? tel_no db 10 dup (?) In this case, all the data is bytes, but that is not necessary. It can be anything. Each separate piece of data is called a FIELD. We have the last_name field, the first_name field, the age field, and the tel_no field. Four fields in all. The structure is 41 bytes long. What if we want to have a list of 100 names in our telephone book? We can allocate memory space with the following definition: address_book db 100 dup ( 41 dup (' ')) {5} Well, that allocates room in memory, but how do we get to anything? First, we need the array itself: mov bx, offset address_book Then we need one specific entry. Let's take entry 29 (which is address_book[28]). Each entry is 41 bytes long, so: mov ax, 28 ; entry (less 1) mov cx, 41 ; entry length mul cx mov di, ax ; move to pointer That gives us the entry, but if we want to get the age, that's not the first byte of the structure, it's the 31st byte (actually address_book[28] + 30 since the first byte is at +0). We get it by writing: mov dl, [bx+di+30] This is the most complex thing we have - two pointers plus a constant. The total code is then: mov bx, offset address_book mov ax, 28 ; entry (less 1) mov cx, 41 ; entry length ____________________ 4 If you don't know about structures or records, now would be a good time to stop and go to a reference book about them. They are not actually covered here. 5 Nesting of dup statements is allowed. Rather than having uninitialized data, this has blanks in all the spaces. The PC Assembler Tutor 110 ______________________ mul cx ; entry offset from array[0] mov di, ax ; move entry offset to pointer mov dl, [bx+di+30] ; total address Though the machine code has only one constant in the code, the assembler will allow you to put a number of constants in the assembler instruction. It will add them together for you and resolve them into one number.{6} Once again, there are a limited number of registers - they are the same registers as before: BX with either SI or DI (but not both) plus constant BP with either SI or DI (but not both) plus constant We can work with structures on the machine level, but it looks like it's going to be hard to keep track of where each field is. Actually, it isn't so bad because of: OUR FRIEND, THE EQU STATEMENT The assembler allows you to do substitution. If you write: somestuff EQU 37 * 44 then every place that the assembler finds the word "somestuff", it will substitute what is on the right side of the EQU. Is that a number or text? Sometimes it's a number, sometimes it's text. Here are four statements which are defined totally in terms of numbers. This is from the assembler listing. (The assembler lists how it has evaluated the EQU statement on the left after the equal sign.) = 0023 statement1 EQU 5 * 7 = 0025 statement2 EQU statement1 + 6 - 4 = 000F statement3 EQU statement2 - 22 = 001F statement4 EQU statement3 + 16 and the assembler thinks of these as numbers (these numbers are in hex). Now in the next set, with only a minor change: = [bp + 3] statement1 EQU [bp + 3] = [bp + 3] + 6 - 4 statement2 EQU statement1 + 6 - 4 = [bp + 3] + 6 - 4 - 22 statement3 EQU statement2 - 22 ____________________ 6 And it does it quite well. The assembler correctly evaluated the following: add ax, (-3*81)+44/8+[si+27]+6+[bx]-7+(43*96)-2 Not bad, huh? Chapter 11 - Addressing Modes 111 _____________________________ = [bp + 3] + 6 - 4 - 22 + 16 statement4 EQU statement3 + 16 the assembler thinks of it as text. Obviously, the fact that it can be either may cause you some problems along the way. Consult the assembler manual for ways to avoid the problem. Now we have a tool to deal with structures. Let's look at that structure again. last_name db 15 dup (?) first_name db 15 dup (?) age db ? tel_no db 10 dup (?) We don't actually need a data definition to make the structure, we need equates: LAST_NAME EQU 0 FIRST_NAME EQU 15 AGE EQU 30 TEL_NO EQU 31 this gives us the offset from the beginning of each record. If we again define: address_book db 100 dup ( 41 dup (' ')) then to get the age field of entry 87, we write: mov bx, offset address_book mov ax, 86 ; entry (less 1) mov cx, 41 ; entry length mul cx ; entry offset from array[0] mov di, ax ; move entry offset to pointer mov dl, [bx+di+AGE] ; total address This is a lot of work for the 8086, but that is normal with complex structures. The only thing that takes a lot of time is the multiplication, but if you need it, you need it.{7} How about a two dimensional array of integers, 60 X 40 int_array dw 40 dup ( 60 dup ( 0 )) These are initialized to 0. For our purposes, we'll assume that the first number is the row number and the second number is the column number; i.e. array [6,13] is row 6, column 13. We will have 40 rows of 60 columns. For ease of calculation, the first array element is int_array [0,0]. (If it is your array, you can ____________________ 7 You will see more of the EQU statement. The PC Assembler Tutor 112 ______________________ set it up any way you want {8}). Each row is 60 words (120 bytes) long. To get to int_array [23, 45] we have: mov ax, 120 ; length of one row in bytes mov cx, 23 ; row number mul cx mov bx, ax ; row offset to bx mov si, 45 ; column offset sal si, 1 ; multiply column offset by 2 (for word size) mov dx, [bx+si] ; integer to dx Using SAL instead of MUL is about 50 times faster. Since most arrays you will be working with are either byte, word, or double word (4 bytes) arrays, you can save a lot of time. Let ELEMENT_NUMBER be the array number (starting at 0) of the desired element in a one-dimensional array. For byte arrays, no multiplication is needed. For a word: mov di, ELEMENT_NUMBER sal di,1 ; multiply by 2 and for a double word (4 bytes): mov di, ELEMENT_NUMBER sal di, 1 sal di, 1 ; multiply by 4 This means that a one-dimensional array can be accessed very quickly as long as the element length is a power of 2 - either 2, 4 or 8. Since the standard 8086 data types are all 1, 2, 4, or 8 bytes long, one dimensional arrays are fast. Others are not so fast. As a quick review before going on, these are the legal ways to address a variable on the 8086: (1) by name. mov dx, variable1 It is also possible to have name + constant. mov dx, variable1 + 27 The assembler will resolve this into a single offset number and will give the appropriate information to the linker. (2) with the single pointers BX, SI, DI and BP (which are enclosed in square brackets). mov cx, [si] ____________________ 8 Bearing in mind that all compiled languages have fixed formats for arrays. If you want your array to interact with C, Fortran, Pascal or Basic, you'd better be sure you have the right format. Chapter 11 - Addressing Modes 113 _____________________________ xor al, [bx] add [di], cx sub [bp], dh (3) with the single pointers BX, SI, DI and BP (which are enclosed in square brackets) plus a constant. mov cx, [si+421] xor al, 18+[bx] add 93+[di]-7, cx sub (54/7)+81-3+[bp]-19, dh (4) with the double pointers [bx+si], [bx+di], [bp+si], [bp+di] (which are enclosed in square brackets). mov cx, [bx][si] xor al, [di][bx] add [bp]+[di], cx sub [di+bp], dh (5) with the double pointers [bx+si], [bx+di], [bp+si], [bp+di] (which are enclosed in square brackets) plus a constant. mov cx, [bx][si+57] xor al, 45+[di+23][bx+15]-94 add [bp]+[di]-444, cx sub [6+di+bp]-5, dh These are ALL the addressing modes allowed on the 8086. As for the constants, it is the ASSEMBLER'S job to resolve all numbers in the expression into a single constant. If your expression won't resolve into a constant, it is between you and the assembler. It has nothing to do with the 8086 chip. The PC Assembler Tutor 114 ______________________ We can consolidate all this information into the following list: All the following addressing modes can be used with or without a constant: variable_name (+constant) [bx] (+constant) [si] (+constant) [di] (+constant) [bp] (+constant) [bx+si] (+constant) [bx+di] (+constant) [bp+si] (+constant) [bp+di] (+constant) This is a complete list. Thus, you can access a variable by name or with one of the eight pointer combinations. There are no other possibilities. One thing that may confuse you about an addressing statement is all the plusses and minuses. As an example: mov cx, -45+27[bx+22]+[-195+di]+23-44 the total address is: -45+27[bx+22]+[-195+di]+23-44 When the 8086 performs this instruction, it will ADD (1) BX (2) DI and (3) a single constant. That single constant can be a positive or a negative number; the 8086 will ADD all three elements. The '+' in front of 'di' is for convenience of the assembler only; [-195-di] is illegal and the assembler will generate an error. If you actually want the negative of what is in one of the registers, you must negate it before calling the addressing instruction: neg di mov cx, -45+27[bx+22]+[-195+di]+23-44 once again, the only allowable forms are +[di], [di] or [+di]. Either -[di] or [-di] will generate an assembler error. If you ever see a technical description of the addressing modes, you will find a list of 24 different machine codes. The reason for this is that: [bx] [bx] + byte constant [bx] + word constant are three different machine codes. Here is a listing of the same machine instruction with the three different styles: Chapter 11 - Addressing Modes 115 _____________________________ MACHINE CODE ASSEMBLER INSTRUCTION 03 04 add ax, [si] 03 44 1B add ax, [si+27] 03 44 E5 add ax, [si-27] 03 84 5BA7 add ax, [si+23463] 03 84 A459 add ax, [si-23463] (27d = 1Bh , 23463d = 5BA7h). The first byte of code (03) is the add (word) instruction. The second byte is the addressing code, and the third and fourth bytes (if any) are the constant (in hex). Addressing code 04 is: (ax, [si]). Addressing code 44 is: (ax, [si] + byte constant). Addressing code 84 is: (ax, [si] + word constant). The fact that there are three different machine codes is of concern to the assembler, not to you. It is the assembler's job to make the machine code as efficient as possible. It is your job to write quality, robust code. SEGMENT OVERRIDES So far, we haven't talked about segment registers. You will remember from the last chapter that the 8086 assumes that a named variable is in the DS segment: mov ax, variable1 If it isn't, the Microsoft assembler puts the correct segment override in the machine code. The segment overrides are: SEGMENT OVERRIDE MACHINE CODE (hex) CS 2E DS 3E ES 26 SS 36 As an example: MACHINE CODE ASSEMBLER INSTRUCTIONS 2E: 03 06 0000 R add ax, variable3 26: 2B 1E 0000 R sub bx, variable2 31 36 0000 R xor variable1, si ; no override 36: 21 3E 00C8 R and variable4, di when the different variables were in segments with different ASSUME statements. If you don't remember this, you should reread the section on overrides in the last chapter. Remember, the colon is in the listing only to tell you that we have a segment override. The colon is not in the machine code. The PC Assembler Tutor 116 ______________________ What about pointers? The natural segment for anything with [bp] is SS, the stack segment.{1} Everything else has DS as its natural segment. The natural segments are: (1) DS variable + (constant) [bx] + (constant) [si] + (constant) [di] + (constant) [bx+si] + (constant) [bx+di] + (constant) (2) SS [bp] + (constant) [bp+si] + (constant) [bp+di] + (constant) where the constant is always optional. Can you use segment overrides? Yes, in all cases.{2} Here is some assembler code along with the machine code which was generated. MACHINE CODE ASSEMBLER INSTRUCTIONS 26: 03 07 add ax, es:[bx] 2E: 01 05 add cs:[di], ax 36: 2B 44 11 sub ax, ss:[si+17] 2E: 29 46 00 sub cs:[bp], ax 3E: 33 03 xor ax, ds:[bp+di] 26: 31 02 xor es:[bp+si], ax 26: 89 43 16 mov es:[bp+di+22], ax 03 04 add ax, [si] 03 44 1B add ax, [si+27] 03 84 A459 add ax, [si-23463] 26: 03 04 add ax, es:[si] 26: 03 44 1B add ax, es:[si+27] 26: 03 84 A459 add ax, es:[si-23463] (17d = 11h, 22d = 16h, 27d = 1Bh, -23463d = 0A459h). The first number (which is followed by a colon) is the segment override that the assembler has inserted in the machine code. Remember, the colon is in the listing to inform you that an override is ____________________ 1 We will see why when we look at subroutines. BP is called the base pointer [bp] and is used in a special way. 2 There are some special instructions for two independent pointers which we will cover at the end of the book. These allow segment overrides but force the override to refer to the first pointer. Chapter 11 - Addressing Modes 117 _____________________________ involved; it is not in the machine code itself. Unfortunately, when you use pointers you must put the override into the assembler instructions yourself. The assembler has no way of knowing that you want an override. This can cause some truly gigantic errors (if you reference a pointer seven times and forget the override once, the 8086 will access the wrong segment that one time), and those errors are extremely difficult to detect. As you can see from above, you put the override in the instructions by writing the appropriate segment (CS, DS, ES or SS) followed by a colon. As always, it is your responsibility to make sure that the segment register holds the address of the appropriate segment before using an override. We have talked about two different types of constants in the chapter, a constant which is part of the address: mov ax, [bx+17] add [si+2190], dx and [di-8179], cx and a constant which is a number to used for an arithmetical or logical operation: add ax, 17 sub dl, 45 add dx, 22187 They are both part of the machine instruction, and are unchangeable (true constants). This machine code is going to be difficult to read, so just look for (1) the constant DATA and (2) the constant in the ADDRESS. All constants in the assembler instructions are in hex so that they look the same as in the listing of the machine code. Here's a listing of different combinations. 1. Pointer + constant as an address: MACHINE CODE ASSEMBLER INSTRUCTIONS 01 44 1B add [si+1Bh], ax 29 85 0A04 sub [di+0A04h], ax 30 5C 1F xor [si+1Fh], bl 20 9E 1FAB and [bp+1FABh], bl 2. Arithmetic instruction with a constant: MACHINE CODE ASSEMBLER INSTRUCTIONS 05 1065 add ax, 1065h 2D 6771 sub ax, 6771h 80 F3 37 xor bl, 37h 80 E3 82 and bl, 82h 3. Pointer + constant as an address; arithmetic with a constant The PC Assembler Tutor 118 ______________________ MACHINE CODE ASSEMBLER INSTRUCTIONS 81 44 1B 1065 add [si+1Bh], 1065h 81 AD 0A04 6771 sub [di+0A04h], 6771h 80 74 1F 37 xor [si+1Fh], BYTE PTR 37h 80 A6 1FAB 82 and [bp+1FABh], BYTE PTR 82h You will notice that the ADD instruction (as well as the other instructions) changes machine code depending on the complete format of the instruction (byte or word? to a register or from a register? what addressing mode? is AX one of the registers?). That's part of the 8086 machine language encoding, and it makes the 8086 machine code extremely difficult to decipher without a table listing all the options. OFFSET AND SEG There are two special instructions that the assembler has - offset and seg. For any variable or label, offset gives the offset from the beginning of the segment, and seg gives the segment address. If you write: mov ax, offset variable1 the assembler will calculate the offset of variable1 and put it in the machine code. It also signals the linker and loader; if the linker should change the offset during linking, it will also adjust this number. If you write: mov dx, seg variable1 The assembler will signal to the linker and the loader that you want the address of the segment that variable1 is in. The linker and loader will put it in the machine code at that spot. You don't need to know the name of the segment. The linker takes care of that. We will use the seg operator later. LEA LEA (load effective address) is a completely different animal. It allows you to use any addressing mode to put an address in a register. One of the addressing modes covered before was for the following code: xor dx, 45+[di+23][bx+15]-94 The 8086 added DI, BX and the constant to calculate the address. It then XOR'ed the variable at that address with DX. If you write: lea dx, 45+[di+23][bx+15]-94 the 8086 will add DI, BX and the constant to calculate the address. It will then put the ADDRESS in DX. LEA can use any Chapter 11 - Addressing Modes 119 _____________________________ addressing mode to calculate an address. The machine code looks almost the same: MACHINE CODE ASSEMBLER INSTRUCTIONS 33 51 F5 xor dx, 45+[di+23][bx+15]-94 8D 51 F5 lea dx, 45+[di+23][bx+15]-94 The first byte of the machine code is the instruction and the second and third byte are the addressing mode. You almost never need LEA. It is slower than: mov dx, offset variable1 However, when the addressing gets complicated (perhaps 1% of the time), it's nice to have. Remember, it will calculate ANY 8086 addressing mode. Let's run a program so we can see what actually happens with LEA ;lea.asm ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE variable1 dw ? ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ; + + + + START CODE BELOW THIS LINE ; reg style mov si_byte, 1 ; signed lea ax, ax_byte call set_reg_style mov bp, 0 ; clear unused registers mov di, 0 ;lea and mov show the two ways to address variable1 lea ax, variable1 ; effective address mov bx, offset variable1 ; offset call show_regs_and_wait lea_loop: mov si, 0 ; clear registers mov dx, 0 mov cx, 0 mov bx, 0 mov ax, 0 call show_regs call get_unsigned ; unsigned for bx mov bx, ax mov ax, 0 ; blank ax call show_regs call get_signed ; signed for si mov si, ax mov ax, 0 ; blank ax The PC Assembler Tutor 120 ______________________ lea cx, [bx+si]+100 ; addresses to cx and dx lea dx, [si+bx-100] call show_regs_and_wait jmp lea_loop ; + + + + END CODE ABOVE THIS LINE The first part of the program shows that LEA and MOV give the same offset address. Then we enter the loop. It gets an unsigned number, puts it in BX, gets a signed number, puts it in SI, then uses LEA to calculate [bx+si+100] and [bx+si-100]. The plus and minus 100 is simply to show you a difference of 200 in the two results. BX and SI could also have contained (1) both signed numbers or (2) both unsigned numbers. It doesn't make any difference. This program has a signed and an unsigned number for variety. Of special interest to you shold be when [bx+si] is within 100 of 65536 (or 0). One of the results will be > 0 while the other result will be < 65536 The address value wraps around from 65535 -> 0. Note that with minor alteration, this program can be used to look at ANY addressing mode that uses pointers. You should make two executable files for this. First: link lea+asmhelp and the second: link asmhelp+lea Give them different names and run them. Note the offset values for: lea ax, variable1 mov bx, offset variable1 With lea+asmhelp you should have an offset of 8 for variable1 since there are 8 bytes in the array (ax_byte, bx_byte, etc.). This array appears before variable1 in the data segment. When you link it the other way (asmhelp+lea), all the data for asmhelp.obj is in front of your data and the offset should be something completely different for variable1. Chapter 11 - Addressing Modes 121 _____________________________ SUMMARY These are the natural (default) segments of all addressing modes: (1) DS variable + (constant) [bx] + (constant) [si] + (constant) [di] + (constant) [bx+si] + (constant) [bx+di] + (constant) (2) SS [bp] + (constant) [bp+si] + (constant) [bp+di] + (constant) Where the constant is optional. Segment overrides may be used. The segment overrides are: SEGMENT OVERRIDE MACHINE CODE (hex) CS: 2E DS: 3E ES: 26 SS: 36 OFFSET The reserved word 'offset' tells the assembler to calculate the offset of the variable from the beginning of the segment. mov ax, offset variable2 SEG The reserved word 'seg' tells the assembler, linker and loader to get the segment address of the segment that the variable is in. mov ax, seg variable2 LEA LEA calculates an address using any of the 8086 addressing modes, then puts the address in a register. lea cx, [bp+di+27] Chapter 12 - Multiple Word Arithmetic I ======================================= 122 Let's review the LOOP instruction. We often want to repeat an action a specific number of times. In a FOR loop, we write: FOR I = 1, 10 That means we want to repeat the code that follows ten times. The 8086 has an instruction for this, called the loop instruction. It looks like this: mov cx, 10 label17: ... (a bunch of code) ... loop label17 The count MUST be in the CX register. This is the only register you can use for this. When the machine sees the loop instruction, it decrements the CX register by one, LEAVING ALL FLAGS ALONE, and if the result is not zero, it loops back to the label. If the result is zero, it falls through the loop. One problem we might have with this instruction is if you enter it with CX = 0, it is going to loop 65,536 times. Intel provided a second instruction to avoid this - JCXZ (jump if CX is zero). You put it right before the loop for insurance. jcxz label19 label17: ... (a bunch of code) ... loop label17 label19: Obviously, in our first example this instruction is not necessary because we set CX to 10 just before entering the loop. If you have seen the list of 8086 instructions, you will have noticed lots of strange looking add and subtract instructions. Why are they there? In this chapter we will look at ADC and SBB. The others will come in later chapters. How do engineers decide what a reasonable size for a number is? When they started making 8 bit machines (the maximum unsigned number is 255) did they go out on the streets and take a poll to find out if 255 was the maximum number that people used? No, they didn't even think about what people needed. It was a question of what the technology would allow at that time. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 12 - Multiple Word Arithmetic I 123 _______________________________________ Similarly, at the time the 8086 was engineered, 16 bits was pushing the limits of the technology. But 65,535 doesn't really cut the mustard. If those are 65,535 pennies, that isn't even your yearly food bill, let alone the cost of housing. The 8086 instructions give us the option of making integers of whatever size we want. Because it is done in software, it is slower, but if we are doing thousands or tens of thousands of additions instead of hundreds of thousands or millions, it won't be a great inconvenience.{1} ASMHELP.OBJ is set up to input 2 word (4 byte) and 4 word (8 byte) numbers so those are what we are going to use. 4 byte numbers are up to +/- 2 billion and 8 byte numbers are up to 9X10exp18. Those should be large enough. The first instruction is ADC, add with carry. When you add by hand, you add everything in the right column, then carry to the next column left, repeating this over and over. We don't need to do it column by column, but it is necessary to do it word by word. In the data section we need: variable1 dq ? variable2 dq ? and the code for a 4 word (8 byte) addition is the following: lea si, variable1 lea di, variable2 mov ax, [di] ; first addition add [si], ax pushf ; save the flags mov cx, 3 add si, 2 ; go to next higher word add di, 2 long_add_loop: ; next three additions mov ax, [di] popf ; restore the flags adc [si], ax pushf ; save the new flags add di, 2 add si, 2 loop long_add_loop popf ; pop the flags off the stack ADC is the same as ADD, but it looks at the carry flag - if the carry flag is 1, it adds 1 to the result; if the carry flag is zero, it does nothing to the result. This works for both signed and unsigned numbers. If you don't believe it, you should go back ____________________ 1 As a benchmark, it took a moderately slow PC 6.5 seconds to do 100,000 eight byte additions. The same PC can do over a million two byte additions in 6 seconds. The PC Assembler Tutor 124 ______________________ to the introductory chapter with the base 10 machine and look at long additions. Notice PUSHF and POPF. These are special instructions called push flags and pop flags. Rather than pushing an arithmetic register on the stack, pushf pushes the register containing the flags. Popf pops them back into the flags register. This is necessary because ADC looks at the carry flag and the ADD instructions in the loop: add di,2 add si,2 might change the carry flag. The last POPF after the loop is to get it off the stack (anything we put on the stack we need to take off the stack). The first addition is a normal addition, the last three take the carry into account. We are moving the pointers a word at a time. Because the 8086 doesn't allow both operands to be in memory, we need to move one into a register. After the addition is performed, the result is in memory. We can discard what is in the register. Notice that the first half of the code looks almost the same as the code inside the loop. If we could only use ADC instead of ADD, we could put the first addition inside the loop. It is possible to do this. There is another instruction, CLC, which clears the carry flag. Recall that if the carry flag is 0, ADC does nothing different from ADD. Therefore, we can have: lea si, variable1 lea di, variable2 mov cx, 4 ; 4 additions in loop clc ; set cf to zero pushf ; save the flags long_add_loop: mov ax, [di] ; word to a register popf ; restore the flags adc [si], ax ; register + memory pushf ; save the flags add di, 2 add si, 2 loop long_add_loop popf ; pop the flags off the stack It's the same code. The number of loops was increased from 3 to 4, and the carry flag was cleared to insure that the first addition would have nothing extra added. Here is the basic program. ; - - - - - PUT DATA BELOW THIS LINE variable1 dq ? variable2 dq ? ; - - - - - PUT DATA ABOVE THIS LINE Chapter 12 - Multiple Word Arithmetic I 125 _______________________________________ ; - - - - - PUT CODE BELOW THIS LINE call show_regs outer_loop: lea ax, variable1 ; get the variables call get_signed_8byte call print_signed_8byte lea ax, variable2 call get_signed_8byte call print_signed_8byte lea si, variable1 ; set the pointers lea di, variable2 mov cx, 4 ; loop 4 times (4 words) clc ; clear the cf pushf ; save the flags long_add_loop: mov ax, [di] ; word to a register popf ; restore the flags adc [si], ax ; register + memory pushf ; save the flags add di, 2 add si, 2 loop long_add_loop popf ; restore the flags lea ax, variable1 call print_signed_8byte jmp outer_loop ; - - - - - PUT CODE ABOVE THIS LINE The calls to get_signed_8byte are followed by print_signed_8byte. This is so you can see what you have actually typed in. It will be neat and with commas, so it will be much easier to read. Everything else is the same as before. As an aside, let's talk about commas. Though we can get along without commas if we have a 4 digit number, There is no reason to do without them when using larger numbers. I find printer output that has 15 digit floating point numbers without commas not only hard to read but also silly. It's not that the computer is incapable of putting in commas, it's that the prople who wrote the programs couldn't be bothered with doing 2 hours of work to save the users lots and lots of aggrivation. Therefore, for all large number input (that is, larger than 1 word) you can use commas. They don't even have to be in the right place, the computer ignores them. All the following are the same number: 723469746 723,4,69746 72,3,46974,6 The PC Assembler Tutor 126 ______________________ 7,2,3,4,6,9,7,4,6 723,469,746 The program strips all commas out of the line and then looks at the input. All large output has the commas in the right spot. In order to enlist you in the crusade to stamp out unreadable input and output, the summary at the end of this chapter has the C code necessary for stripping commas. This is my contribution to world culture. If you have played with the signed addition program, all we need to do to make it unsigned is to change all the get_signed_8byte calls to get_unsigned_8byte calls. Change the print calls to unsigned also. Run the program. Now, let's try subtraction. How much code do we have to alter to make it a subtraction program? The answer is - one word. Just change the ADC (add with carry) to SBB (subtract with borrow). SBB learns from the CF flag whether the last subtraction had to borrow, and tells the next subtraction via the CF flag whether it has had to borrow. Change it, and try it out. (Yes, really do it, don't just pretend that you are going to do it). Why does this program work with exactly 8 bytes? Because we tell the loop via CX that it is 4 words long. If we put 2 in CX, the loop will think that the number is 2 words (4 bytes) long, and if we put 17 in CX, the loop will think that the number is 17 words (34 bytes) long. In fact, this little snippet of code can do either signed or unsigned addition or subtraction of any number of words simply by altering one number and one word of code. The code as it stands has only one shortfall. Remember from our earlier subtraction that we might want to have an INTO (interrupt on overflow) instruction after signed addition and subtraction. It needs to be after the last addition (or subtraction). All we need to do is put it after the POPF just below the loop. At this point the flags show the condition right after the last addition or subtraction: loop long_add_loop popf ; pop the flags off the stack into ; interrupt if overflow is set Chapter 12 - Multiple Word Arithmetic I 127 _______________________________________ SUMMARY ADC (add with carry) is used for multiple word arithmetic. It adds two numbers along with plus the value in CF, the carry flag (1 if the flag is set and 0 if the flag is cleared). CLC (clear the carry flag) should be used to clear CF before the first addition. After the addition, the flags register should be pushed in order to store CF unless it is certain that CF will not be effected by the rest of the loop. popf adc [di], ax pushf SBB (subtract with borrow) is used for multiple word arithmetic. It subtracts one number from the other and subtracts the value in CF to account for any borrows from the right. CLC (clear the carry flag) should be used to clear CF before the first subtraction. After the subtraction, the flags register should be pushed in order to store CF unless it is certain that CF will not be effected by the rest of the loop. popf sbb [di], ax pushf These operations have the typical 5 possibilities: 1) register, register 2) register, memory 3) memory, register 4) memory, constant 5) register, constant You may manually control the value in CF with STC and CLC. STC (set the carry flag) sets the value to 1, while CLC (clear the carry flag) sets the value to 0. These are used for initial settings of multiple word operations. In order to store the values in the flags register you use PUSHF (push the flags) until you need them again. At that time you can get them back with POPF (pop the flags). The PC Assembler Tutor 128 ______________________ STRIPPING COMMAS IN C In order to get rid of commas, you need some discipline in what kind of data entry you have. Specifically, you can't enter large numbers on the same line as text strings because text strings are likely to have commas that should be kept. This is hardly a major restriction. You can have as many numbers as you want on the same line since in C you must have whitespace between pieces of data. We will take all commas out of the line. You can retrofit most old programs with almost no change. The only time that you need this capability is if you are getting something from the keyboard, and what you probably have in the program is: scanf ( "format string", variables ); The method is (1) import a text string as a single string, (2) strip the commas, and (3) use sscanf instead of scanf. char buffer[80] ; fgets (buffer, 80, stdin) ; strip_commas (buffer); sscanf (buffer, "format string", variables) ; Both "format string" and the variables remain unchanged when you switch from scanf to sscanf. You might want to check for EOF with fgets. Heres the program: strip_commas (buffer) char *buffer ; { char *ptr1, *ptr2 ; ptr1 = ptr2 = buffer ; while (1) { if ( *ptr2 == ',') /* skip commas */ { ptr2++ ; continue ; } /*move, increment, and check for 0 */ if (!(*ptr1++ = *ptr2++)) /* this is '=', not '==' */ break ; } return ; } Chapter 13 - Multiple Word Arithmetic II ======================================== 129 We have just done multiple word addition and subtraction, which are easy. Now we have multiplication and division. We are going to multiply and divide long numbers by a one word (2 byte) number. Multiplying multiple-word numbers by multiple-word numbers is complex and time consuming but can be done. Dividing by a multiple-word number is an entirely different ballgame.{1} We'll do unsigned numbers first, then in a later chapter add the code we need for signed numbers. The core routine is the same. UNSIGNED MULTIPLICATION If you multiply an n digit number by an m digit number, there is a possibility of n+m digits in the result. 863 is 3 digits, 4975 is 4 digits, 863 X 4975 = 4,293,425 is 7 digits = 4 + 3. We will be multiplying an 8 byte number by a 2 byte number, so we'll need 10 bytes for the possible maximum result. Here's the code: ; - - - - - - - - ENTER DATA BELOW THIS LINE multiplicand dq ? multiplier dw ? result db 10 dup (?) ; - - - - - - - - ENTER DATA ABOVE THIS LINE ; - - - - - - - - ENTER CODE BELOW THIS LINE outer_loop: lea ax, multiplicand ; load multiplicand call get_unsigned_8byte call print_unsigned_8byte call get_unsigned ; unsigned word to multiplier mov multiplier, ax lea si, multiplicand ; load pointers lea bx, result mov cx, 4 ; number of words sub di,di ; clear di mult_loop: ____________________ 1 For those of you with a hankering for large multiplication and division, I have included subroutines which can multiply and divide numbers of any length in a file called MISHMASH.DOC. It is in \EXTRAFILE. You will need to finish all the chapters before looking at it, since it uses things that you don't know about yet. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 130 ______________________ mov ax, [si] ; multiplicand to ax mul multiplier ; {2} add ax, di ; high word from last multiplication jnc store_result inc dx ; {3} store_result: mov [bx], ax ; store 1 word of result. mov di, dx ; save high word for next multiplication add si, 2 ; increment pointers add bx, 2 loop mult_loop mov [bx], di ; move last word of result mov ax, [bx] call print_hex lea ax, result call print_unsigned_8byte jmp outer_loop ; - - - - - - - - ENTER CODE ABOVE THIS LINE There are two different input calls, an 8 byte one and a 2 byte one. Inside the loop we store the high word from the multiplication in DI and then add it to the next result. This is the same as when you multiply single digits in base 10 (9 X 7 = 63 carry the 6). Note that when you add DI, there can be a carry from AX to DX, but there can be no carry out of DX. After we drop out of the loop, we need to put the last word in result. We take it from DI, but we could take it from DX if we wanted. Finally, the printing. Print_unsigned_8byte can't print the whole result, so we are printing the high word in hex form. If those top two bytes are non zero, what 'print_unsigned_8byte' prints will be incorrect because it is missing the top 2 bytes. Note once again that the only thing constraining this program to an 8 byte number is the 4 that we put in CX - change that number and you can do any size number that you want. Run a bunch of numbers through this, including a couple that have more than a 20 digit result. UNSIGNED DIVISION Division is done the same way in the software as it is done with pencil and paper, starting at the left and working right. On the computer, this means starting with the high order word and working down. ____________________ 2 It would be about 3% faster to have this in a register, but unfortunately we are out of registers. 3 Do we need to check DX for a carry here? No. The maximum multiplication is FFFFh X FFFFh. The result is FFFE 0001h. That means that DX is a maximum FFFEh. If you add one, that's FFFFh, and no carry occurs. Chapter 13 - Multiple Word Arithmetic II 131 ________________________________________ ; - - - - - - - - ENTER DATA BELOW THIS LINE dividend dq ? divisor dw ? quotient dq ? remainder dw ? ; - - - - - - - - ENTER DATA ABOVE THIS LINE ; - - - - - - - - ENTER CODE BELOW THIS LINE outer_loop: lea ax, dividend ; get dividend call get_unsigned_8byte call print_unsigned_8byte call get_unsigned ; get divisor mov divisor, ax lea si, dividend + 6 ; start at the top lea bx, quotient + 6 mov di, divisor mov cx, 4 ; number of words sub dx, dx ; clear dx for first division division_loop: mov ax, [si] ; dividend word to ax div di ; {4} mov [bx], ax ; word of result to quotient sub si, 2 ; decrement the pointers sub bx, 2 loop division_loop mov remainder, dx mov ax, remainder call print_unsigned lea ax, quotient call print_unsigned_8byte jmp outer_loop ; - - - - - - - - ENTER CODE ABOVE THIS LINE That's it? Yup. The division instruction is designed to work effeciently and simply. We start with the most significant digits, divide, put the quotient in the variable "quotient", ____________________ 4 After this division, the quotient is in AX and the remainder is in DX. Say, aren't we going to do anything with the remainder? There's nothing in the code about DX until we get out of the loop. In fact, we ARE doing something with the remainder. Just like division with pencil and paper, when you have a remainder, you bring it down to the left of the next digits you are going to divide. These get divided the next time around. But we don't need to move the remainder because it's already in the right place. Pretty snappy, huh? You don't need to move anything; it all takes care of itself. The PC Assembler Tutor 132 ______________________ DECREMENT the pointers, and get the next word for division. After the final division, we have the remainder left in DX, so we move it to the variable "remainder". The final instructions print the remainder and the quotient. Notice that we don't need to touch the remainder during the entire operation. The 8086 leaves it exactly where it needs to be for the next division. Using the DX register when you have single word division seems screwy, but using DX for multiple word division is both natural and elegant. The Intel people made one instruction do the work of both. Remember from the earlier chapter on division that you can get a zero divide error if the quotient is larger than 65535. Is it possible to get a quotient larger than 65535 in this routine? NO. It is impossible to get a zero divide on anything other than a zero.{5} Run the program and do several examples. You can even do a 0 divide if you feel like interrupting the program. SIGNED NUMBERS For byte or word signed multiplication and division, the 8086 changes the signed numbers into unsigned numbers, does unsigned multiplication/division, then adjusts for sign. For long numbers, we have to do these operations ourselves, so we need three sections of code. (1) change the numbers into unsigned numbers, (2) do unsigned multiplication/division and (3) adjust the signs. The routines that we have here are part two of this scheme. It will be easier to implement this once you know about subroutines, so signed division and multiplication will have to wait till later. ____________________ 5 This is technical, so if you start getting lost, don't worry about it. How do we know that it's impossible? What we are putting in DX is the remainder (R). R is always less than the divisor (D). Let Q be the number in AX the next time around. What we are dividing is: ((R*65536) + Q ) / D <= (( R*65536 ) + 65535 ) /D since Q is less than to or equal to 65535. This is the maximum. ( ((R*(65535 + 1)) + 65535 ) /D = (((R+1) * 65535) + R)/D (huh?) = ((R+1) * 65535)/D + R/D Let's do a few examples: if D = 1, R < D so R = 0 max. = (1*65535)/1 + 0/1 = 65535 rem 0 where rem = remainder. If D = 2, R < D so R = 1 max. = ((1+1)*65535)/2 + 1/2 = 65535 rem 1 If D = 3, R < D so R = 2 max. = (2+1)*65535)/3 + 2/3 = 65535 rem 2 See a pattern here? R/D < 1, so the quotient can never be 65536. The maximum will always be 65535 with the remainder 1 less than the divisor. If you aren't a techie, ignore all this. ________________________________________ SUMMARY For both signed and unsigned numbers, multiple word division and multiplication are based on an unsigned number routine. Signed numbers are changed into unsigned numbers, the operation is performed, and the signs of the results are adjusted. Multiplication is done the same as for single words except that the high word from one result is saved and added to the low word of the next result, thus adding the two partial results. If this addition gives a carry, DX must be incremented. Division operates from left to right. For the first division, DX is zeroed. After that it always contains the remainder from the last division. The quotients in AX are moved to memory one by one. At the end, the final remainder will be in DX. Chapter 14 - Zoom ================= 134 There are only a couple of reasons for working at the assembler level. Perhaps you're curious about how the PC functions at the machine level. Maybe you want to optimize a time consuming section of code. Maybe you want to work easily with the DOS function calls and the BIOS calls (which will be introduced in a later chapter). Or maybe you want raw speed. Every time that I enter: >dir /w and watch DOS meander its way down the screen at a leisurely pace, I think "Can't the computer go any faster?" Let's find out. This is going to be a short chapter. Make a copy of template.asm, and we'll call it zoom.asm. This is going to overwrite the entire screen 200 times.{1} Before we write to the screen, we need to know where the screen is. When IBM designed the structure of the PC, they decided to put the memory for the monochrome card in one place and the memory for the color card in another place - apparently they thought you might want to have both a color monitor and a monochrome monitor running at the same time. The color monitor is in segment 0B800 at offset 0, and the monochrome monitor is in segment 0B000 at offset 0. We need to put the correct number into the program, so you need to know whether you have a color card or a monochrome card. If one segment number doesn't work, you can try the other. TEMPLATE.ASM ; - - - - - - - - - - START CODE BELOW THIS LINE call show_regs_and_wait ; {2} ; we are about to start ; this marks the time mov ax, 0B800h ; color seg. 0B000h is mono seg mov es, ax ; es is at video card segment mov cx, 2 ; do this whole thing twice outer_loop: ____________________ 1 It is now time for you to go out and get a book about the internal structure and i/o interface of the PC. One good book is "The Peter Norton Programmer's Guide To The IBM PC", by guess who? It is clearly written and a good introduction. Another quality book is "DOS Programmer's Reference" by Terry Dettmann. It is very systematically laid out and is more techie oriented. 2 We need to initialize the video card to make sure it is in the right place. This is the easiest way. The registers themselves mean nothing to us. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 14 - Zoom 135 _________________ push cx ; save for outer loop instruction mov cx, 100 ; 100 repeats ; zoom_loop count mov al, '0' ; 100 characters starting at '0' mov ah, 07h ; black background, white letters zoom_loop: ; draw the screen 100 times push cx mov cx, 2000 ; 80 X 25 screen is 2000 words long mov si, 0 ; start at offset 0000. inner_loop: ; inner loop - fill the screen - 2000 words mov es:[si], ax add si, 2 loop inner_loop inc al ; next higher ASCII character pop cx ; zoom_loop count loop zoom_loop pop cx ; outer_loop count loop outer_loop call get_continue ; finished - this is for timing ; - - - - - - - - - - END CODE ABOVE THIS LINE Show_regs_and_wait resets the video card, so we use it to make sure the video memory is set at offset 0000. It then waits for ENTER. At the end, get_continue waits for ENTER.{3} This way you can mark the beginning and the end in order to time it. We set the ES segment to the video segment. This is different depending on whether you have a monochrome card or a color card. The monochrome segment is at 0B000h and the color card is at 0B800h. You need to know which kind of card you have so you can put the right number into ES via AX. Since the normal segment for SI is the DS segment, we need to put in a segment override to use SI with the ES segment. We start with the ASCII character '0' and then do the next 99 characters in increasing ASCII sequence. Technically, ASCII characters end at 127, but the PC extended characters go up to 255, so we will start with ASCII 48d and end with ASCII 147d. The zoom_loop changes the character 100 different times, and the inner_loop fills the screen. That 07h in AH means that we will have white characters on a black background. The outer loop has a count of 2. This should be adjusted. If you have a medium speed machine make it 4 (400 screen fills) and if you have a high speed machine make it 8 (800 screen fills). ____________________ 3 That is its mission in life. It is there so that you can time blocks of code. If you want to find out how long some code takes, repeat it 10,000 (or whatever is appropriate) times and put get_continue in front of it and behind it. That way you can control the start and mark the finish. When this is done, the screen will be filled with characters so you will need to use the DOS command > cls to clear the screen for anything else. When it is all assembled and linked, get a cup of coffee and a wristwatch with a second hand and we'll time it. Divide by 200 (or 400 or 800) to find out how long it takes to fill one screen. If the characters are not on your screen, you probably have the wrong segment address for your video card, so try the other one. Are you ready to time it? Then ready, set, go. Hmmm. On my machine 200 repeats takes about 4 seconds, while the dir command takes about 1 second for one screen. That means that zoom.asm is about 50 times faster. That's the difference between someone running a 4 minute mile and someone running a 3hr. 20min. mile. This is one of the reasons people like to work at the assembler level. Chapter 15 - Subroutines ======================== 137 It is now time to talk about subroutines. If you have only used BASIC this may be difficult for you. It is assumed that you are familiar with subroutines and use them constantly in your programming. You have been using subroutines since the very first program in this manual. When you wrote: call get_num you called a subroutine in ASMHELP.OBJ. Now you are going to write subroutines yourself and have them call each other. There are different template files for programs with subroutines. They are SUBTEMP1.ASM and SUBTEMP2.ASM. We will start with SUBTEMP1. It has the entry subroutine and a space for additional subroutines. The entry subroutine is the subroutine where the operating system starts the program; it does the necessary initialization and has special code for that. You will see some additions to the normal template file. At the top is the line: INCLUDE \pushregs.mac What this is will be explained later, but you must put the file PUSHREGS.MAC in the root directory of your current drive. You will find it in the \TEMPLATE subdirectory. At the end of the SUBTEMP1.ASM is: ; + + + + + + + + + + + + START SUBROUTINES BELOW THIS LINE ; + + + + + + + + + + + + END SUBROUTINES ABOVE THIS LINE This is where you will write all the subroutines except the entry subroutine which is still the same as before. All data for all subroutines still goes in the DATASTUFF segment. Our first program will just call subroutines which will print out messages. Using SUBTEMP1.ASM, it looks like this: ;prog1.asm ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE main_message db "This is the entry routine.", 0 sub1_message db "This is subroutine1.", 0 sub2_message db "This is subroutine2.", 0 sub3_message db "This is subroutine 3.", 0 ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 138 ______________________ ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE mov ax, offset main_message call print_string call sub1 mov ax, offset main_message call print_string ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE ; + + + + + + + + + + + + START SUBROUTINES BELOW THIS LINE ;------------ sub1 proc near push ax mov ax, offset sub1_message call print_string call sub2 mov ax, offset sub1_message call print_string pop ax ret sub1 endp ;------------ sub2 proc near push ax mov ax, offset sub2_message call print_string call sub3 mov ax, offset sub2_message call print_string pop ax ret sub2 endp ;------------ sub3 proc near push ax mov ax, offset sub3_message call print_string pop ax ret sub3 endp ; ---------- ; + + + + + + + + + + + + END SUBROUTINES ABOVE THIS LINE The data consists of messages to be printed by print_string. Print_string prints a zero terminated string (the number zero, not the character '0'), so there must be a zero after each message in the data segment. The entry subroutine prints a message and then calls sub1, the first subroutine, which prints a message and calls sub2 which prints a message and calls sub3. Chapter 15 - Subroutines 139 ________________________ Sub3 prints a message and then returns to sub2 which prints a message and returns to sub1 which prints a message and returns to the entry routine which prints a message and then exits. This program should print 7 messages in all. You will notice that the first thing that each subroutine does is save the value in AX, since it uses the AX register. This is the cardinal rule of robustness at the assembler level. IF YOU USE A REGISTER, YOU MUST SAVE ITS VALUE BY PUSHING IT ON THE STACK; YOU MUST THEN RESTORE THE VALUE JUST BEFORE EXITING. It is impossible to overstress this. The routines which call your routine might rely on the registers remaining unaltered. If you disobey this rule and alter the registers, you'll be sorry. Why doesn't the entry routine push and pop the registers it uses? Well, the operating system assumes the registers will contain trash upon return from the program, so it uses nothing in the data registers. All the subroutines except the entry routine are near routines. We will only use near routines. Assemble this program, link it and run it. If it works ok, it is then time for program 2, which is the same as program1, but is in two files. Often, we want parts of a program in different files. Perhaps parts are standard subprograms which you have already written and assembled, perhaps the total program is too large to be handled comfortably in one file, perhaps different people are writing different parts of the program. Not only must we write the programs, but we must be able to connect them. We will put the entry routine, sub2 and the associated data in subtemp1.asm. We will put sub1, sub3, and the associated data in subtemp2.asm. Take a look at SUBTEMP2.ASM. It is slightly different. First, it does not have the variables that you need for set_reg_style (ax_byte, bx_byte, etc.) but it does have EXTRN statements for them. This means that you can change the register style from this file. SUBTEMP1.ASM has these variables declared PUBLIC so the linker can join them correctly.{1} We will talk about the correct way to declare external data later. SUBTEMP2.ASM has no stack segment, though there could be one. There is no entry subroutine. Therefore at the very end, you have the line: END with nothing after it. In SUBTEMP1.ASM, you have ____________________ 1. The reason for having only one set of variables for the style is so that every time you change one of the style variables, the array is updated. If you had two different arrays you could have two different sets of information for set_reg_style. The PC Assembler Tutor 140 ______________________ END start so the assembler and linker know that the program begins at the label "start". Let's do the two programs. Here are the data, the entry code and the subroutine code from the first file. ;prog1.asm ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE main_message db "This is the entry routine.", 0 sub2_message db "This is subroutine2.", 0 ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE PUBLIC sub2 EXTRN sub1:NEAR, sub3:NEAR mov ax, offset main_message call print_string call sub1 mov ax, offset main_message call print_string ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE ; + + + + + + + + + + + + START SUBROUTINES BELOW THIS LINE sub2 proc near push ax mov ax, offset sub2_message call print_string call sub3 mov ax, offset sub2_message call print_string pop ax ret sub2 endp ; + + + + + + + + + + + + END SUBROUTINES ABOVE THIS LINE Notice that sub1 and sub3 have been declared EXTRN before they were referenced, and the EXTRN statement tells the assembler that they are both near procedures. sub2 has been declared PUBLIC so the assembler will give the address of sub2 to the linker. Here's the data and code for the other file. ;prog2.asm ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE sub1_message db "This is subroutine1.", 0 sub3_message db "This is subroutine 3.", 0 ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ; + + + + + + + + + + + + START SUBROUTINES BELOW THIS LINE Chapter 15 - Subroutines 141 ________________________ PUBLIC sub1, sub3 EXTRN sub2:NEAR ;------------ sub1 proc near push ax mov ax, offset sub1_message call print_string call sub2 mov ax, offset sub1_message call print_string pop ax ret sub1 endp ;------------ sub3 proc near push ax mov ax, offset sub3_message call print_string pop ax ret sub3 endp ; ---------- ; + + + + + + + + + + + + END SUBROUTINES ABOVE THIS LINE Here sub1 and sub3 have been declared PUBLIC and sub2 has been declared EXTRN. Assemble both programs and then link all three. link prog1+prog2+\asmhelp.obj assuming that asmhelp is in the root directory. Run it. You should have the same results as before. We are going to do one more thing with the same two files. Without changing any of the code, we are going to put the data for prog1 in prog2 and the data for prog2 in prog1 like this. ;prog1 ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE sub1_message db "This is subroutine1.", 0 sub3_message db "This is subroutine 3.", 0 ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ;prog2 ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE main_message db "This is the entry routine.", 0 sub2_message db "This is subroutine2.", 0 ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE So far, so good. Obviously we are going to need some more PUBLIC The PC Assembler Tutor 142 ______________________ statements and some EXTRN statements so the linker can link the four messages, but where do they go and what do they look like? The PUBLIC statements are the easiest. Put them in the segment where the message data appears, either before or after the data declaration. The EXTRN statement is a little more complicated. First, all data is declared EXTRN by giving the variable name followed by a colon followed by its data type. The data types are BYTE, WORD, DWORD (4bytes), QWORD (quadword or 8 bytes), and TBYTE (10 bytes). These are the standard 8086/7 data sizes. Therefore we have: EXTRN sub1_message:BYTE, sub3_message:BYTE in prog2.asm and: EXTRN main_message:BYTE, sub2_message:BYTE in prog1.asm. Where do they go? In order to know that, we need to talk about segment overrides again. You will remember from our discussion of the ASSUME statement that every time the assembler writes code with a variable, it checks the ASSUME statements to see which segment register(s) have the address of the segment that that variable is in. If we have: ASSUME cs:SEG1, ds:SEG2, es:SEG3, ss:SEG4 then if variable1 is in SEG2, the assembler will write no override in the code since DS is the 8086 default segment. MACHINE CODE ASSEMBLER INSTRUCTION A1 0000 mov ax, variable1 If variable1 is in SEG1 or SEG3 or SEG4, the assembler will write the appropriate segment override in the code. MACHINE CODE ASSEMBLER INSTRUCTION 2E: A1 0000 mov ax, variable1 26: A1 0000 mov ax, variable1 36: A1 0000 mov ax, variable1 (By the way, those zeros just mean that the variable is at 0000 offset from the beginning of the segment). The same thing happens when you have an EXTRN statement. The assembler associates the externally declared variable with the segment it is declared in. When the variable is used, it then goes through the same actions as if the variable were actually in that segment. Let's declare variable5 external with: EXTRN variable5:WORD If we have: Chapter 15 - Subroutines 143 ________________________ ASSUME cs:SEG1, ds:SEG2, es:SEG3, ss:SEG4 then if variable5 is declared external in SEG2, the assembler will write no override in the code since DS is the 8086 default segment. MACHINE CODE ASSEMBLER INSTRUCTION A1 0000 E mov ax, variable5 If variable5 is declared external in SEG1 or SEG3 or SEG4, the assembler will write the appropriate segment override in the code. MACHINE CODE ASSEMBLER INSTRUCTION 2E: A1 0000 E mov ax, variable5 26: A1 0000 E mov ax, variable5 36: A1 0000 E mov ax, variable5 The "E" after the machine code means that the assembler knows that the variable is external and it will tell the linker so the linker can put the correct offset address at that point in the machine code. Remember, as always, that it is your responsibility to have the correct segment address in the segment register before using a variable. Now we know where it goes. When you declare a variable external, you must put the EXTRN statement in a segment which uses the same segment register as the EXTRN variable is going to use. If the EXTRN variable will use DS, then the segment where the EXTRN statement is must use DS. If the variable uses ES, then the segment the EXTRN statement is in must use ES. In other words, the ASSUME statement for the segment the variable is in must match EXACTLY the ASSUME statement you would write if the variable were internal, not external.{2} Normally, this is DS, but in special circumstances you might want something else. Also, if there is no segment that exactly matches what you want, then you need to create a dummy segment: DUMMY_SEG SEGMENT EXTRN variable7:QWORD DUMMY_SEG ENDS and make the assume statement that you want: ____________________ 2. This means that if the segment with the EXTRN statement has more than one segment register in the assume statement: ASSUME ds:MORESTUFF, es:MORESTUFF then both those registers must be set to the segment of the external variable when using it or your results may be unreliable. The PC Assembler Tutor 144 ______________________ ASSUME es:DUMMY_SEG What segment has DS in an ASSUME statement? DATASTUFF in both files, so that is where the EXTRN declaration goes - in the DATASTUFF segment. ;prog1 ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE PUBLIC sub1_message, sub3_message EXTRN main_message:BYTE, sub2_message:BYTE sub1_message db "This is subroutine1.", 0 sub3_message db "This is subroutine 3.", 0 ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ;prog2 ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE PUBLIC main_message, sub2_message EXTRN sub1_message:BYTE, sub3_message:BYTE main_message db "This is the entry routine.", 0 sub2_message db "This is subroutine2.", 0 ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE Change the data in the two files, assemble them again and link them again: link prog1+prog2+\asmhelp.obj You should get the same results as before. We are now through with these programs. Make sure you understand how to define PUBLIC and EXTRN procedures and PUBLIC and EXTRN data before going on, since we are not going to cover it again. Everything else in this chapter will be done with a single file in order to make life easier. PASSING DATA When you pass data to the routines in ASMHELP.OBJ, you always pass it through the AX register. The reason for this is that you needed to use these routines before you knew much about 8086 assembler language. It is solely for the convenience of beginners and is totally non-standard. In the real world, when you call a subroutine you ALWAYS pass the data on the stack, no matter which language you are using. If you have the C statement: my_procedure (variable1, variable2, variable3) ; then the C compiler will generate the following code: push variable3 push variable2 push variable1 Chapter 15 - Subroutines 145 ________________________ call my_procedure{3} The C language pushes these variables in right to left order. Before the call instruction is executed variable1 is on the top of the stack, variable2 is the next down, and variable3 is third on the stack. Is variable1 still on the stack top after the call instruction is executed? No. The call instruction pushes either one or two words on the stack. Before you go any farther with subroutines you need to know how the call and return instruction operate. Every time you have used show_regs, both CS the code segment address and IP the instruction pointer have been displayed. What does IP do? When the 8086 is ready to execute an instruction, it takes IP, adds it to CS to calculate the total address, and gets the instruction at that address. It then immediately figures out how long the instruction is going to be and adds that amount to IP.{4} What this means is that at any time, IP points to the NEXT instruction, not the current instruction. When you execute a call, the 8086 changes IP to point to the first byte of the called subroutine, so the next instruction executed is the first byte of the called subroutine. There are two different types of procedures, near procedures and far procedures. In a near procedure, you keep CS, the code segment register, the same. In a far procedure you change CS. So, when you call a near procedure you change one thing (IP) and in a far procedure you change two things (IP and CS). When you want to get back from the subroutine, you need to have CS with the segment of the calling routine and IP with the address of the instruction after the call. What are the mechanics of all this? Let's take a near procedure first. In a near call, the 8086 first changes the instruction pointer to point to the next instruction. It then pushes IP on the stack, and puts the address of the called subroutine (which is in bytes 2 and 3 of the call instruction) in IP. IP now points to the called subroutine. There is one more word (2 bytes) on the stack. At the end of the called subroutine, a NEAR return (ret) pops the top word off the stack into IP. IP then points to the instruction after the call instruction. In a far call, the 8086 first changes the instruction pointer (IP) to point to the next instruction. It then pushes CS on the stack, followed by IP. It then loads the offset address of the called subroutine in IP and the segment address of the called subroutine in CS. This new IP is in bytes 2 and 3 of the call instruction and the new CS is in bytes 4 and 5 of the call ____________________ 3. You C fanatics will notice that there are some initial underscores missing. Let's not confuse the issue. 4. Instructions can vary from one byte long to six bytes long, and the 8086 can tell from the first (or first and second) byte(s) how long the total instruction will be. The PC Assembler Tutor 146 ______________________ instruction. IP and CS now have the address of the called subroutine. The stack has two words (4 bytes) more on the stack. The old IP is the stack top and the old CS is next on the stack. At the end of the subroutine, a FAR return (ret) pops the stack top into IP, then pops the next stack item into CS. Now IP and CS point to the instruction after the call instruction. These are two different types of call and they have two different machine codes. These are two different types of returns and they have two different machine codes. MACHINE CODE ASSEMBLER INSTRUCTIONS ; a far routine ;----- far_routine proc far CB ret far_routine endp ;----- ; a near routine ;----- near_routine proc near C3 ret near_routine endp ;----- ; a near and far call E8 0A43 R call near_routine 9A 015C ---- R call far_routine The machine code for a near return is C3; for a far return it's CB. The machine code for a near call is E8; for a far call it's 9A. The near call has the address of the called routine (0A43h) in the following two bytes. The far call has the address of the the called routine (015Ch) in the next two bytes followed by the segment of the called routine. The segment address isn't there yet. It will be put there by the linker and loader, but the assembler has saved the space for the address. That's why the dashes are there. Remember, the R is there because those addresses might be relocated by the linker or the loader. You tell the assembler whether to code a near return or far return by telling it whether it is a near or a far procedure. routine1 proc near routine2 proc far How does the assembler know whether to code a near or far call? If it has already seen the procedure, it knows what type it is. Chapter 15 - Subroutines 147 ________________________ If it hasn't seen it yet, it uses the default type.{5} If it is an external subroutine, the assembler knows because you have written an EXTRN statement. EXTRN routine3:NEAR, routine4:FAR This EXTRN statement should appear before the call. What if the routine appears after the call in the source file but it isn't the default type? You can override the default type. call NEAR PTR routine5 call FAR PTR routine6 This is the same cumbersome syntax that we had with pointers to data, but it's the only game in town. Normally, if the subroutine appears after the call, you don't need to do anything if it is a near call but you need to put a FAR PTR override if it is a far call. ____________________ 5. The default is near for what we are doing. However, Microsoft has something called "simplified" directives and the default changes in these cases. The PC Assembler Tutor 148 ______________________ THE STACK Up to this time we have used the stack for temporary storage. If you want to temporarily save either a register or a value in memory, you push it: push ax push variable1 and if you want to get them back you pop them: pop variable1 pop ax This is always a word (2 bytes) at a time. When you pop the stack, the 8086 gives you back the words in reverse order. Thus if you push the following: push variable1 push variable2 push variable3 push variable4 push variable5 then in order to get the data back in the same place, you need to pop in this order: pop variable5 pop variable4 pop variable3 pop variable2 pop variable1 It pops the last thing that was pushed that hasn't been popped yet. Nothing has been said about where the stack is or how it operates. It's time to change that. When the operating system starts a program, it looks for a stack segment. If the stack segment has been properly defined, the operating system puts the stack segment's segment address in SS (the stack segment register) and sets SP (the stack pointer) to point to the first byte AFTER the end of the stack segment. Exactly where this is depends on how large you have defined your stack segment. SS and SP are set, and there is nothing on the stack. When you push something: push dx the 8086 subtracts 2 from SP (making one word of space) and puts that thing at the new address in SP. SP contains the address of ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 15 - Subroutines 149 ________________________ the last thing pushed. This means that SP is decreasing, and the stack segment is filling up from back to front. In the topsy-turvy world of stacks, when you put things on the stack, the stack grows downward. What makes things especially confusing is that many book writers will picture a stack: variable1 variable2 ax dx and not bother to tell you whether the stack is growing upwards or downwards or where the stack top is. In this book, the stack TOP will always be visually on the BOTTOM. High addresses will be visually up and low addresses will be visually down. You need to get used to SP decreasing as the stack gets larger, and this is the easiest way to do it. So, if you have the instructions: push ax push variable1 push si push di after these instructions, the stack will look like this: VALUE ADDRESS ax sp + 6 variable1 sp + 4 si sp + 2 sp -> di sp + 0 When you pop a value, the 8086 moves the word (2 bytes) at SP to the appropriate location and INCREMENTS SP by 2. pop di You would now have: VALUE ADDRESS ax sp + 4 variable1 sp + 2 sp -> si sp + 0 As long as you are just using PUSH and POP, this is entirely self regulating. SS is set, and SP is modified by the 8086 without you doing anything. It is now time to get more sophisticated. In our C example: my_procedure (variable1, variable2, variable3) ; we generated the code: The PC Assembler Tutor 150 ______________________ push variable3 push variable2 push variable1 call my_procedure What will the stack look like upon entry to my_procedure? That depends on whether my_procedure is a near procedure or a far procedure. If it is a near procedure, you will have: VALUE ADDRESS variable3 sp + 6 variable2 sp + 4 variable1 sp + 2 sp -> old IP sp + 0 If it is a far procedure, you will have: VALUE ADDRESS variable3 sp + 8 variable2 sp + 6 variable1 sp + 4 old CS sp + 2 sp -> old IP sp + 0 Therefore, the variables are in different places relative to SP depending on whether it is a near or a far procedure. All examples will be with near procedures, but they are all valid for far procedures if you adjust for having the old CS on the stack. How do we access these variables? By using a pointer. We could use BX, SI or DI, but they have DS, not SS as their natural segment register. The only pointer with SS as the natural segment register is BP, the base pointer. Since we are going to use BP, we need to push its current value in order to save it: push bp The stack now looks like this: VALUE ADDRESS variable3 sp + 8 variable2 sp + 6 variable1 sp + 4 old IP sp + 2 sp -> old bp sp + 0 This is the standard way to do it and this is what the stack always looks like if you follow the standard method. The standard code for setting up the stack for access is: push bp mov bp, sp Chapter 15 - Subroutines 151 ________________________ We give BP the same value as SP, so BP also points to the top of the stack and we use BP instead of SP. We now have: VALUE ADDRESS variable3 bp + 8 variable2 bp + 6 variable1 bp + 4 old IP bp + 2 bp -> old bp bp + 0 Now, if you want to push and pop things, you can do it to your heart's content. BP will always point to the set of data that you want to work with. Let's take the average of the three variables, and print it. mov ax, [bp+4] ; add the three numbers add ax, [bp+6] add ax, [bp+8] mov dx, 0 ; prepare dx for division mov bx, 3 ; unsigned divide by 3 div bx call print_unsigned ; result is in ax We are using AX, BX, and DX, so we need to push them before doing this: push ax push bx push dx After we are done we need to (1) pop the registers and (2) restore BP. This is also a pop. pop dx pop bx pop ax pop bp ret The whole subprogram now looks like this ;----- my_procedure proc near push bp ; set up base pointer mov bp, sp push ax ; push registers push bx push dx mov ax, [bp+4] ; add the three numbers add ax, [bp+6] add ax, [bp+8] mov dx, 0 ; prepare dx for division mov bx, 3 ; unsigned divide by 3 div bx call print_unsigned ; result is in ax The PC Assembler Tutor 152 ______________________ pop dx ; pop registers pop bx pop ax pop bp ; restore old base pointer ret my_procedure endp ;------ There is only one more improvement to make. If you look at the code, it is not clear what [bp+4] refers to. We know where it is, but what is it? Therefore, we will always use EQU statements to give names to our stack variables. It will be clearer, and if you need to change the code, it is much easier to change the EQU definition than to change the stack references in the code. As usual, we follow the C convention and put EQU names in capital letters. ;----- my_procedure proc near VAR1 EQU [bp+4] VAR2 EQU [bp+6] VAR3 EQU [bp+8] push bp ; set up base pointer mov bp, sp push ax ; push registers push bx push dx mov ax, VAR1 ; add the three numbers add ax, VAR2 add ax, VAR3 mov dx, 0 ; prepare dx for division mov bx, 3 ; divide by 3 div bx call print_unsigned ; result is in ax pop dx ; pop registers pop bx pop ax pop bp ; restore old base pointer ret my_procedure endp ;------ This is a simple example, so it doesn't look that important to use the EQU statements. Just wait till you have more complex subroutines. By the way, this program does no error checking. (If the sum is > 65535 it will give the wrong answer). There is still one thing to do. When we called the subroutine, we pushed the variables on the stack: push variable3 push variable2 push variable1 Chapter 15 - Subroutines 153 ________________________ call my_procedure We now want to take them off. Do we need to pop them? No, this is trash so they go into the Great Bit Bucket. There are two ways of doing this, and this is language dependent.{1} In C, it is the STANDARD that the calling routine takes them off, and it is done this way: push variable3 push variable2 push variable1 call my_procedure add sp, 6 ; 3 variables = 6 bytes we simply INCREASE sp by the number of bytes that we pushed on the stack. Whoof, they're gone. If you use PASCAL or FORTRAN, then the CALLED routine must take the variables off the stack on return. How does it do that? There is yet another type of return statement: ret (6) ; 3 variables = 6 bytes {2} causes the 8086 to increase sp by 6 as the last thing it does before returning from the subroutine. Which method you use is decided by which high-level language you are using. MACHINE CODE ASSEMBLER INSTRUCTIONS ;----- far_routine proc far CA 001A ret (26) ; hex 1A CB ret far_routine endp ;----- ;----- near_routine proc near C2 002C ret (44) ; hex 2C C3 ret near_routine endp ;----- Here are the four different types of returns along with the machine code. Notice that the returns which increment the stack have the increment count coded in the machine code. You may have noticed that even in this first subroutine, pushing and popping the registers takes a lot of space. It is fairly ____________________ 1 And a major reason that is a real pain in keester to have multi-language programs. 2 The parentheses are not necessary. The PC Assembler Tutor 154 ______________________ normal to use 6 registers in a subroutine. This means that you will need to write: push ax push bx push cx push dx push si push di at the beginning of the subroutine and: pop di pop si pop dx pop cx pop bx pop ax before returning. This is a lot of space and it gets boring. Also, you have to remember to pop in the exact reverse order or you will screw things up. Fortunately we have two macros to help us. The file PUSHREGS.MAC has two macros, one called PUSHREGS and the other called POPREGS. A macro is a set of directions for generating additional assembler code before the file is assembled. That's why it is called the Microsoft Macro Assembler. You include \pushregs.mac at the beginning of the file, and then everytime the assembler sees the word PUSHREGS followed by register names it generates push instructions. Every time the assembler sees POPREGS followed by register names, it generates pop instructions. It generates actual text which is assembled later. The form for generating those push instructions above is: PUSHREGS ax, bx, cx, dx, si, di the word PUSHREGS followed by the registers separated by commas. (Make sure there is no comma after the last register). This must all be on one line. PUSHREGS pushes the registers in left to right order. The form for generating those pop instructions above is POPREGS ax, bx, cx, dx, si, di The registers will be popped in the REVERSE order to the way they are listed on the line, that is, in RIGHT TO LEFT order. Notice that the order of registers is the same for both PUSHREGS and POPREGS. This is so that you may write the push part: PUSHREGS ax, bx, cx, dx, si, di and then use your word processor to copy the line to the end of the subroutine, changing PUSHREGS to POPREGS. This insures that Chapter 15 - Subroutines 155 ________________________ the pushes and pops will be in exact reverse order. It saves space and time, and it generates exactly the same code as if you had written all those pushes and pops in the code. Whenever we have subroutines in the future, we will always use it. MOVING A STRING As a final example, we will create a subroutine that moves a Pascal string from one place to another. We'll assume that both strings are in the current DS, so no segments need to be changed. move_string ( from_string, to_string ) ; where from_string and to_string are the ADDRESSES of the strings. The code generated by the Pascal compiler will be: mov ax, offset from_string push ax mov ax, offset to_string push ax call move_string ; this is Pascal, so the CALLED subroutine must ; get rid of the variables on the stack. Notice that Pascal pushes this data in left to right order, exactly the opposite of how C would handle it. After setting up BP, we have: from_string offset bp + 6 to_string offset bp + 4 old IP bp + 2 bp -> old bp bp + 0 Before coding this, you need to know the structure of a Pascal text string. The first byte (string[0]) is not text, but the text count. The second byte is the first piece of text. This means two things. First, the maximum string size in Pascal is 255, the largest count that will fit in one byte. Second, you need to move 'count + 1' bytes. 'count' is how many text bytes there are, but then you need to move the count itself. If the string is empty (count = 0) you need to move 1 byte, the count byte. Here's the code ; - - - - - move_string proc near FROM_ADDRESS EQU [bp + 6] TO_ADDRESS EQU [bp + 4] push bp ; set up bp mov bp, sp PUSHREGS ax, cx, si, di mov si, FROM_ADDRESS ; source mov di, TO_ADDRESS ; destination sub cx, cx ; zero cx The PC Assembler Tutor 156 ______________________ mov cl, [si] ; text byte count of source inc cl ; add 1 for byte count itself move_loop: mov al, [si] ; source to al mov [di], al ; al to destination inc si ; move pointers to next byte inc di loop move_loop POPREGS ax, cx, si, di pop bp ret (4) ; Pascal, so pop offsets. move_string endp ; - - - - - - - We still have some more to do, and we'll do it in part three of the chapter. Chapter 15 - Subroutines 157 ________________________ What if both strings are in memory but we don't know which segments they are in? In that case, the calling subroutine needs to pass both the segment and the offset for both the from_string and the to_string. Let's do this in C. move_string ( from_string, to_string ) ; In C, this will pass the addresses of the arrays, not the arrays themselves. C, once again, pushes these variables in right to left order. If the compiler is set up to move both the segment and the offset, then the generated C code will be: mov ax, seg to_string push ax mov ax, offset to_string push ax mov ax, seg from_string push ax mov ax, offset to_string push ax call move_string add sp, 8 ; 4 pushes = 8 bytes On the 8086, the low two bytes are ALWAYS the offset and the high two bytes are ALWAYS the segment. Remember, SEG gets the segment starting address of the named variable. We will do the subroutine as a near routine. After setting up BP, we will have: to_string segment bp + 10 to_string offset bp + 8 from_string segment bp + 6 from_string offset bp + 4 old IP bp + 2 bp -> old bp bp + 0 In the subroutine we will have to move the segment and the offset for each pointer. Luckily for us, there are two 8086 instructions for doing this: LDS (load DS) loads the first two bytes into the named register and the next two bytes into DS. LES (load ES) loads the first two bytes into the named register and the next two bytes into ES. If we write: LES di, [bp + 8] then the 8086 will load the first two bytes (bp+8 and bp+9) into DI and the next two bytes (bp+10 and bp+11) into ES. This loads ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 158 ______________________ the offset and the segment in the same instruction. If we write: LDS si, [bp + 4] the 8086 will load the first two bytes (bp+4 and bp+5) into SI and the next two bytes (bp+6 and bp+7) into ES, loading both the offset and segment in one instruction. LDS and LES allow you to load the offset into any full arithmetic register (AX, BX, CX, DX, SI, DI, BP or SP) but you can't use AX, CX or DX as addressing registers, so it only makes sense to load BX, SI, DI and BP for use as pointers. The two strings will now be addressed by DS:SI and ES:DI. DS is SI's normal segment, so we don't need to do anything, but we need a segment override for ES:DI. Here is the code for a C subroutine: A C string ends with a 0 byte, that is with a byte having the numeric value 0. It can be any length, but we need to test each byte to find out if it is 0. Notice that we are using (and changing) both DS and ES this time, so we have to PUSH and POP them, just like other registers. ; - - - - - move_string proc near FROM_POINTER EQU [bp+4] TO_POINTER EQU [bp+8] push bp mov bp, sp PUSHREGS ds, es, ax, si, di lds si, FROM_POINTER les di, TO_POINTER move_loop mov al, [si] ; source to al mov es:[di], al ; al to destination inc si ; pointers to next byte inc di and al, al ; is al 0? jnz move_loop POPREGS ds, es, ax, si, di pop bp ret ; calling routine pops variables move_string endp ; - - - - - Basically, the only difference between this and the Pascal move is that (1) here we check for 0 and there we had an actual count, and (2) in Pascal we used "ret (4)" and here the calling routine does the adjustment. Chapter 15 - Subroutines 159 ________________________ DRAWING THE STACK Each time that we have used the stack we have drawn a picture of where everything is on the stack. In case you think that this is some trivial little learning technique, I'm telling you now that at the assembler level, if you are passing variables on the stack and you don't make a diagram on a piece of paper of where everything is, you are guaranteed to consistently reference things by the wrong address. ALWAYS make a paper diagram that includes BP, ALWAYS use EQU statements, and you'll avoid a lot of mistakes. It is now time to get more complex. Everything that follows is more advanced, so it requires some programming experience. Everything that follows is about C modules and recursion. If this gets too complicated or obscure, just skip to the summary at the end of the chapter. I am going to give you a sample subroutine in C and then show you where all the variables go. The following is a complete C file: /* a complete C file - - - - - - - - - - - */ int A, B ; static int C, D ; extern int E, F ; sample_routine ( G, H ) int G, *H ; { int I, J ; static int K, L ; A = I ; /* transfer the words around */ B = G ; J = E ; F = C ; D = K ; *H = I ; return ; } /* end of C file - - - - - - - - - - - - - */ The only thing this routine does is tranfer the words around. This is so you can see where things are stored and how they are accessed. If you don't know C, the only thing you really need to know here is that '*H' means that 'H' is the ADDRESS of an integer, not the integer itself, while '*H' is the actual integer that is being addressed. For those of you who DO know C, you need to know exactly what extern, static and static mean. extern means that it is in an The PC Assembler Tutor 160 ______________________ external file. The first 'static' (which is outside of any subroutine) means that the data is INTERNAL to the file; it won't be shared with other files. Variables A and B, which don't have the word 'static' are GLOBAL and will be shared with other files. The other 'static' (inside the subroutine) means that its address is fixed in memory while I and J, which are not 'static' have their addresses generated every time you call the program. Say what? Yes, that's correct, every time you call the program they can be in a different place. That's what allows recursion, and you'll see how that is implemented in a second. Here is the equivalent program in assembler: ; - - - - - START DATA BELOW THIS LINE PUBLIC A, B EXTRN E:WORD, F:WORD A dw ? B dw ? C dw ? D dw ? K dw ? L dw ? ; - - - - - END DATA ABOVE THIS LINE ; - - - - - START SUBROUTINE BELOW THIS LINE sample_routine proc near ADDRESS_OF_H EQU [bp + 6] G EQU [bp + 4] I EQU [bp - 2] J EQU [bp - 4] push bp mov bp, sp ; set up bp sub sp, 4 ; save space for I and J PUSHREGS ax, si mov ax, I ; A = I mov A, ax mov ax, G ; B = G mov B, ax mov ax, E ; J = E mov J, ax mov ax, C ; F = C mov F, ax mov ax, K ; D = K mov D, ax mov ax, I ; *H = I mov si, ADDRESS_OF_H mov [si], ax Chapter 15 - Subroutines 161 ________________________ POPREGS ax, si mov sp, bp ; readjust sp pop bp ret ; a C routine so calling routine pops sample_routine endp ; - - - - - END SUBROUTINE ABOVE THIS LINE This time the setup is a little longer. It is: push bp mov bp, sp sub sp, 4 PUSHREGS ax, si I and J are not in the data segment, so we need to make space for them somewhere, and we do it on the stack. We subtract 4 from sp to provide ourselves with 4 bytes, two for I and two for J. There are two EQU statements that say exactly where I and J will be. We also push AX and SI because we will use them. After this setup, we have: address of H bp + 6 G bp + 4 old IP bp + 2 bp -> old bp bp + 0 I bp - 2 J bp - 4 old ax bp - 6 sp -> old si bp - 8 We have created space for some variables below bp. This is temporary and will disappear when we leave the subroutine. We do our dummy calculations,{1} and then do the end adjustment. There are two ways to readjust sp before returning: add sp, 4 just adds the amount that we subtracted, so it winds up in the same place. But the place it winds up is ALWAYS the address of the old BP, and bp is now pointing to that, so mov sp, bp does exactly the same thing. ARE YOU REALLY DEMENTED? Yes, for those of you who are truly masochistic, we have - ta-dah! - the Towers of Hanoi in assembler language. ____________________ 1. And that's 'dummy' in more ways than one. The PC Assembler Tutor 162 ______________________ The Towers of Hanoi is a game with three posts and a number of disks which have incrementally smaller diameters. At the beginning of the game, all the disks are on post one, ordered by size with the smallest on top and the largest on the bottom. For five disks it looks like this: (1) (2) (3) X X X X X X XXXXX X X XXXXXXX X X XXXXXXXXX X X XXXXXXXXXXX X X XXXXXXXXXXXXX X X XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX The object is to wind up with all the disks on post three in the same order. The game has only one rule: You may never put a larger disk on top of a smaller disk. The general solution to this problem is that if you have posts A, B and C with N disks on post A and you want to move them to post C, first move (N-1) disks to post B, move the bottom disk from post A to post C, then move (N-1) disks from post B to post C. This works out to be the optimal recursive solution. Try it out with N = 1, N = 2 and N = 3. N = 4 is already complicated. I won't discuss the game further or how to get the solution. If you don't know about it, you can look it up in either Doug Cooper's "Oh! Pascal" or Robert Kruse's "Data Structures and Program Design". The fact that I need to resort to such an unususl example to illustrate recursion underlines the fundamental rule of recursion, which is: YOU SELDOM NEED TO USE RECURSION, BUT WHEN YOU NEED IT YOU REALLY REALLY NEED IT. First, here is the solution in C: /* the C solution - - - - - - - - - - - - - - - */ #define MAXIMUM 9 main () { int count ; while (1) { printf ("Enter a number less than 10.\n" ) ; scanf ( "%d", &count ) ; if ( count > MAXIMUM ) continue ; Chapter 15 - Subroutines 163 ________________________ towers_of_hanoi ( count, 1, 3, 2 ) ; } } /* - - - - - - - - - - - - - - - - - - - - - - - - - */ towers_of_hanoi ( count, from, to, via ) int count, from, to, via ; { if (count <= 0) return ; count-- ; towers_of_hanoi ( count, from, via, to ) ; printf ("Move a disk from %1d to %1d.\n" , from, to ) ; towers_of_hanoi ( count, via, to, from ) ; return ; } Notice that by letting a routine call itself, we have reduced it to just a few lines. And this is a problem that looks very complex. Here comes the assembler equivalent of the C code: ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE MAX_COUNT EQU 9 enter_message db "Enter a number less than 10", 0 make_a_move_message db "Move a disk from " from_byte db "X to " to_byte db "X.", 0 ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE outer_loop: lea ax, enter_message ; count message call print_string call get_unsigned_byte ; count in al sub ah, ah ; zero ah for 'push ax' cmp al, MAX_COUNT ; too big? ja outer_loop ; move from post 1 to post 3 via post 2 mov bx, 2 ; post 2 = 'via' push bx mov bx, 3 ; post 3 = 'to' push bx mov bx, 1 ; post 1 = 'from' push bx push ax ; al = count, ah = 0 call towers_of_hanoi add sp, 8 ; adjust the stack jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE ; + + + + + + + + + + + + START SUBROUTINES BELOW THIS LINE The PC Assembler Tutor 164 ______________________ towers_of_hanoi proc near VIA EQU [bp + 10] TO EQU [bp + 8] FROM EQU [bp + 6] COUNT EQU [bp + 4] push bp ; set up bp mov bp, sp push ax cmp BYTE PTR COUNT, 0 ; if no disks, we are done jbe exit dec BYTE PTR COUNT ; 1 less disk to move ; first half push TO push VIA push FROM push COUNT call towers_of_hanoi add sp, 8 ; adjust the stack ; print the message mov al, FROM ; get 'from' number add al, '0' ; convert to ascii mov from_byte, al ; put into message mov al, TO ; get 'to' number add al, '0' ; convert to ascii mov to_byte, al ; put into message lea ax, make_a_move_message call print_string ; second half push FROM push TO push VIA push COUNT call towers_of_hanoi add sp, 8 ; adjust the stack exit: pop ax pop bp ret towers_of_hanoi endp ; + + + + + + + + + + + + END SUBROUTINES ABOVE THIS LINE The main routine checks that the number you enter is not too big and then calls 'towers_of_hanoi'. After setting up BP, the stack looks like this: VIA post bp + 10 TO post bp + 8 Chapter 15 - Subroutines 165 ________________________ FROM post bp + 6 count bp + 4 old IP bp + 2 bp -> old bp bp + 0 We make some EQU statements to define where each variable is and set up BP. We are using only one register, so we have a single PUSH instead of using PUSHREGS. We then check to see if the count is 0. If it is we are done. If not, we decrement the count by 1 which gives us 'count - 1' and divide the problem into three parts. Part 1 calls 'towers_of_hanoi' and moves 'count - 1' disks from 'from' to 'via'. Part 2 prints a message of where to move the bottom disk. It converts the post numbers into ascii and inserts them in the string where the 'X's are.{2} Part 3 calls towers_of_hanoi again, this time moving the 'count - 1' disks from 'via' to 'to'. Assemble and link it with ASMHELP. When you run it, start with just 1 disk, then 2 then 3 etc. If you use the maximum number (9), it will print 511 lines. For N disks, you need (2 ** N) - 1 moves, so this gets very big very fast. If you still feel no need to draw a picture of the stack or to use EQU statements, why don't you try writing this subroutine using the actual pointer values, i.e: push [bp + 6] and so on. See how long it takes and see how easy it is to read once it's done. Can you get it to work correctly? By the way, how large does the stack get? Well, if you raised the limit to 30 disks and entered 30, It would take your computer about a year, running 24 hours a day, to complete the solution (about 1 billion moves). The maximum stack size would be ((disks+1) * stack_use_per_disk). The extra 1 is for the calling routine. That is 31 * 14 bytes (including pushing IP, BP, and AX), or a mere 434 bytes. ____________________ 2 It uses the fact that for a data declaration, a variable name has the address of the first piece of data on that line. The PC Assembler Tutor 166 ______________________ SUMMARY To call a procedure you use CALL with the procedure name: call subroutine1 Procedures may be either FAR or NEAR. If there is a mismatch between which type the assembler thinks it is and which type it really is, there will be an error, either at the assembler level for internal subroutines or at the linker level for external subroutines. You may override the default type for procedures with PTR: call NEAR PTR subroutine5 call FAR PTR subroutine6 This is normally needed if the procedure comes after the call in the file and is not the default type. To allow other files to use a procedure you declare it PUBLIC within its own segment. PUBLIC subroutine1 To use a PUBLIC procedure from another file you declare it EXTRN, stating which type it is: EXTRN subroutine1:NEAR, subroutine2:FAR You should make this declaration in the code segment and you should make it before the procedure is referenced. A procedure is defined by giving a name followed by the word 'proc' (procedure) followed by either FAR or NEAR subroutine1 proc near subroutine2 proc far and a procedure is ended by giving the procedure name followed by 'endp' (end of procedure): subroutine2 endp One procedure must be ended before another is begun. Data is normally passed from one procedure to another on the stack: push variable1 push variable2 push variable3 call subroutine1 If this is done, then the called procedure references this data by using BP, the base pointer. The standard setup code is: Chapter 15 - Subroutines 167 ________________________ push bp ; save old bp mov bp, sp ; set bp to current top of stack What the stack looks like at this point depends on whether it is a near or a far procedure. For a near procedure, we have: variable1 bp + 8 variable2 bp + 6 variable3 bp + 4 old IP bp + 2 bp -> old BP bp + 0 For a far procedure we have: variable1 bp + 10 variable2 bp + 8 variable3 bp + 6 old CS bp + 4 old IP bp + 2 bp -> old BP bp + 0 Although it is theoretically possible to access these variables by their pointer definition: mov ax, [bp + 10] It is much less error prone and much clearer to use EQU statements: VAR1 EQU [bp + 10] mov ax, VAR1 If you are writing a recursive procedure and you need temporary variables, you can allot space on the stack for these variables: sub sp, 6 ; room for 6 bytes of temp. variables This should be done before any other pushes are done: push bp mov bp, sp sub sp, 6 PUSHREGS ax, bx, cx, dx These variables should also be named with EQU statements, and as always, you should draw a picture of what is on the stack: variable1 bp + 10 variable2 bp + 8 variable3 bp + 6 old CS bp + 4 old IP bp + 2 bp -> old BP bp + 0 VAR4 bp - 2 The PC Assembler Tutor 168 ______________________ VAR5 bp - 4 VAR6 bp - 6 VAR4 EQU [bp - 2] VAR5 EQU [bp - 4] VAR6 EQU [bp - 6] Data which is passed to the procedure is at a positive offset to BP while data that is created in the procedure is at a negative offset to BP. If you have created a data area for yourself on the stack, then you must eliminate it before leaving the procedure. There are two ways of doing this. One way is to add back what you have subtracted: POPREGS ax, bx, cx, dx add sp, 6 pop bp ret The other way is to give SP the value in BP because this is the place where SP will wind up anyway: POPREGS ax, bx, cx, dx mov sp, bp pop bp ret Use whichever one is clearer to you. When you return from a procedure that has had data passed to it, the data must be taken off the stack. There are two ways of doing this. The C standard is that it is the calling program's responsibility to do this: push variable1 push variable2 push variable3 call subroutine1 add sp, 6 ; 3 pushes = 6 bytes The Pascal standard is that you take them off the stack on the return. There is a special return instruction for that: ret (6) ; 3 pushes = 6 bytes DATA Data can be made available to other files and data can be accessed from other files. To make data available, declare it PUBLIC: PUBLIC variable1, variable2, variable3 Chapter 15 - Subroutines 169 ________________________ To access PUBLIC data from other files, use an EXTRN statement which includes the data type: EXTRN variable7:BYTE, variable8:WORD, variable9:DWORD EXTRN variable10:QWORD, variable11:TBYTE This EXTRN statement must be in a segment which has the same ASSUME segment register as will be used when accessing the data. Normally this is DS, but it can be something else. For instance, if the above EXTRN statements were in MORESTUFF and you have: ASSUME es:MORESTUFF then every time you access variable8: mov dx, variable8 the assembler will code an ES segment override. PUSHREGS and POPREGS When writing a subroutine, you should always save any registers that you use by pushing them. push ax push bx push cx They are then popped before returning pop cx pop bx pop ax In order to save a lot of lines of code, there are two macros, PUSHREGS and POPREGS. They are designed so you may use a word processor to copy them. PUSHREGS pushes in left to right order and POPREGS pops in right to left order: PUSHREGS ax, bx, cx, dx POPREGS ax, bx, cx, dx is a matched pair. LES and LDS LDS (load DS) loads the first two bytes into the named register and the next two bytes into DS. LES (load ES) loads the first two bytes into the named register and the next two bytes into ES. les si, [bp+6] lds di, [bp+10] Chapter 16 - Long Signed Multiplication and Division ==================================================== 170 Now that you have some subroutines under your belt, it is time to get back to multiple word arithmetic. This was put on the back burner because we needed to negate long numbers and it is more efficient to do that as a subroutine. First, let's negate a long number and then put the parts together. To negate a number you complement it, then add 1. It looks like this: NUMBER_LENGTH EQU 4 variable1 dq ? mov si, offset variable1 mov cx, NUMBER_LENGTH not_loop: not WORD PTR [si] add si, 2 loop not_loop mov si, offset variable1 mov cx, NUMBER_LENGTH stc ; set carry flag add_loop: adc WORD PTR [si], 0 inc si inc si loop add_loop This is straightforward. First negate, then add 1. The first add will add 1 because the carry flag is set. If there is a carry out, it will be taken care of in the next word with ADC (we add nothing but the carry). We can make this more compact and efficient with: mov si, offset variable1 mov cx, NUMBER_LENGTH stc ; set carry flag negate_loop: not WORD PTR [si] adc WORD PTR [si], 0 inc si inc si loop negate_loop Neither NOT nor INC effect CF, the carry flag, so the correct CF value will be propagated through the whole long number. When we do negation during our multiplication, the multiplicand will be a 4 word negation while the result will be a 5 word ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 16 - Multiple Word Arithmetic III 171 _________________________________________ negation, so we will pass the length as a parameter. The call in C would look like this: negate_it ( &number, length ) ; {1} On entry, the stack will look like this: length bp + 6 address bp + 4 old IP bp + 2 bp -> old BP bp + 0 Here is the entire subroutine: ; - - - - - START SUBROUTINES BELOW THIS LINE negate_it proc near NUMBER_LENGTH EQU [bp+6] NUMBER_ADDRESS EQU [bp+4] push bp mov bp, sp PUSHREGS cx, si mov si, NUMBER_ADDRESS mov cx, NUMBER_LENGTH stc ; set carry flag negate_loop: not WORD PTR [si] adc WORD PTR [si], 0 inc si inc si loop negate_loop POPREGS cx, si pop bp ret ; calling routine adjusts stack negate_it endp ; - - - - - END SUBROUTINES ABOVE THIS LINE Well, so far we have the negation routine and the unsigned multiplication and division routines. What else is necessary? Only the main program, and here it is for multiplication: ; - - - - - START DATA BELOW THIS LINE multiplicand dq ? multiplier dw ? result dt ? result_sign_flag db ? ; - - - - - END DATA ABOVE THIS LINE ____________________ 1. For you non-C people, the '&' stands for the address. The PC Assembler Tutor 172 ______________________ ; - - - - - START CODE BELOW THIS LINE outer_loop: mov result_sign_flag, 0 ; assume positive mov ax, offset multiplicand call get_signed_8byte test WORD PTR multiplicand + 6, 8000h ; is it negative ? jz get_next_number mov ax,4 ; negate 4 word number push ax mov ax, offset multiplicand push ax call negate_it add sp, 4 ; clear 2 pushes off stack not result_sign_flag ; reverse sign of result get_next_number: call get_signed ; get signed multiplier mov multiplier, ax test ax, 8000h ; is it negative jz do_the_multiplication neg multiplier ; negate not result_sign_flag ; reverse sign of result do_the_multiplication: mov ax, offset result push ax mov ax, multiplier ; the number, not the address push ax mov ax, offset multiplicand push ax call multiply_it add sp, 6 ; clear 3 pushes off stack ; is the result negative? test result_sign_flag, 0FFh ; 1111 1111 mask jz print_it mov ax, 5 ; 5 word result push ax mov ax, offset result push ax call negate_it add sp, 4 ; clear 2 pushes off stack print_it: mov ax, WORD PTR result + 8 ; top two bytes call print_hex mov ax, offset result ; the rest of result call print_signed_8byte jmp outer_loop ; - - - - - END CODE ABOVE THIS LINE The driver routine gets an 8 byte signed number. If the number is Chapter 16 - Multiple Word Arithmetic III 173 _________________________________________ negative it negates the number (to make it positive) and switches the sign of the result_sign_flag. The sign flag will either be 00h for positive or FFh for negative. It then gets a two byte signed number. If it is negative, the routine negates it and switches the sign flag. At this point both numbers are positive, so it calls the unsigned multiplication routine. At the end, it checks the result_sign_flag to see if the result should be positive or negative. If it should be negative, the routine calls negate_it one more time. Finally, the routine prints the number. The hex portion will be 0000 for positive or FFFF for negative unless the value is larger than an 8 byte signed number can hold, at which point the value of the 8 byte signed number will be incorrect. Here's the unsigned multiplication routine which has been turned into a subroutine: ; - - - - - multiply_it proc near RESULT_ADDRESS EQU [bp+8] MULTIPLIER_VALUE EQU [bp+6] MULTIPLICAND_ADDRESS EQU [bp+4] push bp mov bp, sp PUSHREGS ax, bx, cx, dx, si, di mov si, MULTIPLICAND_ADDRESS ; load pointers mov bx, RESULT_ADDRESS mov cx, 4 ; number of words sub di,di ; clear di mult_loop: mov ax, [si] ; multiplicand to ax mul WORD PTR MULTIPLIER_VALUE add ax, di ; add high word from last multiplication jnc store_result inc dx store_result: mov [bx], ax ; store 1 word of result. mov di, dx ; save high word for next multiplication add si, 2 ; increment pointers add bx, 2 loop mult_loop mov [bx], di ; move last word of result POPREGS ax, bx, cx, dx, si, di pop bp ret ; calling routine adjusts stack multiply_it endp ; - - - - - - - - - - - ______________________ correct. The multiplication and the negation subroutines go in the subroutine section of SUBTEMP1.ASM. The driver routine is the main routine. If you don't remember how this multiplication routine works, go back to the chapter on unsigned multiple word multiplication since the code is the same. DIVISION Division is the same situation. We need a driver routine, but the division itself will be the unsigned division. In division, the remainder is the same sign as the dividend, and the sign of the quotient is (dividend_sign XOR divisor_sign). If both signs are the same, the quotient is positive; if the signs are different the quotient is negative. Here's the driver routine: ; - - - - - START DATA BELOW THIS LINE dividend dq ? divisor dw ? quotient dq ? remainder dw ? quotient_sign_flag db ? remainder_sign_flag db ? ; - - - - - END DATA ABOVE THIS LINE ; - - - - - START CODE BELOW THIS LINE outer_loop: mov quotient_sign_flag, 0 ; assume positive mov remainder_sign_flag, 0 mov ax, offset dividend call get_signed_8byte test WORD PTR (dividend + 6), 8000h ; is it negative? jz get_next_number mov ax,4 ; negate 4 word number push ax mov ax, offset dividend push ax call negate_it add sp, 4 ; adjust stack not quotient_sign_flag ; switch sign of quotient mov remainder_sign_flag, 0FFh ; remainder is negative get_next_number: call get_signed mov divisor, ax test ax, 8000h ; is it negative jz do_the_division neg divisor not quotient_sign_flag ; switch sign of quotient do_the_division: mov ax, offset remainder push ax Chapter 16 - Multiple Word Arithmetic III 175 _________________________________________ mov ax, offset quotient push ax mov ax, divisor ; the number, not the address push ax mov ax, offset dividend push ax call divide_it add sp, 8 ; clear 4 pushes off stack ; are the remainder and quotient negative? test remainder_sign_flag, 0FFh jz test_the_quotient neg remainder test_the_quotient: test quotient_sign_flag, 0FFh ; 1111 1111 mask jz print_it mov ax, 4 ; 4 word result push ax mov ax, offset quotient push ax call negate_it add sp, 4 ; clear 2 pushes off stack print_it: mov ax, offset quotient call print_signed_8byte mov ax, remainder call print_signed jmp outer_loop ; - - - - - END CODE ABOVE THIS LINE We get the dividend and check the sign. If it is negative, we (1) negate the number, (2) switch the sign of the quotient, and (3) set the remainder sign flag to negative. We get the divisor, check for negative; if it is negative we negate it and switch the sign of the quotient. We now have two unsigned numbers and do unsigned division. After division, both the quotient and remainder are adjusted for sign. The division routine is the same as the unsigned routine before except it is now a subroutine: ; - - - - - - - - ENTER SUBROUTINE BELOW THIS LINE divide_it proc near REMAINDER_ADDRESS EQU [bp+10] QUOTIENT_ADDRESS EQU [bp+8] DIVISOR_VALUE EQU [bp+6] DIVIDEND_ADDRESS EQU [bp+4] push bp mov bp, sp PUSHREGS ax, bx, cx, dx, si, di The PC Assembler Tutor 176 ______________________ mov si, DIVIDEND_ADDRESS mov bx, QUOTIENT_ADDRESS add si, 6 ; start at the top word add bx, 6 mov di, WORD PTR DIVISOR_VALUE mov cx, 4 ; number of words sub dx, dx ; clear dx for first division division_loop: mov ax, [si] ; dividend word to ax div di mov [bx], ax ; word of result to quotient sub si, 2 ; decrement the pointers sub bx, 2 loop division_loop mov bx, REMAINDER_ADDRESS ; store remainder mov [bx], dx POPREGS ax, bx, cx, dx, si, di pop bp ret ; calling routine adjusts the stack divide_it endp ; - - - - - - - - ENTER SUBROUTINE ABOVE THIS LINE Draw a picture of the stack to verify that the EQU statements are correct for a NEAR routine. The division and negation subroutines go in the SUBROUTINES section of SUBTEMP1.ASM. The driver is the main program. If you don't remember how this division works, go back to the division chapter and look it over. Try out a few numbers to make sure that it is working the way it should. DATA INTEGRITY One thing that may have been annoying some of you is that when the programs sent us numbers for multiplication and division we sometimes negated them, effectively changing the data in memory, but never changed them back when we were done. In an operational subroutine, you would have to do it differently. The logic would be: NUMBER NEGATIVE? no everything's o.k. yes make copy negate reset pointer If the number is positive we won't change it. If the number is negative, we make a copy, negate the copy and use the copy for the operation. Chapter 17 - Iinterrupts ======================== 177 Your word processor will work on a Compaq, an IBM, an AST or any other type of PC compatible computer. Every time it wants to read a file from disk or write to a printer, it calls a DOS or BIOS function that takes care of it.{1} Yet every computer has its own versions of these subroutines, and they not only are different, they are in different places in memory. How does the word processor know where to find them? It uses interrupts.{2} An interrupt is a glorified subprogram call. The first 1024 bytes of your computer's memory (that's from 0000:0000 to 0000:03FF) contain the addresses of all the DOS and BIOS interrupts. Which address contains the address of which subprogram was decided by the triumverate of Intel/IBM/Microsoft. We have two different sets of addresses here, so let's keep them straight. Starting at memory address 0000 there are 4 byte addresses called interrupt vectors. There is a different vector at 0000d, at 0004d, at 0008d, at 0012d at 0016d etc.; there are 256 of them in all. Each of these 256 places can hold the address of a subprogram somewhere in memory (although not all of them are used). When you call an interrupt, the 8086 goes to the appropriate place in low memory (from 0000 to 3FFh) and finds the address of the subprogram that you want, loads it into CS and IP, and goes to that subprogram. When that subprogram is done, it goes back to the next instruction in your program after the interrupt. How did those addresses get into the first 1024 bytes? The computer put them there when you started it up. It is one of the first things that the computer did when you turned it on. These subprograms do EVERYTHING, and you can't run the computer without them. If you want to scroll the screen, you put the appropriate information in the 8086 registers and use: int 10h ; decimal 16 {3} The 8086 goes to the interrupt 16 address (4*16 = 64), gets the address of the program that contains the video subprograms, and ____________________ 1. BIOS stands for basic input/output services. 2. If you haven't gotten it yet, it's time to get either "DOS Programmer's Reference" or "The Peter Norton Programmer's Guide to the IBM PC." I'm serious. They contain the information you need to make use of this chapter. 3. It is normal to use hex numbers for the interrupts, so if you read about an interrupt make sure you know if the numbers are hex or decimal. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 178 ______________________ goes to it. If you want to write to the printer, you put the appropriate information in the 8086 registers and call: int 21h ; decimal 33 The 8086 goes to address 132 (4*33 = 132), gets the address of the DOS program, puts it in CS and IP, and starts it. If you want to get input from the keyboard, you put the appropriate information in the 8086 registers and call: int 21h ; decimal 33 That's right, it is the same program as the one that does the printer. The 8086 goes to 132 (4*33), gets the program address, and goes to it. The lowest interrupt is int 0 (address = 4*0 = 0000). The highest interrupt is int 255 (address = 4*255 = 1020). This is an intelligent way to handle the situation. As long as everyone agrees which interrupt contains the address of which subprogram, our programs will work on any PC compatible. This is one of the things that is meant by PC compatible. On my computer, here is a section of these addresses starting with int 1Eh (30d). High memory is at the top, low memory is at the bottom. INT # DATA LOCATION 0724h cs 146 36 04A8h ip 144 cs 0724h 142 35 ip 01BDh 140 0724h cs 138 34 01B0h ip 136 cs 019Fh 134 33 ip 05EBh 132 019Fh cs 130 32 05E7h ip 128 cs 0000h 126 31 ip 0000h 124 0070h cs 122 30 0EB8h ip 120 If we call int 30d, the 8086 goes to 120 (4*30), puts 0EB8h into the IP, and puts 0070h (from the next higher location) into CS. The next instruction it does will be 0070:0EB8. If we call int 35d, the 8086 goes to 140 (4*35), puts 01BDh in IP, puts 0724h (from the next higher location) in CS. The next instruction the 8086 does will be 0724:01BD. If we call int 33d, the 8086 goes to 132 (4*33), and puts 05EBh in IP, 019Fh (from the next higher location) in CS. The next instruction executed will be 019F:05EBh. The 0000:0000 for int 31d indicates that there is no int 31d (address 0000:0000 contains data, [the vectors for int 0], not a program). Make sure that you understand how this is working before you go on. Chapter 17 - Interrupts 179 _______________________ On the PC, information for the interrupts is always passed through registers, not on the stack. Each interrupt type has a specific register for each piece of information it needs. We will do a couple to see how they work. DISPLAYING A CHARACTER We can print a character at a time on the monitor, so we'll input a string, and then print out each character of the string individually. We will stop the printout when we see the 0 at the end of the string. ; - - - - - - - - - - START DATA BELOW THIS LINE buffer db 80 dup (?) ; - - - - - - - - - - START DATA BELOW THIS LINE ; - - - - - - - - - - START CODE BELOW THIS LINE call show_regs outer_loop: mov ax, offset buffer call get_string mov si, offset buffer inner_loop: mov al, [si] cmp al, 0 je next_string mov ah, 14 ; ah contains function number mov bh, 0 ; where in memory int 10h ; 33d inc si jmp inner_loop next_string: call get_continue jmp outer_loop ; - - - - - - - - - - START CODE BELOW THIS LINE The program is simple. It gets a string, then checks each character for 0 (end of string), before shipping it off to the screen. I'll explain AH in a second. There are several places this character can be displayed{4}, so use show_regs to force the video screen to a certain place in memory, then put BH = 0 to tell the interrupt that the screen memory is in that place. Finally, with all the information in place, we do the interrupt. You will not print a carriage return, only the data, so we use get_continue to give us a carriage return. It's a little sloppy, but much easier. We have lots and lots of subprograms for the disks, printer, screen, etc. There are only 255 interrupts, so it was decided at the beginning to make most of the interrupts groups of programs ____________________ 4. Cf. one of those two books. The PC Assembler Tutor 180 ______________________ instead of a single program. Int 10h (16d) contains about two dozed different video subprograms. Each program is distinguished by a specific number in AH. For int 10h (16d): ah = 2h Get cursor position ah = 6h Scroll window up ah = Eh (14d) Write character to screen For int 21h (33d): ah = 1h Keyboard input ah = 5h Printer output ah = 17h Rename a file ah = 2Ch Get the time As you can see, int 21h is a potpourri of subprograms. We'll do the same program as above, but with printer output. Everything is the same except that the inner loop should be changed to look like this: ; - - - - - mov si, offset buffer inner_loop: mov dl, [si] cmp dl, 0 je next_string mov ah, 5 ; ah contains function number int 21h ; 33d inc si jmp inner_loop ; - - - - - There is practically no change. We use DL instead of AL, have "int 21h" instead of "int 10h", and change the function number in AH to 5. Also, BH is not needed. Leave your printer off at the beginning to see what happens. Turn your printer on and enter a string of 10 or 20 letters. Probably nothing happened. Enter another 20 or 30 letters. Nothing again. Try 50 letters this time. This time it should work. Lots of printers won't print anything unless (1) they get a carriage return (which you haven't sent) or (2) they have a backlog of more than 80 characters. If you ever use the printer interrupt, you'll have to remember that. These interrupts which are in your program are called software interrupts. They are your interface with the peripheral devices on your machine, doing everything from disk i/o to handling memory allocation. They do all your housekeeping for you.{5} If you are going to work at this level then you should buy one of those two books. They contain all the software interrupts (and there are about a hundred of them), tell how they work and which registers to set for each interrupt. If you don't have access to ____________________ 5. But they don't do Windows. Chapter 17 - Interrupts 181 _______________________ these interrupts it's like having your arms cut off - you're unable to do any i/o at all. HARDWARE INTERRUPTS The interrupts that we write in our programs aren't the only interrupts there are. The hardware uses interrupts to take temporary control of the computer. On a Macintosh, if you insert a disk, the program stops and the operating system reads in the disk directory. A modem that is in use will request time for doing i/o. Also, if the 8086 detects a zero divide, it will trigger a special interrupt. You have met int 4 already. It is the INTO instruction, which will trigger an interrupt if the overflow flag is set. Your interrupts are in specific places in your code, but these machine interrupts can happen at any time. There are two lines (wires) into the 8086. One is for serious problems that need to be taken care of NOW, and it has non-maskable interrupts. The other line is for interrupts that need to be taken care of in a timely fashion, and they are maskable interrupts. A non-maskable interrupt (NMI) is when the hardware detects that it is in deep doo doo. It sends a signal on the NMI line that says "Hey! I need an interrupt." The 8086 finishes the instruction it is processing and then IMMEDIATELY gives over control. The NMI uses the same 1024 bytes in low memory for interrupt vectors, but has its own interrupt numbers. Normally this is for very serious errors, so the interrupt program may decide to abort your program and return to the operating system; if it makes sense to, it will return to your program where your program left off. A maskable interrupt is when a piece of hardware has some work to do. It sends a signal to the 8086 (on the INTR line), and the 8086 takes care of it when it is ready. When the 8086 is ready depends on you. In the flags register is the IEF, the interrupt enable flag. It should always be set to 1 unless you are doing something critical. Basically, the only things that are critical are interrupts themselves and context switches. Context switches are done by the operating system in multitasking environments, so they don't concern you, and you are not writing interrupts, so they don't concern you. Therefore, always keep the IEF set. Just for your information, you set the IEF with: sti ; set interrupt flag (interrupts enabled) and clear it with: cli ; clear interrupt flag (interrupts disabled) Why do interrupt programs clear the interrupt flag? Because you could have interrupts interrupting other interrupts and wind up with scads of half finished interrupts lying around. This way, The PC Assembler Tutor 182 ______________________ one interrupt finishes before another can take over. Suppose you are in the middle of the following two instructions when an interrupt hits: cmp al, 7 jne some_label If the hardware interrupt takes over after the CMP instruction but before the JNE instruction, the correctness of the result will depend on the zero flag staying the same. Can you trust it? The answer is yes. The first three things an interrupt request does (in the microcode) are: push flags ; push the flags push old CS ; push the code segment push old IP ; push the instruction pointer On return, it pops the flags back in place, so they are exactly the same as just before the interrupt. You will notice that there are 3 things on the stack instead of the usual 2 in a far call. Therefore, there is a special RET instruction called IRET (return from interrupt) which pops not only IP and CS, but the flags as well. You can only use this instruction in an interrupt because it assumes that the flags register is right after CS on the stack. DEBUGGING Finally, if you have a debugger installed, the debugger uses two interrupts, int 1 and int 3. You code int 3 into the program by placing: int in the program with no number.{6} The interrupt will call the debugger. The reason for this int with no number is that afterwards, you can replace it with NOP, since they are both 1 byte instructions. Int 3 is a 2 byte instruction (they are coded differently, even though they wind up at the same place). Int 1 is the trap instruction. In the flags register is the trap flag (TF). If TF is set, the 8086 will interrupt after the next instruction. That way, the debugger can single step through sections of code. You put Int 3 to set a breakpoint and then set TF to single step from there. The only way to set TF from your own program is: ____________________ 6. Though the Microsoft assembler considers this an error. You must code it as: int 3 Chapter 17 - Interrupts 183 _______________________ pushf pop ax or ax, 0100h push ax popf but you should let the debugger do it. Just so you can get a feel for how the debugger system works, there is a pseudo-debugger called PSEUDDBG.COM in \XTRAFILE. It is not a debugger. It does nothing but tell you whether you have generated a breakpoint interrupt (int 3) or a single_step interrupt (int 1). It is memory resident. That means that once you have loaded it, it will stay in memory until you shut the machine off or reset the machine. You load it by executing: >pseuddbg It will tell you that it has been loaded and then sit there waiting for interrupts. When you generate a breakpoint interrupt, PSEUDDBG will allow you to set the trap flag or just continue. When you generate a single step interrupt, PSEUDDBG will allow you to clear TF or to continue single stepping. All you need to try it out is a simple program: ; - - - - - ENTER CODE BELOW THIS LINE outer_loop: call get_continue mov cx, 4 int 3 inner_loop: mov ax, 5 add ax, 3 add ax, 7 add ax, 10 loop inner_loop loop outer_loop ; - - - - - FINISH CODE ABOVE THIS LINE Each time you start the outer loop, it will generate a breakpoint interrupt. At this point you can set TF to single step or continue. If you single step, watch IP. It will be changing, cycling through the loop till it gets to get_continue. If you start trapping through get_continue, things will get strange because get_continue stores the flags. This means that after you quit trapping, get_continue will POP some flags that have TF set and it will start trapping again. If this happens, just continue hitting 0 until you get out of the single step mode. If you like this method of single stepping through code, there is a memory resident version of ASMHELP called HELPMEM.COM which contains show_regs and allows single stepping. It can be used in conjunction with ASMHELP.OBJ. For details consult APP1.DOC in the appendix. The PC Assembler Tutor 184 ______________________ SUMMARY Software interrupts send the 8086 to the first 1024 bytes of memory where it finds the address of the DOS or BIOS subprogram that you want to go to. The correspondence between the interrupt number and the location of the address is: address location = 4 * interrupt number The address is stored in the next 4 bytes; first IP, then CS. The first 5 interrupts are: int 0 divide by zero int 1 trap flag induced single stepping for the debugger int 2 non_maskable_interrupt (NMI) int 3 1 byte interrupt for the debugger int 4 interrupt on overflow There are two types of hardware interrupts. NMI (non maskable interrupt) interrupts are for serious problems that need to be taken care of IMMEDIATELY. They have priority and take over right after the current instruction is finished. Maskable interrupts are hardware interrupts that should be taken care of in a timely fashion. If IEF (the interrupt enable flag) is set the maskable interrupts will go through. If IEF is cleared, the maskable interrupts must wait until it is set again. Set the IEF with: sti ; set interrupt flag and clear it with: cli ; clear interrupt flag The IEF should always stay set unless there is a specific reason for not allowing interrupts to happen. If you are writing an interrupt routine, then instead of RET you use IRET (return from interrupt) which not only pops off IP and CS, but pops the flags register as well. Chapter 18 - Ports ================== 185 In order to communicate with the outside world, the outside world being your printer, your disk drives, your modem etc., the 8086 uses ports. A port is an address distinct from a memory address where the i/o device can be reached.{1} When IBM set up the first PC, they decided what these port addresses would be and what the function of each bit at each address would be. Any VGA, EGA, CGA or monochrome card has to have it's ports at the same addresses as those set by IBM, and these ports have to do the same thing. Normally, an i/o device has more than one port address. The device not only has to transfer the data, it has to tell the computer whether it is ready, find out if the computer wants to send information, confirm that the transfer was successful etc. COM1 is port addresses 3F8 - 3FF. COM2 is 2F8 - 2FF. The CGA adapter is ports 3D0 - 3DF. In the CGA, port 3D9 sets the different colors. The 8086 writes to it, but can't read it. Port 3DA, it's neighbor, tells certain status information and is read only. All control information for the video cards is passed back and forth through the ports. However, the video card comes into memory 50 times a second to see what it should write to the screen. The characters on the screen don't pass through the ports. The mountain comes to Mohammed. The data does pass through a port on the way to your serial printer. The 8086 sends your printer control information through one port address and sends the data through a different port address. The port instructions can be either with AL or AX. AL and AX are the only registers you can use, AL for a byte transfer and AX for a word transfer. There are two forms to the input instruction: in al, port_address or in al, dx port_address is a constant and is limited to 0 - 255. It is a one byte constant. DX is the only register than can be used with the second form. DX can contain any number from 0 - 65535. The ____________________ 1. There is memory address 0000 and there is port address 0000; they are two entirely different things. You can reach memory address 0000 with the DS:SI pair 0000:0000, but DS:SI can't get to port address 0000. The instruction "out 0, al" writes the contents of al to port address 0000, but the OUT instruction can't get anywhere near memory address 0000. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 186 ______________________ following two pieces of code do the same thing: in al, 155 mov dx, 155 in al, dx This code moves one byte of data from port 155 to register AL. The output instruction is similar. We have: out port_address, al or out dx, al where port_address is a constant 0-255 and DX can contain a number 0 - 65535. out 97, ax moves a word from AX to port addresses 97-98. Notice that both the IN and the OUT instructions follow the DESTINATION, SOURCE ordering found in all the other 8086 instructions. The "in ax, port_address" form is not all that useful. If you look at the addresses for the video ports, for COM1 and COM2, you will see that they are all larger than 255. Also, most equipment can be at several different addresses. A modem can be at COM1, COM2, COM3, or COM4. If the port address is hard coded into the instructions, you need 4 subprograms to handle the 4 different addresses. It is easier to find out where the modem is, then use: modem_data_address dw ? mov dx, modem_data_address in al, dx and mov dx, modem_data_address out dx, al If you aren't planning on writing device drivers this is for background information only, since all standard i/o is done through DOS or BIOS interrupts. One last thing is the parity flag. It has been sitting on the screen with all the other flags. What is it for? When you send information over a modem, there is a large possibility of line interference. If it is text, and an occasional screw up is not too bad, using a parity check may be enough.{2} Parity can be even or odd. Even means that an even number of bits are set to 1; odd means that an odd number of bits are set to 1. This is not whether the number is even or odd, it is whether the number of 1 ____________________ 2. It is NEVER enough if you are transferring binary data or executable files. Chapter 18 - Ports 187 __________________ BITS is even or odd. Here's a short list. NUMBER BINARY PARITY 6 0110 even 7 0111 odd 8 1000 odd 9 1001 even 10 1010 even 8 is an even number but has odd parity, 9 is an odd number but has even parity. 6 has two 1 bits (even), 7 has three 1 bits (odd), 8 has one 1 bit (odd), 9 has two 1 bits (even) and 10 has two 1 bits (even). How does this help us? If there is a chance of 1/100 of screwing up a single bit in a byte, there is only a chance of 1/10,000 of screwing up two bits in a byte. What this means is that of every 100 errors, 99 of them will will be detectable because of a change in parity (by changing a bit from 0 to 1 or from 1 to 0) and only 1 of them will go undetected because two changes will keep the same parity. This is a little obscure, so make sure you understand why before continuing. This doesn't help us much yet, because we don't know what the parity was originally. 9 has even parity, 7 has odd parity. What communications programs do is FORCE the parity. They can either force it even or force it odd. The standard ASCII characters end at 127 - that is, 0111 1111 (7F) is the highest legal number you can transmit. The left bit is unused, so we can use it to force the parity. We are going to force it even, but forcing it odd uses the same technique. The steps are: (1) Find the parity of the byte. (2) If it is even, leave it alone, if it is odd, put a 1 in the left hand bit. The parity is now even. If both the sending and receiving program have agreed on even parity, then on the receiving end, the program: (1) Checks for even parity. If the parity is odd, it is an error. (2) Puts a 0 back in the left hand bit. The original number is restored. We are going to make a loop with both parts and use show_regs to watch the parity flag. Here's the program: ; - - - - - - - - - - ENTER DATA BELOW THIS LINE error_banner db "Whoa! We have a screw up.", 13, 10, 0 The PC Assembler Tutor 188 ______________________ ; - - - - - - - - - - ENTER DATA ABOVE THIS LINE ; - - - - - - - - - - ENTER CODE BELOW THIS LINE mov ax_byte, 0A3h ; al binary mov bx_byte, 0A2h ; unsigned half registers mov cx_byte, 0A2h ; unsigned half registers mov dx_byte, 0A3h ; dl binary lea ax, ax_byte call set_reg_style comm_loop: mov ax, 0 call set_count ; reset count to 0 mov bx, 0 ; clear registers mov cx, 0 mov dx, 0 call show_regs ; (1) ; the sending program call get_unsigned_byte and al, 7Fh ; 0111 1111 - make sure al < 128 mov bl, al ; copy to bl call show_regs_and_wait ; (2) and al, al ; check parity call show_regs_and_wait ; (3) jpe do_nothing or al, 80h ; if parity odd, set left bit do_nothing: and al, al ; check parity for show_regs call show_regs_and_wait ; (4) ; the receiving program mov dl, al ; transmit from al to dl mov cl, dl ; copy to cl call show_regs_and_wait ; (5) and dl, dl jpe zero_left_bit ; is parity even? mov ax, error_banner call print_string zero_left_bit: and dl, 7Fh ; 0111 1111 binary, left bit 0 mov cl, dl ; copy to cl call show_regs_and_wait ; (6) jmp comm_loop ; - - - - - - - - - - ENTER CODE ABOVE THIS LINE First, we set the register style so AL and DL are binary, BL and CL are unsigned. We'll use AL and BL for sending, CL and DL for receiving. Next, we reset the counter to 0 so we can see where we Chapter 18 - Ports 189 __________________ are, and zero AX, BX, CX, and DX. We get a byte and make sure that it is 127 or less,{3} then send a copy to BL. BL stays unchanged for the rest of the loop. We test the parity. and if it is odd, change it. Finally, we send it off to DL. On the receiving end, we test the parity. If it is odd, we print an error message. Then we zero the left hand bit. It is faster to zero the left hand bit than to check to see if it needs to be zeroed, so we do it for all data. The result is put in CL for comparison. AL and DL are shown in binary form so you can count the number of 1 bits in the byte. ____________________ 3. The following could happen if we didn't take this step. We get a number > 128 with odd parity. 131 (1000 0011 83h) is an example. The sending program sees that it is odd parity, so it puts a 1 in the left hand bit. But there is already a 1 in the left hand bit, so the parity doesn't change, it is odd. The receiving program gets the byte, tests it, notices that it is odd, and sends back a transmission error. Since the sending program didn't see anything wrong, it just sends the same byte again. It will never get through correctly. In real life, the sending program would probably test the byte to see if it was greater than 127 and print an error if it was. The PC Assembler Tutor 190 ______________________ SUMMARY POSSIBLE I/O INSTRUCTIONS in ax, constant (= port address 0 - 255) in ax, DX in al, constant (= port address 0 - 255) in al, DX out constant, ax (constant = port address 0 - 255) out DX, ax out constant, al (constant = port address 0 - 255) out DX, al In these instructions DX holds a port address and can be from 0 - 65535. PARITY Parity is whether the number of 1 bits in a byte (or word) is even or odd. The 8086 sets the parity flag after most arithmetic and all logical operations. If parity is even, the flag is 1, if parity is odd, the flag is 0. Chapter 19 - Strings ==================== 191 Sometimes we want to deal with long strings of information. Here long means hundreds or thousands of bytes, not tens of bytes. The 8086 provides a group of instructions to move and compare strings. These instructions have a rigid structure, but with a little bit of effort we can get them to work easily for us. We will start with SCAS, since it is simple, yet embodies all the rigid features of these instructions. SCAS (scan string) compares either a byte to AL or a word to AX. The byte or word must be in memory, and the register must be AL or AX. SCAS also increments or decrements the pointer. First, the size: scasb compares a byte to AL, while: scasw compares a word to AX. But where's the pointer? You have no choice, it's DI. Not only is it DI, but it MUST be ES:DI. The ES segment is coded into the 8086 microcode; the DI register is coded into the 8086 microcode; there is nothing you can do to change it. What about incrementing or decrementing? In the flags register, there is a flag called the direction flag. It is set manually by the program. If DF = 0, SCAS increments DI; if DF = 1, SCAS decrements DI.{1} The equivalent software for the instruction would be: (scasb) (scasw) DF = 0 cmp al, es:[di] cmp ax, es:[di] pushf pushf add di, 1 add di, 2 popf {2} popf DF = 1 cmp al, es:[di] cmp ax, es:[di] pushf pushf sub di, 1 sub di, 2 popf popf ____________________ 1. Every time you have called show_regs DF has been there; it doesn't show 0 and 1, it shows + and - (+ = 0 , - = 1). 2. The microcode doesn't really push and pop the flags. This is only to indicate that the order of operations is (1) get the byte (word) from the string, (2) compare and set the flags, and finally (3) increment (decrement) the pointer without changing any of the flags. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 192 ______________________ Thus, at the end of the end of the instruction, DI is in a new place and the flags are set according to the compare result. DI is incremented by a byte for the byte instructions; it is incremented by a word for the word instructions. The same pattern holds true for decrementing DI. We set the direction flag with the instruction STD (set direction flag) and we clear the direction flag by using CLD (clear direction flag). It only needs to be set or cleared once, and this should be done before starting the operation. DF is only changed by those specific instructions from the program - it can't be changed by any arithmetical or logical operation on the chip. If you have a string and you are looking for a specific number, (27 for instance), you simply put that number in AL (or AX) and run a loop. If: long_string db 5000 dup (?) contains data and we want to look for a 27d then the operation is: lea di, long_string{3} mov al, 27 cld search_loop: scasb jne search_loop on exiting, DI will point 1 PAST the matching byte (word). You move back one byte (word) to find the match. Why would anyone want this instruction? With a 0 in AL, it will find the end of a C (0d terminated) string quickly. Also, that number 27 is no accident. 27d is the ASCII escape character. For a lot of hardware, 27d indicates that the bytes that follow are not ASCII characters but are technical information. For instance, on my printer the sequence (27d, 65d, 0d, 11d) sets tabs every 11 columns. SCAS can find where these substrings are so the program can operate on them. In order to use string instructions, we need strings to work on. The one we will use is called CH1STR.OBJ. It is in \XTRAFILE. It is an object file that contains one data segment containing a string of lower case characters. The string is several thousand bytes long, it is terminated by 0, and it contains ONLY the lower case letters (a-z). It is the first draft of part of chapter 0 with all punctuation, numbers, spaces, carriage returns etc. deleted. All upper case letters have been converted to lower case so we don't have to worry about the difference between A and a, Q and q. The name of the array in CH1STR.OBJ is CH1STR and it is defined: ____________________ 3. Assuming that long_string's segment address is in ES. Chapter 19 - Strings 193 ____________________ PUBLIC ch1str so that you can access it with the SEG and OFFSET operators. First, let's find out how long the string is.{4} MYPROG1.ASM ; - - - - - - - - - - STRINGSTUFF SEGMENT PUBLIC 'DATA' EXTRN ch1str:BYTE STRINGSTUFF ENDS ; - - - - - - - - - - ;- - - - - - - - - - PUT CODE BELOW THIS LINE mov ax, seg ch1str ; segment address of ch1str mov es, ax mov di, offset ch1str ; offset address of ch1str mov al, 0 ; try to match zero cld ; increment (DF = 0) string_end_loop: scasb jne string_end_loop dec di ; back up one mov ax, di sub ax, offset ch1str call print_unsigned ;- - - - - - - - - - PUT CODE ABOVE THIS LINE In all these string operations, we need to be careful about boundary conditions. What if there is no valid data? What if there is one valid item? What if the string is empty? On exiting the loop, DI will point 1 past the first 0d, so we need to back up one to point to the first 0d. Then subtracting the starting position will give us the count.{5} Try it out and find out how long it is. Since we now have 3 object modules, the link instruction must read: link myprog+ch1str+asmhelp ; assuming that you name your program myprog.asm. Save the result because we will need to use this number several times. ____________________ 4. Just to keep it from being too easy, I have put garbage both in front of the string and behind the string. That means that the string length is shorter than the length of the object file and the string does not start with the first byte of the object file. 5. If the first byte in the string is 0d, we move one, then move back one which gives the length zero. The PC Assembler Tutor 194 ______________________ You will notice that we have gotten the segment address by using the SEG operator. You don't need to know the name of the segment. The segment doesn't even have to be PUBLIC. As long as the VARIABLE is either in the same file or is in another file and PUBLIC, the linker will find the correct segment address and put it there. To make things a little more complicated, we will make another infinite loop. This time you will enter a character, and the program will find the first occurance of that character. We need to add some error checking here. Since you will probably be dreaming about taking your next vacation in Hawaii while you are entering the data, a few characters that don't exist in the string (things like G $ ? ~ ) might creep in. It would be possible to run way past the end of the string before you found that character. We'll put the length of the string (from the last program) in CX, have a regular loop so we can't go too far, and jump out of the loop if we find a match. MYPROG2.ASM ; - - - - - - - - - - STRINGSTUFF SEGMENT PUBLIC 'DATA' EXTRN ch1str:BYTE STRINGSTUFF ENDS ; - - - - - - - - - - ;- - - - - - - - - - PUT CODE BELOW THIS LINE mov ax, seg ch1str mov es, ax outer_loop: call get_ascii_byte ; returns character in al mov cx, $$$$$$$ ; enter string length here mov di, offset ch1str cld ; increment (DF = 0) string_end_loop: scasb je after_loop ; if equal, we found the char loop string_end_loop mov ax, 0 ; we fell through the loop call print_unsigned jmp outer_loop after_loop: mov ax, di ; move for printing sub ax, offset ch1str ; number of bytes call print_unsigned jmp outer_loop ;- - - - - - - - - - PUT CODE ABOVE THIS LINE Chapter 19 - Strings 195 ____________________ Those dollar signs are the place to enter the exact length of the string that you got from the last program. This time we jump out of the loop if we find a match; DI will be 1 past the matching character, but this will give us the right count (if we find the character in the first space, we increment once). If we can't find a match we fall through the loop and print a 0. Remember to link all 3 modules when you run the program. Run the program and then we'll move forward. This type of thing is so common with string operations that there is a special prefix for SCAS and all other string operations which makes the coding simpler. It has several forms: rep decrement cx ; repeat if cx is not zero repe decrement cx ; repeat if cx not zero and zf = 1 repz decrement cx ; repeat if cx not zero and zf = 1 repne decrement cx ; repeat if cx not zero and zf = 0 repnz decrement cx ; repeat if cx not zero and zf = 0 REP is for the move instructions which we will see later - it won't work here. For each prefix, if either (or both) of the conditions is not true, the repitition stops. For instance, with REPE, if cx is zero, and/or if the comparison was not equal (so the zero flag was not set), the instruction will stop. For our program, the coding is: repne scasb That's it. That replaces the whole inner loop. Here is our new coding of the last program. MYPROG3.ASM ; - - - - - - - - - - STRINGSTUFF SEGMENT PUBLIC 'DATA' EXTRN ch1str:BYTE STRINGSTUFF ENDS ; - - - - - - - - - - ;- - - - - - - - - - PUT CODE BELOW THIS LINE mov ax, STRINGSTUFF mov es, ax cld ; increment (DF = 0) outer_loop: call get_ascii_byte ; returns character in al mov cx, $$$$$$$ ; enter string length here lea di, ch1str ; address of string repne scasb je found_the_char ; an equal comparison mov ax, 0 ; we didn't find a match call print_unsigned jmp outer_loop found_the_char: mov ax, di ; move for printing The PC Assembler Tutor 196 ______________________ sub ax, offset ch1str ; number of bytes call print_unsigned jmp outer_loop ;- - - - - - - - - - PUT CODE ABOVE THIS LINE There are two possibilities for exiting the 'repne scasb' instruction. Either we found an equal comparison or we exhausted all the characters in ch1str. If we found an equal comparison, JE will send us to the print routine. Otherwise we print a 0 because we finished the loop without finding anything. STOS We can ask the operating system to allocate memory for us while the program is running.{6} When you get it, however, it will contain trash. The fast way to clear it is to use STOS (store to string). The instruction is: stosb or: stosw The equivalent action (not counting changing the value of DI) is: mov es:[di], ax ; or AL for byte moves Once again (1) the pointer is the ES:DI pair, which is mandatory, and (2) DI is incremented or decremented (by 1 for byte, by 2 for word) depending on the status of DF, the direction flag. The instruction moves a byte (a word) from the AL (AX) register to the memory address pointed to by ES:DI. We can use the REP{7} instruction to speed things up a bit. If we have a 11,872 word block of memory, we can clear it with the following instructions: ; - - - - - - - DATASTUFF SEGMENT my_bufferdw 11872 dup (?) DATASTUFF ENDS ; - - - - - - - mov ax, seg my_buffer mov es, ax cld ; increment (DF = 0) mov ax, 0 ; clear the buffer with 0s mov di, offset my_buffer mov cx, 11872 rep stosw ____________________ 6. Cf. You-know-who's Programmer's Guide to You-know-what or "DOS Programmer's Reference." 7. There is no comparison here, so REPE or REPNE doesn't make any sense. Chapter 19 - Strings 197 ____________________ That's as fast as it gets. Why does the STOS instruction use AX? Because that's the register that port i/o uses. If you are writing a communications program, you need speed. You can have the following: ; - - - - - - - - - DATASTUFF SEGMENT port_address dw 0F2A8h ; this address is legal but ; there's nothing there. input_buffer db 4000h dup (?) output_buffer db 4000h dup (?) DATASTUFF ENDS ; - - - - - - - - - mov ax, DATASTUFF mov es, ax cld ; increment (DF = 0) mov di, offset input_buffer mov dx, port_address input_loop: in al, dx stosb jmp input_loop ; - - - - - - - - - A real program would be much more complicated because we would have to check to see if data was ready to come in and we might need to check the data for errors. Also we would occasionally have to clear the buffer. The port address F2A8h is just an arbitrary address. It's a legal address but there's nothing there. We should write a program, so let's input a character and have it fill the screen. We'll leave the last line of the screen alone so you can see your input. Move your cursor to the last line before beginning the program. ; - - - - - ENTER CODE BELOW THIS LINE mov ax, 0B800h ; or 0B000h for a monochrome card mov es, ax cld ; increment (DF = 0) outer_loop: call get_ascii_byte ; AL = fill char from input mov ah, 07h ; black background, white letters sub di, di ; set di to zero mov cx, 1920 ; 24 lines X 80 chars rep stosw jmp outer_loop ; - - - - - ENTER CODE ABOVE THIS LINE If you have a monochrome card, the segment address is 0B000h. If The PC Assembler Tutor 198 ______________________ you have a color card and are in text mode, the segment address should be 0B800h. This fills the first 24 lines with the input character. The STOS instruction has no effect on the cursor. LODS The opposite of STOS is LODS (load string) It moves a byte (word) from the string to the AL (AX) register. This time, for a change, we use the SI register as a pointer, and the default register is DS.{8} As always, SI is incremented or decremented by a byte (word) depending on the setting of DF, the direction flag. The two possibilities are: lodsb and lodsw The equivalent action (not counting changing the value of SI) is: mov ax, [si] ; or AL for byte moves This is an instruction for people that write device drivers. You could use it if you are sending a string of characters to the printer, but that's about it. Code for doing that would have the following form: ; - - - - - - - buffer db 1000 dup (?) ; - - - - - - - lea si, buffer ; the buffer must be in the ds segment cld ; increment out_loop: lodsb and al, al ; if 0, end of string jz quit_loop mov dl, al ; move character to dl {9} mov ah, 5 ; int 21h function 5 int 21h ; print a character jmp out_loop quit_loop: ; continue with the program If you actually run this program, many printers will not print anything until they get an end of line signal ( 10d, 13d). ____________________ 8. Register DS can be overriden. We'll talk about that in the second part of this chapter. 9. Int 21h (AH = 5) prints one character from DL to the printer. Why it's DL and not AL is a mystery. 199 By this time you may have become annoyed by the fact that there is no instruction for moving data from one place in memory to another. That is, you can't have: mov variable2, variable1 Instead you have to have: mov ax, variable1 mov variable2, ax There is one instruction, however, where you can move a BLOCK of data from one place in memory to another. It is called MOVS. As usual with these instructions, there are two forms. movsb moves a byte from DS:SI to ES:DI and increments or decrements SI and DI by one, depending on the setting of DF, the direction flag. Notice that either both are incremented or both are decremented. You can't have one pointer incrementing while the other one is decrementing. movsw moves a word from DS:SI to ES:DI and increments or decrements SI and DI by two, depending on the setting of DF, the direction flag. This requires the same amount of setup as all the other routines we have looked at so far, so it is not efficient to use it for just a few bytes. For 30 or so, it is very efficient. This has the equivalent effect (except for changing DI and SI) as: mov WORD PTR es:[di], ds:[si] ; or BYTE PTR for bytes We are going to write some subroutines which copy strings from one place in memory to another.{1} But first we need to review text strings. The text string world is divided into Pascal and C. A Pascal string has its length in the first byte and the first character in the second byte. Since the length is in one byte, the string length may only be 0 to 255. You read the first byte of the string to get the length. A C string can have any length. The end of a C string is marked by a byte with the value 0d. This is not the character '0', it is the number 0 (0hex). In order to find the end of a C string, you need to check each character to see if it is 0h. We'll do the Pascal string first. We are going to pass the ____________________ 1. For the technically minded, these routines will be only half of a real life subroutine, since they assume that the two strings do not overlap. In robust subroutines, these routines would be the section for when the destination address is lower than the source address. The PC Assembler Tutor 200 ______________________ addresses of the strings. If you have the Pascal call: move_pascal_string (from_string, to_string) ; The first thing you need to know is that Pascal pushes things on the stack from left to right. In other words, Pascal will generate the following code: lea ax, from_string push ax lea ax, to_string push ax call move_pascal_string We will start by assuming both near data (all data is in DS), and near subroutines. After setting up BP, the stack will look like this:{2} from_string address bp + 6 to_string address bp + 4 old IP bp + 2 bp -> old BP bp + 0 Here's the subroutine. Remember, PUSHREGS and POPREGS are macros: ; ---------- move_pascal_string proc near FROM_PTR EQU [bp+6] TO_PTR EQU [bp+4] push bp ; set up bp mov bp, sp pushf ; push the flags PUSHREGS cx, si, di, es ; push the registers push ds ; move ds to es pop es mov si, FROM_PTR ; load pointers mov di, TO_PTR cld ; clear DF (increment) sub cx, cx ; zero cx mov cl, [si] ; length to cl inc cx ; increment count by one rep movsb ; the actual move POPREGS cx, si, di, es popf ; pop the flags pop bp ret (4) ; pop pointers and return ____________________ 2. If you forgot about BP, go back to the chapter on subroutines. Chapter 19 - Strings 201 ____________________ move_pascal_string endp ; ---------- The count is increased by one since we need to move not only the text, but the count itself. If the length is 0, we still need to move one byte - the count byte. The value in DS is moved to ES with a PUSH and a POP. You cannot move directly from one segment register to another. Also, at the return we POP 4 bytes (2 words) to get the pointers off the stack. Remember, in Pascal, it is the subroutine's responsibility to get rid of the arguments from a subroutine call. You will notice that this time we saved the flags register. Why? Because we are clearing DF. When we return from the subroutine, we want DF to be exactly the same as it was on entry to the subroutine. The calling program may have DF set in some special way and we don't want to interfere with that. There are three flags which I will call 'hard' flags. Once they are set they do not change. These are (1) TF, the trap flag, (2) IEF, the interrupt enable flag, and (3) DF, the direction flag. The 'soft' flags are CF, OF, ZF, etc. If you call a subroutine you expect CF, OF, ZF etc. to be unreliable, but you expect these three 'hard' flags to remain the same. TF is the domain of a debugger, so it is none of your business. IEF is only of interest to you if you are writing an interrupt procedure.{3} The third one, DF, is your concern. If you use DF in a subroutine, you MUST save the flags to ensure that the DF flag has the same value at the return that it had on entry. Now for the C subroutine. If we have a C subroutine call: move_c_string ( from_string, to_string ) ; C pushes things on the stack from right to left (the exact opposite of Pascal). The C complier will generate the following code. lea ax, to_string push ax lea ax, from_string push ax call move_pascal_string add sp, 4 After setting up BP, the stack will look like this: to_string address bp + 6 from_string address bp + 4 old IP bp + 2 bp -> old BP bp + 0 ____________________ 3. If you do an interrupt procedure you don't have to worry because INT automatically saves the flags while clearing IEF, and IRET restores the flags on exiting. The PC Assembler Tutor 202 ______________________ Here's the C subroutine: ; ---------- move_c_string proc near FROM_PTR EQU [bp+4] TO_PTR EQU [bp+6] push bp ; set up bp mov bp, sp pushf ; push the flags PUSHREGS ax, si, di, es ; push the registers push ds ; move ds to es pop es mov si, FROM_PTR ; load pointers mov di, TO_PTR cld ; clear DF (increment) move_loop: lodsb ; source to al stosb ; al to destination and al, al ; check for 0 jnz move_loop POPREGS ax, si, di, es popf ; pop the flags pop bp ret move_c_string endp ; ---------- We set up the routine the same way, but we cannot use MOVSB. We need to check each individual byte to see if it is 0 hex, so we move it to AL, move it from AL to the destination, and then check AL for 0. Also note that we did not pop the addresses off the stack with the return statement, since in C it is the calling program's responsibility to do that. If you look at the calling code above, you will see: add sp, 4 which gets rid of the two pointers from the stack. Remember, the stack grows downward, so you ADD to decrease the size of the stack. Let's do the same Pascal program again, but this time use long pointers, that is, give both the segment and offset of the string. This means that we will be able to move from any place in memory to any place in memory. Here is the calling code. mov ax, segment from_string Chapter 19 - Strings 203 ____________________ push ax mov ax, offset from_string push ax mov ax, segment to_string push ax mov ax, offest to_string push ax call move_pascal_string We will still keep it a near subroutine. After setting up BP, the stack will look like this: from_string segment bp + 10 from_string offset bp + 8 to_string segment bp + 6 to_string offset bp + 4 old IP bp + 2 bp -> old BP bp + 0 Here's the subroutine: ; ---------- move_pascal_string proc near FROM_PTR EQU [bp+8] TO_PTR EQU [bp+4] push bp ; set up bp mov bp, sp pushf ; push the flags PUSHREGS cx, si, di, ds, es ; push the registers lds si, FROM_PTR ; load pointers les di, TO_PTR cld ; clear DF (increment) sub cx, cx ; zero cx mov cl, [si] ; length to cl inc cx ; increment count by one rep movsb ; the actual move POPREGS cx, si, di, ds, es popf ; pop the flags pop bp ret (8) ; pop pointers and return move_pascal_string endp ; ---------- This takes slightly less code since we load SI and DS at the same time (with LDS -load DS) and we load DI and ES at the same time (with LES - load ES). Remember, 8086 instructions which move an offset:segment pair always have the offset in low memory and the segment in high memory; the offset is the first two bytes and the segment is the next two bytes. The PC Assembler Tutor 204 ______________________ We changed the EQU statements, and the return statement is now: ret (8) so we take 8 bytes (4 words) off the stack, but the rest is the same. CMPS The final instruction in this group is CMPS, and as usual, it comes in two varieties. cmpsb compares the byte addressed by DS:SI to the byte addressed by ES:DI. It is the same as the CMP instruction. It moves both bytes into the 8086, subtracts the DI byte from the SI byte and sets the flags. The two bytes in memory remain unchanged. You can look at the flags to see which byte is larger, or if they are equal. As usual, both SI and DI are incremented or decremented by one, depending on the setting of DF, the direction flag. cmpsw compares the word addressed by DS:SI to the word addressed by ES:DI. It is the same as the CMP instruction. It moves both words into the 8086, subtracts the DI word from the SI word and sets the flags. The two words in memory remain unchanged. You can then look at the flags to see which word is larger, or if they are equal. Both SI and DI are incremented or decremented by two, depending on the setting of DF, the direction flag. This instruction has the same effect on the flags as: push ax mov ax, ds:[si] ; or AL for bytes cmp ax, es:[di] ; performs ( DS:[si] - ES:[DI] ) pop ax What use is this instruction? It is possible to use this for word find, and we will do that later, but it is a little unsophisticated for that. It is great for data verification, however. When you use the DISKCOMP utility in DOS which compares two floppy disks, it reads each of the disks sector by sector, and then compares them. A sector is 512 bytes. The code for this utility looks like this: ; - - - - - DATA - - - - - error_message db "Sectors are not the same", 0 disk1_buffer db 512 dup (?) disk2_buffer db 512 dup (?) ; - - - - - CODE - - - - - Chapter 19 - Strings 205 ____________________ get_next_sector: ; the code for reading one sector from each disk goes here. ; then we have the code to compare the two sets of data. mov si, offset disk1_buffer mov di, offset disk2_buffer mov cx, 256 ; 512 / 2 = 256 repe cmpsw je get_next_sector lea ax, error_message ; we had an unequal comparison call print_string jmp get_next_sector ; - - - - - - - - - - We do a word compare since it takes only half as many steps. If there is an unequal comparison at any time, the REPE instruction will terminate the loop. We can test for this inequality with JE or JNE. In this example we assume that DS and ES have the same segment address. Any time you need to verify data, this is the instruction to use. We are going to build a word search program. It is not very valuable since 'a' will not match 'A', but it is a good exercise to look at CMPS. We will use ch1str.obj, the file we used at the beginning of the chapter, as the text file and you can try to find individual words in the file. Remember, the file is continuous characters (no spaces), and all characters are small. If you didn't save the file length, you will have to run that program again to find the length of the file. Here's the word_search program: ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE EXTRN ch1str:BYTE entry_banner db 13,10, "Enter a word for a word search", 0 no_match_banner db "There was no match", 0 input_buffer db 80 dup (?) letter_count dw ? ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE mov ax, seg ch1str ; load es register mov es, ax cld ; clear DF (increment) big_loop: ; get a word for the word search mov ax, offset entry_banner call print_string mov ax, offset input_buffer The PC Assembler Tutor 206 ______________________ call get_string ; find the end of string mov al, 0 ; compare with 0 mov bx, offset input_buffer mov cx, 0 ; letter count letter_count_loop: cmp al, [bx] ; compare to 0 je end_of_count_loop inc cx ; increment count inc bx ; increment pointer jmp letter_count_loop end_of_count_loop: jcxz big_loop ; if cx = 0, string is empty so redo mov letter_count, cx ; store our count ; look for word match mov di, offset ch1str mov cx, $$$$ ; $$$$ = length of ch1str sub cx, letter_count ; calculate last possible match word_search_loop: push di ; start of search push cx ; count for ch1str mov si, offset input_buffer mov cx, letter_count repe cmpsb ; the actual comparison je found_it ; if equal, we have a match ; no match. are we finished? pop cx pop di inc di ; move to next starting address loop word_search_loop ; we fell through. finished, but no match mov ax, offset no_match_banner call print_string jmp big_loop found_it: pop cx ; clear cx off the stack pop di ; start of the match mov si, offset input_buffer mov cx, 25 ; move 25 characters to buffer transfer_loop: mov al, es:[di] mov [si], al inc si inc di loop transfer_loop mov BYTE PTR [si], 0 ; end of a C string mov ax, offset input_buffer call print_string jmp big_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE Chapter 19 - Strings 207 ____________________ The code is so long that the whole assembler file has been put on disk so you don't have to do all the typing. The pathname is \XTRAFILE\COMPARE.ASM. All you need to do is enter the length of ch1str in the MOV instruction where the dollar signs are: mov cx, $$$$ ; $$$$ = length of ch1str Link with 'link compare+ch1str+\asmhelp'. You enter a text string and the program looks for an exact match in ch1str. Here is how the program is structured. First, the program prompts you to enter a string. The program then counts the number of bytes in the string. It must have a non-zero length or the program will prompt you again for a string. The program then starts at the beginning of the text. It saves a copy of the pointer to the start of the comparison so if we fail we can start over again at the next character. The actual comparison is: repe cmpsb If that makes it through all the letters in the search string, REPE will quit because CX = 0, not because we have an unequal character. If the comparison failed we pop DI (the text pointer) and start at the next character. If there is a match, we move 25 characters (starting with the matching characters) from the text to the buffer. It is necessary to move these because when you call print_string, the string must be in the DATASTUFF segment, and ch1str isn't. We haven't used MOVSB here because ES and DS are in the wrong place. For 25 characters there is only a marginal advantage to setting up for MOVS. Finally, the 25 characters are printed. If there is no match, a message to that effect is printed. The text in ch1str is the first draft of chapter 1, but just for interest, I have hidden eight C keywords and eight of your favorite Middle English words in the text.{4} See if you can find them. SEGMENT OVERRIDES Here are the string instructions and the override rules for each one. LODS moves a byte or word from DS:[si] to AL or AX. You may use CS:[si], SS:[si] or ES:[si]. STOS moves a byte (or a word) from AL (or AX) to ES:[di]. NO ____________________ 4. Two hints. You might find four of these Middle English words in the name of a boutique. The other four of the Middle English words are some of your favorite monosyllabic words. The PC Assembler Tutor 208 ______________________ OVERRIDES ARE ALLOWED. SCAS compares AL (or AX) to the byte (or word) pointed to by ES:[di]. NO OVERRIDES ARE ALLOWED. MOVS moves a byte (or a word) from DS:[si] to ES:[di]. You may use CS:[si], SS:[si] or ES:[si], but you MAY NOT OVERRIDE ES:[di]. CMPS compares the byte (or a word) from DS:[si] to ES:[di]. You may use CS:[si], SS:[si] or ES:[si], but you MAY NOT OVERRIDE ES:[di]. Looking at the whole group, you may override DS:[si], but you may not override ES:[di]. The form of the override is strict. We will take MOVS as an example. Till now, the instructions were written: movsb ; byte move movsw ; word move If you want to do an override, the syntax is: movs BYTE PTR ES:[di], SS:[si] movs WORD PTR ES:[di], SS:[si] If you write: movsb ES:[di], SS:[di] you will get an assembler error. Here are all the legal forms: LODS lodsb lodsw lods BYTE PTR SS:[si] ; or CS:[si], DS:[si], ES:[si] lods WORD PTR SS:[si] ; or CS:[si], DS:[si], ES:[si] STOS stosb stosw stos BYTE PTR ES:[di] ; no override allowed stos WORD PTR ES:[di] ; no override allowed SCAS scasb scasw scas BYTE PTR ES:[di] ; no override allowed scas WORD PTR ES:[di] ; no override allowed MOVS movsb movsw movs BYTE PTR ES:[di], SS:[si] ;or CS, DS, ES:[si] movs WORD PTR ES:[di], SS:[si] ;or CS, DS, ES:[si] CMPS cmpsb Chapter 19 - Strings 209 ____________________ cmpsw cmps BYTE PTR SS:[si], ES:[di] ;or CS, DS, ES:[si] cmps WORD PTR SS:[si], ES:[di] ;or CS, DS, ES:[si] Just because you can do overrides with these instructions doesn't mean that you should. In fact, there is a problem. If you are using the REP instruction with an override: rep movs WORD PTR ES:[di], SS:[si] and the 8086 gets a hardware interrupt,{5} the 8086 forgets the override. What this means is that one moment you are moving data from the SS segment, and the next moment you are moving data from the same offset, but in the DS segment. This just won't do. Thus the rule is: NEVER USE AN OVERRIDE WITH A REP/REPE/REPNE INSTRUCTION This actually is no hardship. Using the override adds time to the instruction. All you need to do is change the segment addresses for the duration of the string instruction, and the code will run faster. Of course, there is the setup time, but the break even point is say, 20 repeats. Here is what you would do if you needed an SS segment override: push ds ; save old DS push ss ; move SS to DS pop ds ; the same as an SS:[di] override rep movsb pop ds ; get old DS back The other possibility is to use LOOP instead of REP. It is slower, but better slower and reliable than faster and unreliable. rep movs BYTE PTR ES:[di], SS:[si] is the same as: repeat_loop: movs BYTE PTR ES:[di], SS:[si] loop repeat_loop There are even three forms of the LOOP instruction: LOOP, LOOPE, LOOPNE which are the exact counterparts to REP, REPE, REPNE. ____________________ 5. Which can be caused by such rare occurances as your pressing a key on the keyboard or one of the 18 timer interrupts that happen each second. The PC Assembler Tutor 210 ______________________ SUMMARY LODS (load from string) moves a byte or word from DS:[si] to AL or AX, and increments (or decrements) SI depending on the setting of DF, the direction flag (by 1 for bytes and by 2 for words). You may use CS:[si], SS:[si] or ES:[si]. This performs the same action (except for changing SI) as: mov ax, DS:[SI] ; or AL for bytes The allowable forms are: lodsb lodsw lods BYTE PTR SS:[si] ; or CS:[si], DS:[si], ES:[si] lods WORD PTR SS:[si] ; or CS:[si], DS:[si], ES:[si] STOS (store to string) moves a byte (or a word) from AL (or AX) to ES:[di], and increments (or decrements) DI depending on the setting of DF, the direction flag (by 1 for bytes and by 2 for words). NO OVERRIDES ARE ALLOWED. This performs the same action (except for changing DI) as: mov ES:[DI], ax ; or AL for bytes The allowable forms are: stosb stosw stos BYTE PTR ES:[di] ; no override allowed stos WORD PTR ES:[di] ; no override allowed SCAS compares AL (or AX) to the byte (or word) pointed to by ES:[di], and increments (or decrements) DI depending on the setting of DF, the direction flag (by 1 for bytes and by 2 for words). NO OVERRIDES ARE ALLOWED. This sets the flags the same way as: cmp ax, ES:[DI] ; or AL for bytes The allowable forms are: scasb scasw scas BYTE PTR ES:[di] ; no override allowed scas WORD PTR ES:[di] ; no override allowed MOVS moves a byte (or a word) from DS:[si] to ES:[di], and increments (or decrements) SI and DI, depending on the setting of DF, the direction flag (by 1 for bytes and by 2 for words). You may use CS:[si], SS:[si] or ES:[si], but you MAY NOT OVERRIDE Chapter 19 - Strings 211 ____________________ ES:[di]. Though the following is not a legal instruction, it signifies the equivalent action to MOVS (not including changing DI and SI): mov WORD PTR ES:[DI], DS:[SI] ; or BYTE PTR for bytes The allowable forms are: movsb movsw movs BYTE PTR ES:[di], SS:[si] ;or CS, DS, ES:[si] movs WORD PTR ES:[di], SS:[si] ;or CS, DS, ES:[si] CMPS compares the byte (or a word) at DS:[si] to the one at ES:[di], and increments (or decrements) SI and DI, depending on the setting of DF, the direction flag (by 1 for bytes and by 2 for words). You may use CS:[si], SS:[si] or ES:[si], but you MAY NOT OVERRIDE ES:[di]. Although the following is not a legal action, it signifies the equivalent action to CMPS (not including changing DI and SI): cmp WORD PTR DS:[SI], ES:[DI] ; or BYTE PTR for bytes The allowable forms are: cmpsb cmpsw cmps BYTE PTR SS:[si], ES:[di] ;or CS, DS, ES:[si] cmps WORD PTR SS:[si], ES:[di] ;or CS, DS, ES:[si] The string instructions may be prefixed by REP/REPE/REPNE which will repeat the instructions according to the following conditions: rep decrement cx ; repeat if cx is not zero repe decrement cx ; repeat if cx not zero AND zf = 1 repz decrement cx ; repeat if cx not zero AND zf = 1 repne decrement cx ; repeat if cx not zero AND zf = 0 repnz decrement cx ; repeat if cx not zero AND zf = 0 Here, 'e' stands for equal, 'z' is zero and 'n' is not. These repeat instructions should NEVER be used with a segment override, since the 8086 will forget the override if a hardware interrupt occurs in the middle of the REP loop. 'HARD' FLAGS IEF, TF and DF are 'hard' flags. Once they are set they remain in the same setting. If you use DF, the direction flag, in a subroutine, you must save the flags upon entry and restore the flags on exiting to make sure that DF has not been altered. Chapter 20 - Control Structures =============================== 212 Control structures are things like 'for', 'do', 'while' and 'if' that you have in high level languages. It is possible to do these things in assembler language if you want. They are included here so if you ever want to use them at this level, you will have a template on how to set them up. All the examples will use integers only. If you have real numbers: if ( variable < 3.891 ) x = 4 * y ; it is too difficult to implement on the 8086. Of course, if you have an 8087, it's a piece of cake. IF IF is the easiest and it will illustrate how we are going to set up our labels. Let's take that first example. Since it is a waste of space to define variables over and over, we will assume that all variables are words, not bytes. The examples will all be written in C. if (variable < 7) { x = 4 + y ; } The assembler code looks like this: if1: ;---------------- cmp variable, 7 jge end_if1 mov x, 4 mov ax, y add x, ax end_if1: ;---------------- All labels will look like variations of this. The control words will be the labels.{1} They will have a unique number for each block so you can write as many different blocks as you want; the label that signals the end of the block will be the word 'end_' followed by the control word. The semicolon with the line is a ____________________ 1. The words 'if' and 'endif' are Microsoft macro directives, so if you use them without a tag number, you will get some bizarre code indeed. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 20 - Control Structures 213 _______________________________ comment, and it is used for visual separation. By using a different number each time, you may nest as deeply as you want, though the code will be pretty unreadable: if (variable < 7 ) { if ( x > 9 ) { if ( z == 0 ) { y = 12 ; } q = 5 ; } r = 7 ; } becomes: if1: ;--------------- cmp variable, 7 jge end_if1 if2: ;--------------- cmp x, 9 jle end_if2 if3: ;--------------- cmp z, 0 jne end_if3 mov y, 12 end_if3: ;----------- mov q, 5 end_if2: ;----------- mov r, 7 end_if1: ;----------- With the Microsoft assembler, you don't need to have the labels start at the beginning of the line, but they MUST be the first thing on the line. You can indent them: if1: ; -------------- cmp variable, 7 jge end_if1 if2: ;------------- cmp x, 9 jle end_if2 if3: ;------------ cmp z, 0 jne end_if3 mov y, 12 The PC Assembler Tutor 214 ______________________ end_if3: ;-------- mov q, 5 end_if2: ;--------- mov r, 7 end_if1: ;--------- You will notice that they are all compares and jumps using reverse logic. If the condition is NOT true, then we jump. If it is true, we just go on. What about IF ELSE? It looks almost the same. if ( j < 12 ) { y = 19 ; } else { z = 25 ; } We get: if16: ;---------------- cmp j, 12 jge else16 mov y, 19 jmp end_if16 else16: mov z, 25 end_else16: end_if16: ;--------------- We jump to another part inside the block, and throw in a JMP after the IF part. WHILE WHILE is the same as IF except that at the end of the block, we jump back to the beginning. while ( j < 20 ) { j = j + 1 ; } while25: ;------------ cmp j, 20 jge end_while25 inc j jmp while25 end_while25: ;----------- Chapter 20 - Control Structures 215 _______________________________ DO WHILE DO WHILE always goes through the code once. The decision process is at the end. do { k = k + 7 ; } while (k < 0) ; do_while74: ;------------ add k, 7 cmp k, 0 jl do_while74 end_do_while74: ;------------ Notice that here the jump is a positive decision. If the condition is fulfilled, we jump back to the start, otherwise we fall through. What do you do with 'break' and 'continue'? {2} Let's put them in: do { k = k + 7 ; if ( y < 9 ) { continue ; } j = j - 6 ; if ( z == 5 ) { break ; } y = y * 2 ; } while ( j < 17) ; In assembler language, this becomes: do_while143: ;---------- add k, 7 if144: ;----------- cmp y, 9 jge end_if144 jmp continue143 end_if144: ;----------- sub j, 6 ____________________ 2. For those of you who are not C people, BREAK breaks you out of the innermost loop. CONTINUE skips all the code till the end of the loop. The PC Assembler Tutor 216 ______________________ if145: ;----------- cmp z, 5 jne end_if145 jmp break143 end_if145: ;----------- sal y, 1 ; shift left 1 = multiply by 2 continue143: cmp j, 17 jl do_while143 end_do_while143: break143: In DO WHILE, CONTINUE is the label just before the decision process. BREAK is always the first label after the block. We can just use the label that marks the end of our block for the break. Instead of: jmp break143 we can have: jmp end_do_while143 From now on, we'll put in the continue and break, even if they aren't used. Once you are used to it, you can drop the break label and use the end of block label. Going back to the WHILE statement, we have: while25: ;------------ cmp j, 20 jge end_while25 inc j continue25: jmp while25 end_while25: ;----------- break25: FOR The FOR statement in C is more sophisticated than in other languages, but we will keep the example simple. It has 3 parts: for ( i = 11 ; i < 20 ; i = i + 1 ) can be looked at as: for ( initialize ; test ; update ) The reason that it is sophisticated is that it can have any number of things in the first and third parts. Chapter 20 - Control Structures 217 _______________________________ for ( i = 11, j = 27 z = 4 ; i < 20 ; k = k + 1, j = j/2) is legitimate. Therefore we need to set aside a block for the initialize part, a block for the test part, and a block for the update part. One question is whether we want the initialization inside the FOR label. I'm not doing it, but it is a possibility. Here's the C code. for ( i = 11 ; i < 20 ; i = i + 1 ) { k = k + 9 } And its corresponding assembler code: init217: ;--------------- mov i, 11 end_init217: ;------ for217: ; --------------- test217: cmp i, 20 jge end_for217 end_test217: ;----- add k,9 continue217: update217: inc i end_update217: ; ---- jmp test217 end_for217: ;------------- break217: My that's a lot of labels. In this example we would get rid of most of them because there is so little code, but if the block contained a lot of code they would be a help, not a hindrance. Remember, a label generates no code, so this makes no difference in the size of the object file. Some of the labels which are not targets of jumps could be made into comments. end_update217: ;---- can become: ; end_update217 ---- Whichever way looks better to you. Using comments instead of labels gives you a slightly faster compile time. Just make sure you don't change one that IS a target of a jump. SWITCH Finally there is the switch statement. If you don't know what it is, skip this section because it will probably be confusing. The PC Assembler Tutor 218 ______________________ switch ( k ) { case 'A': x = x + 9 ; break ; case '&': y = y * 2 ; break ; case 'j': z = z - 7 ; break ; } We get the following: switch82: ;----------------- cmp k, 'A' jne test82_2 jmp case82_1 test82_2 cmp k, '&' jne test82_3 jmp case82_2 test82_3 cmp k, 'j' jne end_test82 jmp case82_3 end_test82: jmp deafult82 ; ---------- case82_1: add x, 9 jmp break82 case82_2: sal y, 1 ; multiply by 2 jmp break82 case82_3: sub z, 7 jmp break82 default82: jmp break82 end_switch82: ;------------ break82: It may look like there are unnecessary jumps at the beginning, but it is likely that in a real program some of the case statements would be more than +127 bytes away from the conditional jumps, so you need JMP, which can go anywhere in the segment. Chapter 21 - .COM Files ======================= 219 All the programs that we have made so far have been .EXE files. That means that the file extension has always been .EXE after linking. When you have and .EXE file, the program loader makes certain adjustments to the machine code at run time. These adjustments are the actual segment addresses of the segments. There is another type of executable file, and that is a .COM file. When the loader puts a .COM file into memory, it makes no adjustments, it simply reads the file directly from disk into memory. Therefore, a .COM file loads faster. However, there is a restriction: All code, data, and the stack must be in a single segment. This effectively limits a .COM program to 65536 bytes of code+data+stack. {1} In general, it is easier for program development to keep code and data separate, but we can mix them together. Let's look at the template for a .COM file. It is called COMTEMP.ASM. ; com file template ; put name here ; * * * * * * * * * * * * * * * INCLUDE \PUSHREGS.MAC COMSEG SEGMENT PUBLIC 'CODE' ASSUME cs:COMSEG, ds:COMSEG, es:COMSEG, ss:COMSEG ; - - - - - - - - - - main proc NEAR ORG 100h start: ; - - - - START CODE BELOW THIS LINE ; - - - - END CODE ABOVE THIS LINE ret main endp ; - - - - START SUBROUTINES BELOW THIS LINE ; - - - - END SUBROUTINES ABOVE THIS LINE ____________________ 1. Actually, it is possible to get around this restriction with suitable fiddling, but by that time you have made the program much more complicated so you have lost any advantage that you had by not using an .EXE file. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 220 ______________________ ; - - - - START DATA BELOW THIS LINE ; - - - - END DATA ABOVE THIS LINE Stack_area dw 500 dup (?) COMSEG ENDS ; * * * * * * * * * * * * * * * END start We will take the things in order. First, there is the line: ORG 100h This effectively tells the assembler to put 100h (256d) bytes of zeros at the beginning. Also notice that the setup code that we had in a .EXE file is missing; we start with the code immediately. This all has to do with the PSP (program segment prefix).{2} PSP You will remember from the chapter describing the .EXE template file that we wrote: push ds sub ax, ax push ax because upon entry to an .EXE file, DS contains the segment address of the PSP, and at offset 0000 (that is, the first byte of the segment) there is a machine instruction for an orderly exit from the program. In an .EXE file, the PSP is somewhere in memory, put there by the loader. In a .COM file, the PSP is the first 100h (256d) bytes of the segment. The loader fills in the PSP and then reads the file directly from disk. That means that the machine instruction for an orderly exit is at 0000 of the current segment. Notice that we have a NEAR procedure. A .COM file has a near return which will stay in the same segment. Normally we would have to write: sub ax, ax push ax to put the return address 0000 on the stack, but for a .COM file the loader pushes 0000 on the stack before giving control to the program. Why the loader provides this service for a .COM file but not for an .EXE file is a mystery. In any case, with a .COM file, you don't need to push the return address on the stack, since it's there already. ____________________ 2. If you want to know what the PSP (program segment prefix) is exactly, consult either of the two books on hardware and interrupts. The PSP is always exactly 256 bytes long. Chapter 21 - .COM Files 221 _______________________ We have the assembler directive: ASSUME cs:COMSEG, ds:COMSEG, es:COMSEG, ss:COMSEG The reason for including all the segments is (1) the loader actually loads the segment address into all four segments and (2) the assembler will use the natural segment for all instructions. The BP instructions will refer to SS, the data instructions will refer to DS, the string move instructions (SCAS, CMPS, MOVS) will refer to ES. That means that the assembler will not use segment overrides. This helps avoid something called phase errors which will be explained at the end. Right after the ORG instruction is start: (the first instruction executed in the program.) ORG 100h ; 256d start: This is inflexible. After reading the program into memory, the loader ALWAYS starts the program at 100h. All .COM files start execution at 100h. Period. There is space for subroutines, space for data, and finally space for the stack. The order of subroutines and data can be changed. The stack space is technically not necessary, but it is a good reminder to leave it there. All .COM files take up a whole 65536 byte segment in memory, no matter how short they are.{3} The stack is at the very end of the segment, so if your program is 200 bytes long, you have 65336 bytes of stack space. Let's make a simple program. It is the famous "Hello,World" program. ; - - - - START CODE BELOW THIS LINE mov dx, offset mr_happy_face ; int 21h, function 9 mov ah, 9 int 21h ; - - - - END CODE ABOVE THIS LINE ; - - - - START DATA BELOW THIS LINE mr_happy_face db "Hello, world!", 13, 10, "$" ; - - - - END DATA ABOVE THIS LINE This program prints 'Hello World!' and a new line. The dollar sign signifies the end of the string for this interrupt. It is simple enough, and it gives us the chance to look at the extra step needed to make a .COM file. Assemble it, and link it. When you link it, you will get a warning that there is no stack segment. For .COM files, this warning is unimportant. We now have an .EXE file (which, by the way, won't run correctly). How do we ____________________ 3. This is a slightly abridged explaination. For the real details, consult Microsoft's "The MS-DOS Encyclopedia". The PC Assembler Tutor 222 ______________________ make it a .COM file? Among the programs that you got with DOS is one called EXE2BIN. It takes an .EXE file, and if possible, converts it into a .COM file. You simply write the name of the file you want converted and the name of the converted file. Both of these names must have the full file extension: exe2bin programA.exe programA.com You will now have programA.com as a .COM file. If we now write: C> programA Will the loader load the .COM file or the .EXE file? The DOS order for execution is .COM files first, then .EXE files, then .BAT files. DOS will execute the .COM file. Try it. That was pretty easy. Now, as a technical exercise, we are going to link together three different files. Here's the first file: ; prog1.asm ; * * * * * * * * * * * * * * * INCLUDE \PUSHREGS.MAC COMSEG SEGMENT PUBLIC 'CODE' ASSUME cs:COMSEG, ds:COMSEG, es:COMSEG, ss:COMSEG PUBLIC message1 EXTRN subroutine_a:NEAR, message3:BYTE ; - - - - - - - - - - main proc NEAR ORG 100h start: mov ah, 9 ; int 21h, ah = 9 mov dx, offset message1 int 21h mov ah, 9 ; int 21h, ah = 9 mov dx, offset message3 int 21h call subroutine_a ret main endp message1 db "This is from the main program.", 13, 10, "$" COMSEG ENDS ; * * * * * * * * * * * * * * * END start ; ---------- Here is the second file: ; prog2.asm Chapter 21 - .COM Files 223 _______________________ ; * * * * * * * * * * * * * * * INCLUDE \PUSHREGS.MAC COMSEG SEGMENT PUBLIC 'CODE' ASSUME cs:COMSEG, ds:COMSEG, es:COMSEG, ss:COMSEG PUBLIC message2, subroutine_a EXTRN subroutine_b:NEAR, message3:BYTE ; - - - - - - - - - - subroutine_a proc NEAR PUSHREGS ax, dx mov ah, 9 ; int 21h, ah = 9 mov dx, offset message2 int 21h mov ah, 9 ; int 21h, ah = 9 mov dx, offset message3 int 21h call subroutine_b POPREGS ax, dx ret subroutine_a endp message2 db "This is from subroutine A.", 13, 10, "$" COMSEG ENDS ; * * * * * * * * * * * * * * * END ; ---------- And here is the third file: ; prog3.asm ; * * * * * * * * * * * * * * * INCLUDE \PUSHREGS.MAC COMSEG SEGMENT PUBLIC 'CODE' ASSUME cs:COMSEG, ds:COMSEG, es:COMSEG, ss:COMSEG PUBLIC message3, subroutine_b EXTRN message1:BYTE, message2:BYTE ; - - - - - - - - - - subroutine_b proc NEAR PUSHREGS ax, dx mov ah, 9 ; int 21h, ah = 9 mov dx, offset message1 int 21h mov ah, 9 ; int 21h, ah = 9 mov dx, offset message2 int 21h POPREGS ax, dx The PC Assembler Tutor 224 ______________________ ret subroutine_b endp message3 db "This is from subroutine B.", 13, 10, "$" COMSEG ENDS ; * * * * * * * * * * * * * * * END ; ---------- If you look at them, you will see that all they do is print messages; sometimes from their own file, sometimes from external files. The subprograms all have different names since they call each other. Of course, they have both PUBLIC and EXTRN statements. Only prog1 has: start: since that is where the program execution will start, and only prog1 has: ORG 100h since having it in the other files would leave unnecessary blank spaces in the other programs. Assemble the programs. When you link the programs, prog1 MUST be the first on the line: link prog1+prog2+prog3 link prog1+prog3+prog2 are both ok, but: link prog2+prog1+prog3 will not work since the linker, exe2bin, and the loader are counting on the starting instruction being at 100h. If you change the order, you will either get complaints from EXE2BIN or the program won't run correctly. Now with: exe2bin prog1.exe prog1.com you have a .COM file. Try it out. Any .COM file can also be made into an .EXE file. We will make a simple program in .COM format, and then add the necessary things to make it an .EXE format. First, here's the .COM format. ; commode.asm ; * * * * * * * * * * * * * * * COMSEG SEGMENT PUBLIC 'CODE' ASSUME cs:COMSEG, ds:COMSEG, es:COMSEG, ss:COMSEG main proc NEAR Chapter 21 - .COM Files 225 _______________________ ORG 100h start: ; get video mode int 10h, function 0Fh mov ah, 0Fh int 10h ; al = display mode mov bl, 10 ; divide by 10 mov si, offset ones ; right hand digit mov cx, 2 ; 2 digit answer division_loop: mov ah, 0 ; clear ah div bl ; al/bl, remainder in ah add ah, '0' ; change to ascii mov [si], ah dec si ; one byte to the left loop division_loop ; display string mov dx, offset message ; int 21h, service 9 mov ah, 9 int 21h ret main endp ; - - - - START DATA BELOW THIS LINE message db "The current video mode is " ones db ? db ".", 13, 10, "$" ; - - - - END DATA ABOVE THIS LINE COMSEG ENDS ; * * * * * * * * * * * * * * * END start ; - - - - - - - - - - This program gets the video mode which is a number which tells you what mode the monitor is operating in. To find out what the number means, consult either of those two hardware books. It gets the mode through an interrupt. The mode is returned in AL. It then puts the number in a string and prints the string. We cannot link with asmhelp.obj, so it takes all this work is simply to output a number. We'll call this commode.asm. Assemble, link, use exe2bin, and run it. Now let's make the .EXE counterpart. Here it is. ; exemode.asm ; * * * * * * * * * * * * * * * STACKSEG SEGMENT STACK 'STACK' dw 20 dup (?) STACKSEG ENDS The PC Assembler Tutor 226 ______________________ ; - - - - - - - - - - COMSEG SEGMENT PUBLIC 'CODE' ASSUME cs:COMSEG, ds:COMSEG, es:COMSEG main proc FAR start: push ds ; same as before sub ax, ax push ax push cs ; ds = cs pop ds ; get video mode int 10h, service 15 mov ah, 15 int 10h ; al = display mode mov bl, 10 ; divide by 10 mov si, offset ones ; right hand digit mov cx, 2 ; 2 digit answer division_loop: mov ah, 0 ; clear ah div bl ; al/bl, remainder in ah add ah, '0' ; change to ascii mov [si], ah dec si ; one byte to the left loop division_loop ; display string mov dx, offset message ; int 21h, service 9 mov ah, 9 int 21h ret main endp ; - - - - START DATA BELOW THIS LINE message db "The current video mode is " ones db ? db ".", 13, 10, "$" ; - - - - END DATA ABOVE THIS LINE COMSEG ENDS ; * * * * * * * * * * * * * * * END start This is almost the same. We have put in a small stack segment. Here's the different part: ;---------- ASSUME cs:COMSEG, ds:COMSEG, es:COMSEG main proc FAR Chapter 21 - .COM Files 227 _______________________ start: push ds ; same as before sub ax, ax push ax push cs ; ds = cs pop ds ;---------- SS is no longer in the ASSUME statement, since it now refers to the stack segment. CS, DS, and ES still refer to COMSEG. The procedure is now a FAR procedure. We have taken the ORG out, since that would simply waste 100h (256) bytes of space. Then we have the normal .EXE startup except the data is now in COMSEG, so we move CS to DS. That's it. We'll call this one exemode.asm. Assemble, link and run it. It should give you the same result. Here is the listing for both executable files: COMMODE COM 65 10-21-89 11:11p EXEMODE EXE 631 10-21-89 11:10p Notice how much bigger the .EXE file is. That is because the .EXE file has a bunch of information for the loader. Also, if you run both programs, the .COM file will start a little quicker. Those are the only advantages. PHASE ERRORS The assembler generates code in two steps. On the first pass, it calculates the address of each variable and machine instruction without actually writing code. For instance, if there is the instruction: mov ax, variable1 The assembler will allocate 4 bytes for the instruction. If variable1 has already been defined, but is in ES, then the assembler will allocate 5 bytes; 1 for the segment override and 4 for the instruction itself. If, however, variable1 has not been defined yet (it appears later in the code), then the assembler will assume that it is in DS and allocate 4 bytes. If it turns out that it is later defined to be in ES, then when the assembler generates code, it will write 5 bytes, 1 for the override and 4 for the instruction. But this means that EVERYTHING after this instruction will have been shifted one byte, so EVERYTHING after the instruction will be at the wrong address. The assembler will detect this and print out a PHASE ERROR. This means that the machine code is garbage. By having all four segment registers in the ASSUME statement of a .COM file, you guarantee that the assembler will not generate segment overrides. The PC Assembler Tutor 228 ______________________ add dx, [bp] will have BP relative to SS. sub variable1, si will have variable1 relative to DS. This will go a long way towards eliminating errors in a .COM file. Chapter 22 - BCD Numbers ======================== 229 This chapter covers numbers which are in BCD format, both packed and unpacked. You will probably never need to write any programs on the 8086 that need these instructions, so you can either do this chapter because it is the only time that you will run into them, or you can skip the chapter because you will never see them again. My advice would be to go through it anyways so you know what the capabilities of the 8086 are. The programs in this chapter are more advanced so will be more of a challenge to your understanding. BCD stands for binary coded decimals. In their unpacked form, each byte stands for a single decimal digit. If we take a number like 831974 and put it in memory, the bytes will look like this: 08h 03h 01h 09h 07h 04h With high memory at the top and low memory at the bottom. Notice that the top number is not ASCII '8' (hex 38), but the number 8 (hex 08). The same holds true for all the bytes; they are not ASCII characters, they are numbers. Why would we want to have numbers like this? They use up more space (about 2 1/2 times as much), and the arithmetic operations are slower (a multiplication on the 8086 can be several HUNDRED times slower). They are not used for scientific operations. They are used in business for billing. In a typical billing operation, the number is entered from a terminal and stored. At the end of the month, the number is printed on one line of the bill, and then added to the total. If it were converted to an integer, it would be necessary to do one conversion during data entry, then another conversion during printing. This way, the only time it will be converted is if it is used in some arithmetic. One excuse for using BCD numbers is that they are more accurate. It is true that they have no rounding errors, but neither do integers. Integers and BCD numbers can have the same accuracy. The typical form for BCD numbers is 18 digits. If you want to convert an 18 digit number to a standard integer, it takes about 25 multiplications. That is a lot of multiplications to convert one number. Similarly, it takes about 25 divisions to get a number out of integer form back into a BCD number. This is a big waste of time for a business situation. We keep the numbers in decimal form and use them occasionally for arithmetic. Actually, you would have to be crazy to use an 8086 for BCD ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 230 ______________________ numbers. If you have a PC and use BCD numbers habitually, an 8087 coprocessor will work on BCD numbers at lightning speed. An investment of $175 will save you countless hours of grief. Also, it is next to impossible to do BCD division on the 8086, but takes no longer than normal to do BCD division on the 8087. That being said, just consider this chapter an academic exercise. It will give you the information so if the question should arise at a party, you will be able to say how BCD arithmetic works on the 8086. All 8086 numbers have the least significant digit in low memory and the most significant digit in high memory. This includes packed and unpacked BCD numbers. In the unpacked form, each byte represents a digit, and has a value from 0 to 9. Here are some numbers and their unpacked BCD encoding (in hex): 3986149 27 961728 74610 03h 02h 09h 07h 09h 07h 06h 04h 08h 01h 06h 06h 07h 01h 01h 02h 00h 04h 08h 09h This wastes a lot of space. Someone early on noticed that the upper half byte is completely unused. You can cut the space consumption in half if you put two digits in each byte - one in the lower half byte, and the other in the upper half byte. This packing always starts from the least significant digit and goes to the most significant digit. Here are the same numbers in packed form. 3986149 27 961728 74610 03h 27h 96h 07h 98h 17h 46h 61h 28h 10h 49h This is a considerable saving in space. The 8087 (not 8086) standard for these numbers is 10 byte long numbers. The low order 9 bytes contain 18 digits. The last byte is 00h if the number is positive and 80h if the number is negative. Here are an 18 digit positive number and an 18 digit negative number, along with their 10 byte BCD encoding. Chapter 22 - BCD Numbers 231 ________________________ +137486298374691552 -581726405829645298 00 80 13 58 74 17 86 26 29 40 83 58 74 29 69 64 15 52 52 98 Let's work with the packed BCD numbers first. PACKED BCD NUMBERS The 8086 can manipulate packed BCD numbers. There is a subprogram in asmhelp.obj that gets a BCD number from the keyboard and one that prints a BCD number on the screen. Let's do some input and output first. ; - - - - - - - - - - START DATA BELOW THIS LINE variable1dw ? ; first four bcd digits variable2dw ? ; second four bcd digits variable3dw ? ; third four bcd digits variable4dw ? ; fourth four bcd digits variable5dw ? ; last two bcd digits and sign ; - - - - - - - - - - END DATA ABOVE THIS LINE ; - - - - - - - - - - START CODE BELOW THIS LINE outer_loop: lea ax, variable1 call get_bcd mov ax, variable5 call print_hex mov ax, variable4 call print_hex mov ax, variable3 call print_hex mov ax, variable2 call print_hex mov ax, variable1 call print_hex lea ax, variable1 call print_bcd loop outer_loop ; - - - - - - - - - - END CODE ABOVE THIS LINE This gets a 10 byte BCD number, prints it out in hex (with high memory on top), and then prints the BCD number out again. Commas The PC Assembler Tutor 232 ______________________ are allowed. Printing the number in hex form allows you to see what the numbers look like internally. There are two instructions for BCD numbers - DAA (decimal adjust for addition), and DAS (decimal adjust for subtraction). We'll use each one on individual bytes to see how they work. Let's try DAA. ; - - - - - - - - - - START CODE BELOW THIS LINE mov ax_byte, 0C4h ; half registers, hex mov bx_byte, 094h ; half regs signed, hex mov dx_byte, 91h ; half registers, signed lea ax, ax_byte call set_reg_style sub cx, cx ; clear cx for clarity outer_loop: mov ax, 0 ; clear the registers mov bx, 0 mov dx, 0 call set_count call show_regs call get_hex_byte ; byte for al mov dx, ax ; copy to dx push ax ; save al call get_hex_byte ; byte for bl mov bl, al mov bh, bl ; copy to bh pop ax ; restore al call show_regs_and_wait add al, bl ; normal add mov dx, ax ; copy to dx call show_regs_and_wait daa ; make adjustment mov dx, ax ; copy to dx call show_regs_and_wait jmp outer_loop ; - - - - - - - - - - END CODE ABOVE THIS LINE Since the program works with both BCD adjustments and integer arithmetic, DX shows a signed integer copy of AX and BH shows a signed integer copy of BL. A copy of AX is moved to DX every time an operation is performed on AL. The idea of this subprogram is to enter hex numbers that look like decimal numbers - e.g. 65h, 78h, 08h, 29h. You can enter numbers that contain A-F if you want to see what happens with bad data. Each half byte of a BCD number should look like a decimal number. You will notice that the actual addition is done by ADD. The alteration is done by DAA, which assumes that you have just added two legitimate BCD numbers, and the result is in AL. The register MUST be AL. What DAA does will be discussed in a moment. Chapter 22 - BCD Numbers 233 ________________________ The DAA instruction looks at AL as being two half bytes. If the result from the lower half byte addition is 10 or over, the 8086 subtracts 10 and then adds 1 to the upper half byte.{1} For instance, if the result is 17, it subtracts 10, to leave 7 in the lower half byte, and adds 1 to the upper half byte. After it has done this, it looks at the upper half byte. If the upper half byte is now 10 or over, it subtracts 10 and sets the carry flag for future additions.{2} While the whole thing sounds a little confusing, if you look at the bytes in hex, they will act in the same way as if you were doing normal decimal addition with pencil and paper except all the carries are done at one time. CF will be set if the hundreds digit is 1 and will be cleared if the hundreds digit is 0. Use the previous program a few times to get the hang of what is going on. When you feel confident, we'll move on to subtraction. SUBTRACTION DAS (decimal adjust for subtraction) is similar to DAA. It makes an adjustment after the subtraction itself. We'll use the same program as before, making two alterations. Where you have ADD, change it to SUB ; where you have DAA, change it to DAS. That's all. add -> sub daa -> das Run the program, and put in a number of examples. If the lower number is larger than the top number, there will be a borrow. DAS works the opposite of DAA. If there has been a borrow into the low half byte, it adds 10 to the low half byte and subtracts 1 from the high half byte. It then looks at the high half byte. ____________________ 1. The technical description is a little confusing, so if you get muddled up, just forget it. Here it goes. (AF is the auxilillary flag). The 8086 checks for either (1) the low half byte > 9 (that is, between 10 and 15) or (2) AF = 1. AF is set on a byte addition when there is a carry out of the low half byte, that is, the result is greater than 16 for the low half byte. If either of these events has occured, the 8086 ADDS 6 to the low half byte. Why? Let 10 + x represent the result of the addition, where x < 10. We then have: 10 + x + 6 = 10 + 6 + x = 16 + x but 16 is 0001 0000 (10h), the first bit in the high byte, so this has the effect of leaving the part less than 10 in the low half byte and adding 1 (the carry) to the high half byte. 2. This is exactly the same logic as in the last footnote, except that the 8086 adds 60h to the high byte. This has the effect of shifting the excess out of AL altogether. The 8086 then sets the carry flag. The PC Assembler Tutor 234 ______________________ If there has been a borrow into it, the 8086 adds 10 to the high half byte and sets the carry flag. Do a few more examples with the program to make sure you understand what's happening. It will look just like what you do with pencil and paper, except that with pencil and paper you do the borrow before you do the subtraction and here the borrow is done afterwards if it is needed. ADDITION AND SUBTRACTION We have a number of possibilities with addition and subtraction. Here they are. The sign of the numbers is in parentheses and the operation is in between. (+) + (+) (+) + (-) (+) - (+) (+) - (-) (-) + (+) (-) + (-) (-) - (+) (-) - (-) The subroutines we use are going to assume that (1) both numbers are the same sign, and (2) for subtraction, the larger number is on top and the smaller number is on the bottom. There should be a section of the program that decides whether addition or subtraction should be used, and which number is on top. Then comes the addition or subtraction. We will write the first part later. For now, when you input numbers, both numbers must have the same sign, and for subtraction, the larger number must be first. Here's the addition subroutine. ; - - - - - START SUBROUTINE BELOW THIS LINE _bcd_addition proc near RESULT_ADDRESS EQU [bp + 8] BOTTOM_ADDRESS EQU [bp + 6] TOP_ADDRESS EQU [bp + 4] push bp mov bp, sp PUSHREGS ax, bx, cx, si, di mov si, TOP_ADDRESS mov bx, BOTTOM_ADDRESS mov di, RESULT_ADDRESS mov cx, 9 ; 9 bytes of BCD numbers clc ; clear the carry flag add_loop: mov al, [si] ; move and add bytes adc al, [bx] ; add two numbers and the carry daa ; bcd adjust mov [di], al ; store result inc si ; increment pointers inc bx inc di loop add_loop jnc continue_addition ; BCD overflow if CF = 1 Chapter 22 - BCD Numbers 235 ________________________ lea ax, overflow_message call print_string continue_addition: mov al, [si] ; sign of top addend to result mov [di], al POPREGS ax, bx, cx, si, di pop bp ret _bcd_addition endp ; - - - - END SUBROUTINE ABOVE THIS LINE AL contains the first number. We use the carry flag (ADC) for carrying from byte to byte. The carry flag is cleared before the first addition since we don't want a carry there. The INC instruction was designed by Intel so that it would not alter the carry flag just so we could use it in situations like this. If CF = 1 upon exiting the loop, the result was too large and we had overflow. In this case we print a message to that effect. The result [di] can be stored in the same place as one of the numbers. The addition of each byte is completed before the result for that byte is stored, so this won't interfere with the addition. We assume that the sign of the result is the sign of the top addend. (If this goes into a working program, it will only be used if both numbers have the same sign). This is designed as a C subroutine. The calling program pushes things on the stack and it has the responsibility of popping them off the stack on return from the call. Here's the calling routine: ; - - - - - ENTER DATA BELOW THIS LINE bcd_num1 dt ? bcd_num2 dt ? bcd_num3 dt ? overflow_message db "We had overflow.", 0 ; - - - - - ENTER DATA ABOVE THIS LINE ; - - - - - ENTER CODE BELOW THIS LINE outer_loop: lea ax, bcd_num1 ; get 2 numbers for addition call get_bcd lea ax, bcd_num2 call get_bcd lea ax, bcd_num3 ; push addresses for subroutine push ax lea ax, bcd_num2 push ax lea ax, bcd_num1 push ax The PC Assembler Tutor 236 ______________________ call _bcd_addition add sp, 6 ; 3 pushes = 6 bytes lea ax, bcd_num1 ; print both numbers and result call print_bcd lea ax, bcd_num2 call print_bcd lea ax, bcd_num3 call print_bcd loop outer_loop ; - - - - - ENTER CODE ABOVE THIS LINE This is the main program. The other one should be in the subroutine section. This program is straightforward. We get two numbers, call the subroutine, and print the numbers and the result. Try some numbers. The results you get will always have the sign of the top number and will have the same numerical result as adding two unsigned numbers. The subtraction routine is the same as the addition but we need to make some changes because it is subtraction: ADC -> SBB DAA -> DAS (decimal adjust for subtraction) add_loop: -> subtract_loop loop add_loop -> loop subtract_loop jmp continue_addition -> jmp continue_subtraction continue_addition: -> continue_subtraction: Also, we want to change the subroutine name and call _bcd_addition -> _bcd_subtraction call _bcd_addition -> call _bcd_subtraction Do all these changes, run the program, but make sure the top number is larger or you will get strange results. We now have an addition routine that only works with the right numbers and a subtraction routine that is very touchy about the numbers that are put in. How can these things help us? In fact they are almost all we need. The only thing else we need is a preliminary subroutine to organize what we do. To see why we have some orginizational problems, take out a pencil and paper and do the following additions and subtractions. Do these with a pencil and paper, not a pocket calculator: (+15) + (+27) (+15) + (-27) (+15) - (+27) (+15) - (-27) (-15) + (+27) (-15) + (-27) (-15) - (+27) (-15) - (-27) (+27) + (+15) (+27) + (-15) (+27) - (+15) (+27) - (-15) (-27) + (+15) (-27) + (-15) (-27) - (+15) (-27) - (-15) Chapter 22 - BCD Numbers 237 ________________________ There are only four possible answers: +12, -12, +42 and -42. We had 16 different additions and subtractions, yet we got only 4 possible answers and only 2 possible absolute values. We could have a different subroutine for each one, but 16 subprograms is a LOT of code, so it's easier to do it with 2 subprograms. We merely need to order the numbers and pick the correct subroutine. Here is the BCD driving routine. We'll get a number and then we'll get an operation, either addition or subtraction. If it is subtraction, when we get the second number, we REVERSE the sign. We don't even check what it is; plus becomes minus and minus becomes plus. What we have now is the ADDITION of two signed numbers. We XOR the two signs. If the result is 0, they both are the same sign and we can go to the addition subroutine. If the signs are different we need to subtract the smaller from the larger. We find out which one is larger and then call the subtraction routine. It is going to take you a little time to read this code. ; - - - - - START DATA BELOW THIS LINE top_number dt ? bottom_number dt ? result dt ? sign_mask db ? sign_message db "Enter either + or -.", 0 overflow_message db "We had overflow.", 0 ; - - - - - END DATA ABOVE THIS LINE ; - - - - - START CODE BELOW THIS LINE outer_loop: lea ax, top_number call get_bcd sign_loop: ; get either a '+' or a '-' lea ax, sign_message ; prompt for operation call print_string call get_ascii_byte ; operation type in al cmp al, '+' jne check_for_minus mov sign_mask, 00h ; for XOR of bottom number sign jmp get_second_number check_for_minus: cmp al, '-' jne sign_loop ; if not a minus then redo mov sign_mask, 80h ; for XOR of bottom sign get_second_number: lea ax, bottom_number call get_bcd ; XOR bottom sign with sign mask mov al, sign_mask xor BYTE PTR (bottom_number + 9), al ; sign byte ; same sign or different signs? mov ah, BYTE PTR (top_number + 9) ; sign byte The PC Assembler Tutor 238 ______________________ xor ah, BYTE PTR (bottom_number + 9) ; different? jnz which_is_larger ; if different, subtract ; same sign, so add lea ax, result ; push parameters and add push ax lea ax, bottom_number push ax lea ax, top_number push ax call _bcd_addition add sp, 6 ; adjust the stack jmp print_the_numbers which_is_larger: lea si, top_number + 8 ; top digits lea di, bottom_number + 8 mov cx, 9 ; 9 bytes of digits check_for_greater_loop: mov al, [si] ; top number cmp al, [di] ; bottom number ja top_is_more jb bottom_is_more dec si ; equal, so continue dec di loop check_for_greater_loop ; we fell through so they are the same ; leave the top number on top top_is_more: lea ax, result push ax lea ax, bottom_number push ax lea ax, top_number push ax call _bcd_subtraction add sp, 6 ; adjust the stack jmp print_the_numbers bottom_is_more: lea ax, result push ax lea ax, top_number push ax lea ax, bottom_number push ax call _bcd_subtraction add sp, 6 ; adjust the stack print_the_numbers: lea ax, top_number call print_bcd lea ax, bottom_number call print_bcd lea ax, result call print_bcd Chapter 22 - BCD Numbers 239 ________________________ jmp outer_loop ; - - - - - END CODE ABOVE THIS LINE The sign_mask is either 00h if it is addition or 80h if it is subtraction. 00h XOR sign_byte will leave the sign byte unchanged. 80h XOR sign_byte will reverse the sign of the sign byte. Thus, as we did in the earlier multiplication and division routines, we sometimes change the sign of a number (bottom_number). In a real routine we would either have to make a copy of it when we change the sign or change the sign back at the end. Here we leave it with the sign change (if any) so you can see what is happening internally. If you want to restore the number, you can alter the code a little: print_the_numbers: mov al, sign_mask ; add this xor BYTE PTR (bottom_number + 9), al ; add this lea ax, top_number This will restore the bottom number to its original form ASSUMING THAT THE RESULT HAS NOT BEEN STORED THERE. It is possible to get -0 as a result. This is a legally defined number: +0 = -0. There are three places where we have 'BYTE PTR'. This is because the variables are defined as 10 byte long objects and the assembler will complain if we don't put in the 'BYTE PTR'. Finally, if we want to use the typical 'destination, source' style for the 8086, it is built in. Here it is for addition: lea ax, top_number push ax lea ax, bottom_number push ax lea ax, top_number push ax We put the address of 'top_number' where the result address goes. The routines store a byte only after all calculations are done on that particular byte, so there is no interference from doing this. 240 The first thing to talk about is the 8086 mnemonics. The four instructions for unpacked BCD numbers are: AAA ASCII adjust for addition AAD ASCII adjust for division AAM ASCII adjust for multiplication AAS ASCII adjust for subtraction. Even though all four instructions have ASCII as part of their mnemonic, they have NOTHING to do with ASCII numbers. These instructions operate on unpacked BCD numbers. They always give results which are unpacked BCD numbers. Because of side effects of the 8086 microcode, the add and subtract instructions do something unusual, but this will be covered a little later. Just like the packed BCD instructions, these four unpacked instructions use the normal arithmetic operations and adjust the results to compensate for their being unpacked BCD numbers. Let's start with addition. The following program is like the BCD program except that we use AAA (ascii adjust for addition) instead of DAA (decimal adjust for addition). ; - - - - - - - - - - START CODE BELOW THIS LINE mov ax_byte, 0C4h ; half registers, hex, hex mov bx_byte, 94h ; half regs, signed, hex mov dx_byte, 91h ; half regs, signed, signed lea ax, ax_byte call set_reg_style mov cx, 0 ; clear cx for clarity outer_loop: mov ax, 0 ; clear the registers mov bx, 0 mov dx, 0 call set_count call show_regs call get_hex_byte ; byte for al mov dx, ax ; copy for dx push ax ; save al call get_hex_byte ; byte for bl mov bl, al mov bh, bl ; copy to bh pop ax ; restore al call show_regs_and_wait add al, bl ; normal add mov dx, ax ; copy to dx call show_regs_and_wait aaa ; make adjustment mov dx, ax ; copy to dx call show_regs_and_wait jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 22 - BCD Numbers 241 ________________________ Since this is a mixture of unpacked BCD instructions and normal integer instructions, a copy of AX (which is showing hex) is moved to DX (which is showing signed) each time AX is changed. Also, a copy of BL is put in BH. For the rest of this chapter, I will refer to the one's digit, the ten's digit and the hundred's digit. In the number 346, 3 is the hundred's digit, 4 is the ten's digit, and 6 is the one's digit. In the number 928, 9 is the hundred's digit, 2 is the ten's digit, and 8 is the one's digit. If two valid unpacked BCD numbers are added, the result will be between 0 and 18. This result MUST be in AL. AAA sets CF if this result is greater than 9 ( i.e. there was a carry). AAA leaves the one's digit in AL and adds the ten's digit to AH. (The ten's digit can only be 0 for 0 - 9 or 1 for 10 - 18). Run this. The standard input should be hex numbers between 00h and 09h. When you feel comfortable with this, stop for a minute and read on. We now come to the peculiarity of this instruction. It isn't operating on the whole byte, only the lower half byte. Technically, if the LOWER HALF BYTE of each of the two numbers which are added is a legitimate BCD digit, then at the end of AAA the above results will be true and the UPPER HALF BYTE of AL will be set to 0. If you put anything in the upper half byte, it will be added by ADD, but will be blanked out (zeroed) by AAA.{1} As a side effect of blanking out (zeroing) the upper half byte, it is possible to use the ASCII numbers '0' (30h) to '9' (39h) as addends. The result (after AAA) will still be a number between 00h and 09h with the carry flag either set or cleared, and AH incremented if appropriate. Don't use it this way. The multiplication and division instructions REQUIRE that the numbers be between 00h and 09h, so all numbers should be changed into unpacked BCD on data entry. Here is a partial list of things you can add to get the result 07h: 03h + 04h 'c' + 'd' '+' + 't' '3' + '4' 's' + 't' 'c' + '4' 'C' + 'D' '#' + '$' 'S' + '$' As you can see, the addition instruction is pretty indiscriminate about what kind of data you can put in and still get a legitimate unpacked number out. It is your job to make sure that you have ____________________ 1. They used the same microcode as for the beginning of DAA. If the low half byte of AL is greater than 9, or if there was a carry out of the low half byte (if AF, the auxillary flag was set), then the 8086 adds 6 to AL, which shifts the excess out of the low half byte. It then zeros the high half byte, sets the carry flag, and increments AH. If you don't understand this, go back and give the footnote on DAA a try. The PC Assembler Tutor 242 ______________________ legitimate unpacked data upon data entry. If this is just a side effect of blanking the upper half byte, why did they name the instruction "ascii adjust for addition"? It's just that impish sense of humor of those Intel engineers. If you think of these instructions as having anything to do with ASCII numbers, you will only confuse yourself. These are instructions on unpacked BCD numbers, period. The other thing to notice is that this instruction increments the AH register if there is a carry, so whenever you use AAA, it is going to trash the AH register. Count on it. If AH contains important data, store it somewhere else. SUBTRACTION When you have gotten used to how this works, look at subtraction. The subtraction routine is the same as the addition routine except that (1) ADD is replaced by SUB and (2) AAA is replaced by AAS (ascii adjust for subtraction). If you generate a borrow, the carry flag will be set and AH will be decremented by 1. If you have two legitimate unpacked BCD numbers, then after subtraction the result will be from +9 to -9. This result must be in AL. After AAS (ascii adjust for subtraction), (1) if AL was from 0 to +9, it will stay the same and CF will be cleared (CF=0) or (2) if AL was -1 to -9, AAF will borrow 1 from AH (considering it a ten's digit), and add 10 to AL. This will give a number from +1 to +9. It will also set the carry flag (CF=1) to signal a borrow. This is what you do with pencil and paper except that you always do the borrow BEFORE the subtraction and AAS always does the borrow AFTER the subtraction. Once again, this operates on the LOW HALF BYTE, so what is in the upper half byte of the numbers is irrelevant. It will be zeroed by AAS. There is all sorts of data which will generate a legitimate unpacked result; some of it is actually legitimate input. It too trashes the AH register, so beware. Run this program till you see what is going on with individual numbers. MULTIPLICATION There is also an instruction for multiplication. It assumes that AL contains the result of the multiplication of two unpacked BCD numbers. That is, 0 <= AL <= 81. After AAM (ascii adjust for multiplication), AH will contain the 10's digit and AL will contain the 1's digit. If AL is 75, then after AAM, AH will be 7 and AL will be 5. If AL is 48, then after AAM, AH will be 4 and AL will be 8. (If AL is 183, an illegal result, then after AAM, AH will be 18 and AL will be 3). We are going to use a similar program for multiplication, but AX and BL will be half byte unsigned; BH and DX will be half byte hex. Every time we change AX, we will copy it to DX. Here is the program: Chapter 22 - BCD Numbers 243 ________________________ ; - - - - - - - - - - START CODE BELOW THIS LINE mov ax_byte, 0A2h ; half registers, unsigned mov bx_byte, 0C2h ; half regs, hex, unsigned mov dx_byte, 0C4h ; half regs, hex lea ax, ax_byte call set_reg_style mov cx, 0 ; clear cx for clarity outer_loop: mov ax, 0 ; clear the registers mov bx, 0 mov dx, 0 call set_count call show_regs call get_hex_byte ; byte for al mov dx, ax ; copy to dx push ax ; save al call get_hex_byte ; byte for bl mov bl, al mov bh, bl ; copy to bh pop ax ; restore al call show_regs_and_wait mul bl ; unsigned multiplication mov dx, ax ; copy to dx call show_regs_and_wait aam ; make adjustment mov dx, ax ; copy to dx call show_regs_and_wait jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE All you need to change from the addition program is the register style, ADD -> MUL and AAA -> AAM. Try this out. Notice that after MUL, what we get in DX is garbage. This number is a pure unsigned number, not a BCD number. Just to underline how important it is to have legitimate unpacked BCD data, try a few multiplications where the upper half byte is non-zero and see what happens. DIVISION AAD (ascii adjust for division) is slightly different. What we want to do is divide a number from 0 - 99 by a number from 1 to 9 (0 gives a zero divide interrupt). Therefore, AAD multiplies what is in AH by 10 and adds it to what is in AL. It clears AH for the coming unsigned division.{2} If AH contains 7 and AL contains 2, then after AAD, AL will be 72 and AH will be 0. If AH is 4 and AL is 6, then after AAD, AL will be 46 and AH will be 0. (If AH is 14 and AL is 7 - an illegal situation - then after AAD, AL will ____________________ 2. Remember that for unsigned byte division, we set AH to 0. This was back in the first chapter on division. The PC Assembler Tutor 244 ______________________ be 147 and AH will be 0) After you have done AAD, you are ready for regular unsigned division. After division, the quotient will be in AL and the remainder will be in AH. Here's the program: ; - - - - - - - - - - START CODE BELOW THIS LINE mov ax_byte, 0A2h ; half registers, unsigned mov bx_byte, 0C2h ; half regs, hex, unsigned mov dx_byte, 0C4h ; half regs, hex lea ax, ax_byte call set_reg_style mov cx, 0 ; clear cx for clarity outer_loop: mov ax, 0 ; clear the registers mov bx, 0 mov dx, 0 call set_count call show_regs call get_hex ; word mov dx, ax ; copy to dx push ax ; save al for later call get_hex_byte ; byte for bl mov bl, al mov bh, bl ; copy to bh pop ax ; get back al call show_regs_and_wait aad ; adjust for division mov dx, ax ; copy to dx call show_regs_and_wait div bl ; unsigned division mov dx, ax ; copy to dx call show_regs_and_wait jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE The differences from the multiplication routine are (1) we get a TWO byte hex number for the dividend and (2) AAD comes first, then DIV. That's all. We need to enter a TWO byte unpacked BCD number in order to get grist for the mill. 87 is 0807h, 93 is 0903h, 42 is 0402h. This will give us two unpacked digits so we have something to put in AH. Run this program and enter legitimate (and if you want, illegitimate) data. Notice that even with legitimate data, it is possible to get a quotient that is larger than 9. (47 / 3 is 15, remainder 2). Don't worry about this. We will avoid this problem in the real program. PACKING AND UNPACKING Chapter 22 - BCD Numbers 245 ________________________ Making an i/o routine is such a bother that we will enter data with 'get_bcd' and output data with 'print_bcd'. To do this, we need to pack and unpack the data. The calls in C would be: unpack_bcd (&packed_number, &unpacked_number) ; pack_bcd (&unpacked_number, &packed_number) ;{3} The thing to be operated on is the first variable and the result is the second variable. First, here's the unpacking routine: ; - - - - - - - - - - unpack_bcd proc near UNPACK_UNPACKED_ADDRESS EQU [bp + 6] UNPACK_BCD_ADDRESS EQU [bp + 4] push bp mov bp, sp PUSHREGS ax, bx, cx, si, di mov si, UNPACK_BCD_ADDRESS mov di, UNPACK_UNPACKED_ADDRESS ; start at top to unpack add si, 9 ; bcd sign add di, 18 ; unpacked sign mov al, [si] ; move sign to unpacked mov [di], al dec si ; move down to next byte dec di mov cx, 9 ; 9 bcd data bytes ; unpack high byte first, then low byte unpack_loop: push cx ; save counter mov al, [si] ; bcd byte to al mov bl, al ; copy to bl mov cl, 4 ror al, cl ; high half byte to low half and al, 0Fh ; blank upper half byte mov [di], al ; al is high half of bcd byte dec di ; result pointer and bl, 0Fh ; blank upper half byte mov [di], bl ; bl is low half bcd byte dec di ; result pointer dec si ; source pointer pop cx ; restore counter loop unpack_loop POPREGS ax, bx, cx, si, di pop bp ret ____________________ 3. That '&' means that we are passing the addresses, not the values. The PC Assembler Tutor 246 ______________________ unpack_bcd endp ; - - - - - - - - - - - - - - - - This is pretty straightforward. First it moves the sign. Then, taking a BCD byte, the routine divides it in two parts, rotating the high half byte to the proper position, storing it, DECREMENTING the pointer and storing the low half byte. The upper half byte of each unpacked byte is zeroed. Notice that by starting at the top, it is possible to unpack a number in place. Diagram what is happening to make sure you believe that. Here's the packing routine: ; - - - - - - - - - - - - - - - - pack_bcd proc near PACK_BCD_ADDRESS EQU [bp + 6] PACK_UNPACKED_ADDRESS EQU [bp + 4] push bp mov bp, sp PUSHREGS ax, bx, cx, si, di mov si, PACK_UNPACKED_ADDRESS mov di, PACK_BCD_ADDRESS ; start at bottom to pack mov cx, 9 ; 9 bytes of bcd data pack_loop: push cx ; save counter mov al, [si + 1] ; high unpacked byte mov cl, 4 ror al, cl ; move to high half byte or al, [si] ; OR low half with high half mov [di], al ; move to BCD inc di ; adjust pointers add si, 2 pop cx ; restore counter loop pack_loop ; si and di are now in the right place for the sign mov al, [si] mov [di], al POPREGS ax, bx, cx, si, di pop bp ret pack_bcd endp ; - - - - - - - - This does the reverse process, rotating the high byte to the proper place, then ORing the two together to form a packed BCD byte. Notice that by starting at the BOTTOM it is possible to pack a number in place. Diagram the action to make sure you believe this. Chapter 22 - BCD Numbers 247 ________________________ These two routines allow us to use 19 byte unpacked BCD numbers. Here's the multiplication routine for a 19 byte (18 data bytes and 1 sign byte) unpacked number by a 1 byte unpacked number: ; - - - - - unpacked_multiply proc near PUSHREGS ax, bx, cx, dx, si, di mov si, offset multiplicand mov di, offset result mov dh, 0 ; clear dh for carry mov bl, multiplier_copy mov cx, 18 ; 18 data bytes mult_loop: mov al, [si] ; multiplicand to al mul bl ; multiply add al, dh ; add old carry aam ; adjust al mov [di], al ; partial result mov dh, ah ; save carry inc si ; adjust pointers inc di loop mult_loop mov extra_byte, ah ; extra byte POPREGS ax, bx, cx, dx, si, di ret unpacked_multiply endp ; - - - - - This is a clone of what we had in the chapter on multiple word multiplication but is byte multiplication instead of word multiplication. Refer back to that chapter to understand the mechanics of the process. We store the partial result and save the high byte for addition with the next multiplication. At the end we save the 19th byte for printout. We are cheating a little. we have: mul bl ; multiply add al, dh ; add old carry aam ; adjust al when we should have: mul bl ; multiply aam ; adjust al add al, dh ; add old carry aaa ; adjust al The reason this works is that the maximum multiplication is 9X9=81. The maximum addition is 81+9 = 90. AAM will work correctly with any number 99 or less, so we save a step; we do one adjustment instead of two. The PC Assembler Tutor 248 ______________________ Now, let's look at the driver for the program: ; + + + + + START DATA BELOW THIS LINE multiplier_message db "Enter a number from -9 to +9", 0 bcd_in dt ? bcd_out dt ? multiplicand db 19 dup (?) multiplier db ? multiplier_copy db ? result db 19 dup (?) extra_byte db ? result_sign db ? ; + + + + + END DATA ABOVE THIS LINE ; + + + + + START CODE BELOW THIS LINE outer_loop: mov ax, offset bcd_in call get_bcd call print_bcd ; reprint for clarity lea cx, multiplicand push cx ; unpacked address push ax ; bcd_in address call unpack_bcd add sp, 4 ; adjust the stack mov al, multiplicand + 18 ; sign byte mov result_sign, al ; either 00h or 80h enter_multiplier: lea ax, multiplier_message call print_string call get_signed_byte ; check for valid multiplier cmp al, 9 ; > +9 ? jg enter_multiplier cmp al, -9 ; < -9? jl enter_multiplier ; adjust multiplier sign mov multiplier, al mov multiplier_copy, al and al, 80h ; sign bit set? jz do_the_multiplication ; negative multiplier, so adjust neg multiplier_copy ; make positive xor result_sign, 80h ; reverse sign of result do_the_multiplication: call unpacked_multiply mov al, result_sign ; transfer sign to result mov result + 18, al ; pack the result mov ax, offset bcd_out ; bcd number push ax Chapter 22 - BCD Numbers 249 ________________________ mov ax, offset result ; unpacked number push ax call pack_bcd add sp, 4 ; adjust the stack ; print multiplicand, multiplier, extra byte and result mov ax, offset bcd_in call print_bcd mov al, multiplier call print_signed_byte mov al, extra_byte call print_hex_byte mov ax, offset bcd_out call print_bcd jmp outer_loop ; - - - - - END CODE ABOVE THIS LINE We enter a multiplicand, unpack it, enter a number from -9 to +9 and save a copy of its absolute value as well as the sign the result will be. We adjust the sign of the result after the multiplication. No matter what you enter, the sign will be correct, and if you put the 19th byte in front of the BCD number which you have printed, the absolute value will be correct. We are multiplying with a COPY of the multiplier, so we can reprint the actual multiplier; it hasn't changed. At the end we pack the result and print everything. The order of printout is multiplicand, multiplier, extra byte and result. Notice that we are not passing the parameters for unpacked_multiply on the stack. This is pure whim. A rule for when to pass on the stack is (1) If the subroutine doesn't know where the parameters will be located in memory, you MUST pass them on the stack but (2) if the subroutine knows for certain where the parameters will be, it can fetch them itself. DIVISION Here is the division routine for 19 byte unpacked numbers: ; - - - - - - - - unpacked_divide proc near PUSHREGS ax, bx, cx, si, di mov bl, divisor_copy mov si, offset dividend + 17 ; start at the top mov di, offset quotient + 17 mov cx, 18 ; 18 numeric bytes mov ah, 0 ; clear ah for division div_loop: mov al, [si] ; dividend byte to al aad ; adjust for unpacked number div bl ; bl is divisor mov [di], al ; move partial quotient dec si ; decrement pointers The PC Assembler Tutor 250 ______________________ dec di loop div_loop mov remainder, ah ; final remainder POPREGS ax, bx, cx, si, di ret unpacked_divide endp ; - - - - - It is pretty simple; it too is a clone of the multiple word division process. Go back to multiple word division if you don't understand it. We keep the previous remainder for the next division. Here is the driver: ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE divisor_message db "Enter a number from -9 to +9 (but not 0).",0 bcd_in dt ? bcd_out dt ? quotient_sign db ? remainder_sign db ? divisor db ? divisor_copy db ? dividend db 19 dup (?) quotient db 19 dup (?) remainder db ? ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE outer_loop: mov ax, offset bcd_in call get_bcd call print_bcd ; reprint for clarity lea cx, dividend push cx ; unpacked address push ax ; bcd_in address call unpack_bcd add sp, 4 ; adjust the stack mov al, dividend + 18 ; sign byte mov quotient_sign, al ; either 00h or 80h mov remainder_sign, al ; ditto enter_divisor: lea ax, divisor_message call print_string call get_signed_byte ; check for valid divisor cmp al, 0 ; 0? je enter_divisor cmp al, 9 ; > +9 ? jg enter_divisor cmp al, -9 ; < -9? jl enter_divisor ; adjust divisor, quotient sign mov divisor, al Chapter 22 - BCD Numbers 251 ________________________ mov divisor_copy, al and al, 80h ; sign bit set? jz do_the_division ; negative divisor, so adjust neg divisor_copy ; make positive xor quotient_sign, 80h ; reverse sign of quotient do_the_division: call unpacked_divide mov al, quotient_sign ; transfer sign to quotient mov quotient + 18, al test remainder_sign, 0FFh ; positive or negative? jz pack_the_quotient neg remainder ; make the remainder negative pack_the_quotient: mov ax, offset bcd_out ; bcd number push ax mov ax, offset quotient ; unpacked number push ax call pack_bcd add sp, 4 ; adjust the stack ; print dividend, divisor, quotient and remainder mov ax, offset bcd_in call print_bcd mov al, divisor call print_signed_byte mov ax, offset bcd_out call print_bcd mov al, remainder call print_signed_byte jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE Enter a BCD number, unpack it. Enter a signed number, do range checking, and if it is ok, store it (along with a copy, which we make sure is positive). Calculate the sign of the quotient and remainder. Do the division, then print the dividend, divisor, quotient and remainder. (which have been sign adjusted). The remainder is a signed byte. Like unpacked_multiply, unpacked_divide fetches the parameters directly from memory rather than from the stack. The PC Assembler Tutor 252 ______________________ SUMMARY PACKED BCD INSTRUCTIONS DAA (decimal adjust for addition) adjusts AL, assuming that it contains the result of a legitimate packed BCD addition. It treats AL as two independent half-bytes. If the result of the lower half-byte is 10 or over, it subtracts 10 from the lower half-byte and adds the carry to the upper half byte. It then looks at the upper half byte. If its result is 10 or over, DAA subtracts 10 from the upper half byte and sets the carry flag. Otherwise the carry flag is cleared. DAS (decimal adjust for subtraction) adjusts AL, assuming that it contains the result of a legitimate packed BCD subtraction.It treats AL as two independent half-bytes. If the result of the lower half-byte is -1 or less, it adds 10 to the lower half-byte and borrows 1 from the upper half byte. It then looks at the upper half byte. If its result is -1 or less, DAS adds 10 to the upper half byte and sets the carry flag to indicate a borrow. Otherwise the carry flag is cleared. UNPACKED BCD INSTRUCTIONS AAA (ascii adjust for addition) adjusts AL, assuming that it contains the result of a legitimate unpacked BCD addition. If the lower half-byte has generated a result 10 or over, it subtracts 10, carries 1 to AH, and sets the carry flag. If the result is 9 or less, it clears CF. In either case it zeroes the upper half-byte of AL. AAD (ascii adjust for division) PREPARES AL and AH for division. It assumes that AH contains the 10's digit and AL contains the 1's digit of a two byte unpacked BCD number. It multiplies AH by 10 and adds it to AL, thus making a single integer between 0 and 99. It zeroes AH in preparation for division. AAM (ascii adjust for multiplication) adjusts AL, assuming that it contains the result of a legitimate BCD multiplication. It divides the result by 10, putting the quotient in AH and the remainder in AL. AAS (ascii adjust for subtraction) adjusts AL, assuming that it contains the result of a legitimate unpacked BCD subtraction. If the lower half-byte has generated a result -1 or less, it borrows 1 from AH, adds 10 to AL, and sets the carry flag. If the result is 0 or more, it clears CF. In either case it zeroes the upper half-byte of AL Chapter 23 - XLAT ================= 253 The 800 pound gorilla in the computer field is, of course, IBM. It can go its own way and other companies have to adjust to keep themselves in line with what IBM is doing. You have been using ASCII characters since the first time you used BASIC (or whatever your first high-level language was). Every character has a unique number which represents it. character ASCII encoding A 65d a 97d ? 63d 0 48d IBM has its own encoding for mainframe computers. It is called EBCDIC (pronounced ebb'-sih-dick).{1} It is a spinoff of the coding on punch cards. You remember punch cards? This coding is entirely different from ASCII. Here are some examples. character ASCII code EBCDIC code a 97d 129d ? 63d 111d 0 48d 240d H 72d 200d I 73d 201d J 74d 209d K 75d 210d You can see that there is no relationship between the two encodings. Also, notice that while the alphabet is a continuous section of ASCII coding, there are breaks in the EBCDIC code (I=201, J=209). All PCs use ASCII, so if we want to transfer text from a PC to an IBM mainframe computer, we need to change ASCII -> EBCDIC going to the mainframe and change EBCDIC -> ASCII coming from the mainframe. This is the responsibility of the communications program that runs the modem, so you will never have to do it yourself. Intel has provided an instruction to help the communications program do this translation. It is called XLAT. In order to use XLAT, you need a translation table. This is a 256 byte array where each element of the array contains the result you want. Looking at the data above: ____________________ 1. Which stands for Extended Binary Coded Decimal Interchange Code. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor 254 ______________________ CHARACTER ASCII TO EBCDIC TABLE EBCDIC TO ASCII TABLE a array1 [97] = 129 array2 [129] = 97 ? array1 [63] = 111 array2 [111] = 63 0 array1 [48] = 240 array2 [240] = 48 H array1 [72] = 200 array2 [200] = 72 I array1 [73] = 201 array2 [201] = 73 J array1 [74] = 209 array2 [209] = 74 K array1 [75] = 210 array2 [210] = 75 We have two different tables here. Array1 takes the ASCII encoding and gives back the EBCDIC encoding. Array2 takes the EBCDIC encoding and gives back the ASCII encoding. For each character, the appropriate table gives the correct translation from one encoding to another. All we need now is the translation instruction. Put the address of the translation table in BX. This table should be in the DS segment, but DS may be overriden: mov bx, offset ascii_to_ebcdic_table Put the character you want translated in al: mov al, character translate: xlat To translate a 20 byte string of ASCII data into EBCDIC, you might have the following code: ;---------- mov di, offset ebcdic_string mov ax, seg ebcdic_string mov es, ax mov si, offset ascii_string mov bx, offset ascii_to_ebcdic_table mov cx, 20 ; translate 20 bytes cld ; clear DF (increment) translation_loop: lodsb ; ascii to al xlat ; translate stosb ; al to ebcdic loop translation_loop ; ---------- Since this is ASCII to EBCDIC, if AL contained 63 before XLAT, then after XLAT AL would contain 111. If AL contained 73 before XLAT, then after XLAT it would contain 201. If AL contained 97 before XLAT, after XLAT it would contain 129. If we wanted to go the other direction we would have to make the EBCDIC string the source string, make the ASCII string the Chapter 23 - Xlat 255 _________________ destination string, and use the other table: mov bx, offset ebcdic_to_ascii_table The rest of the code would be the same. Since this is done by the communications program, we won't concern ourselves with ASCII <-> EBCDIC any more, but we will use XLAT in two slightly different ways. First, let's categorize characters. Some things are Whitespace (that is, tabs, newlines, spaces, form feeds, etc.) Some characters are octal, decimal, punctuation, hex, etc. There is a pre-existing table called translation_table in the subdirectory XTRAFILE. Its pathname is \xtrafile\transtbl.obj. It has all 256 ascii characters coded in the following way: WHITESPACE EQU 80h ; 1000 0000 PUNCTUATION EQU 40h ; 0100 0000 ALPHABETIC EQU 20h ; 0010 0000 OCTAL EQU 10h ; 0001 0000 DECIMAL EQU 08h ; 0000 1000 HEX EQU 04h ; 0000 0100 BOX_CHAR EQU 02h ; 0000 0010 GREEK_CHAR EQU 01h ; 0000 0001 If the character is whitespace, then the leftmost bit is set. If it is a greek character (ascii 224 - 239 on the PC) then the rightmost bit is set. If it is more than one thing, then the appropriate bits are set. For instance, '6' is octal, decimal and hex, so it's encoding is: '6' 0001 1100 'a' is both alphabetic and hex, so it's encoding is: 'a' 0010 0100 The following program inputs a character, and finds out whether it is punctuation, a letter, etc. If it is none of the eight things, then the program prints that nothing was found. It is the same block of code over and over, so you might want to do only part, or you might want to cut it out with a word processor and insert it in the template file (don't forget to delete the page headers and page numbers). ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE EXTRN translation_table:BYTE ;\xtrafile\transtbl.obj whitespace_banner db "It is whitespace." , 0 punctuation_banner db "It is punctuation." , 0 alphabet_banner db "It is alphabetic." , 0 octal_banner db "It is octal." , 0 decimal_banner db "It is decimal." , 0 The PC Assembler Tutor 256 ______________________ hex_banner db "It is hex." , 0 drawing_banner db "It is a box drawing character." , 0 greek_banner db "It is a Greek character." , 0 nothing_banner db "No match was found." , 0 dirty_flag db ? ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE WHITESPACE EQU 80h PUNCTUATION EQU 40h ALPHABETIC EQU 20h OCTAL EQU 10h DECIMAL EQU 08h HEX EQU 04h BOX_CHAR EQU 02h GREEK_CHAR EQU 01h ; set up the xlat table mov ax, seg translation_table mov es, ax mov bx, offset translation_table outer_loop: mov dirty_flag, 0 ; marker for success call get_ascii_byte ; input a byte to al xlat es:[bx] ; do the translation test al, WHITESPACE jz punct_check push ax ; save translation in al mov ax, offset whitespace_banner call print_string pop ax mov dirty_flag, 1 ; set the dirty flag punct_check: test al, PUNCTUATION jz alpha_check push ax ; save translation in al mov ax, offset punctuation_banner call print_string pop ax mov dirty_flag, 1 ; set the dirty flag alpha_check: test al, ALPHABETIC jz octal_check push ax ; save translation in al mov ax, offset alphabet_banner call print_string pop ax mov dirty_flag, 1 ; set the dirty flag octal_check: test al, OCTAL jz decimal_check Chapter 23 - Xlat 257 _________________ push ax ; save translation in al mov ax, offset octal_banner call print_string pop ax mov dirty_flag, 1 ; set the dirty flag decimal_check: test al, DECIMAL jz hex_check push ax ; save translation in al mov ax, offset decimal_banner call print_string pop ax mov dirty_flag, 1 ; set the dirty flag hex_check: test al, HEX jz drawing_check push ax ; save translation in al mov ax, offset hex_banner call print_string pop ax mov dirty_flag, 1 ; set the dirty flag drawing_check: test al, BOX_CHAR jz greek_check push ax ; save translation in al mov ax, offset drawing_banner call print_string pop ax mov dirty_flag, 1 ; set the dirty flag greek_check: test al, GREEK_CHAR jz nothing_check push ax ; save translation in al mov ax, offset greek_banner call print_string pop ax mov dirty_flag, 1 ; set the dirty flag nothing_check: cmp dirty_flag, 0 ; was anything found? je print_nothing_banner jmp outer_loop print_nothing_banner: mov ax, offset nothing_banner call print_string jmp outer_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE you need to: link prog1+transtbl+\asmhelp ; The PC Assembler Tutor 258 ______________________ The program is long, but straightforward. Input a character and get its encoding. Test for each characteristic. If it is found, print the appropriate message and set the dirty_flag to indicate something was printed. At the end, if nothing was printed, print the failure message. Notice that the translation table is in ES and we are using a segment override for it. If you look at the EXTRN statement for 'translation_table', you will see that even though we are using ES, it is declared EXTRN in a segment with an: ASSUME ds:DATASTUFF statement. How can we get away with this? The assembler never deals with 'translation table' directly. The only thing it does is put the offset in BX. We put the segment override in ourselves with: xlat es:[bx] so the assembler never has to decide whether a segment override is necessary or which segment override to use. WORD SEARCH When doing the mock word search program in the chapter on string instructions, I mentioned that it really wouldn't cut the mustard when it comes to real word searches. Why? If we are looking for "when" we also want to find "When". If we are looking for " searches ", we also want to find " searches,", that is, punctuation should not interefere unless we want it to, and capitals should not interefere unless we want them to. With the aid of a translation table, we will make a word search program which uses the following rules. In the SEARCH string (the string that defines what you are looking for): (1) Any small letter will match either a small or large letter. (2) A capital letter will match only a capital letter. (3) A blank will match any whitespace or punctuation. (4) A punctuation mark will only match itself. With these rules "Why" must start with a capital 'W' to be a match, but 'h' and 'y' may be either capital or small. " some," may have any whitespace (including a carriage return) in front, but must hava a comma ',' at the end. This program has two data files. \XTRAFILE\SRCHTBL.OBJ contains the translation table. It is called "wordsearch_table" and is in DATASTUFF, so will be in our normal DS segment. In order to have text to search I have included an object file that is the text of a chapter from a book. (The object file text includes carriage returns). The text is a C string - it is terminated by a 0. The book was written by C.D. Huffam, and is the autobiographical account of his dual life as a writer and lecturer. The book is Chapter 23 - Xlat 259 _________________ called "A Tale of Two C.D.s". The object file with the text is \XTRAFILE\TWOTALE.OBJ. It is in a private segment and will use ES as a segment register. There is also a straight text file which you can print out so you can see what is in the object file. It is \XTRAFILE\TWOTALE.DOC. Here's the program. The explaination is at the end. ; + + + + + + + + + + + + + + + START DATA BELOW THIS LINE EXTRN tale_text:BYTE, wordsearch_table:BYTE entry_message db 13,10, "Enter a word for a word search", 0 no_match_message db "There was no match", 0 input_buffer db 80 dup (?) text_file_length dw ? letter_count dw ? ; + + + + + + + + + + + + + + + END DATA ABOVE THIS LINE ; + + + + + + + + + + + + + + + START CODE BELOW THIS LINE ; find the length of the text file mov ax, seg tale_text ; load es register mov es, ax mov di, offset tale_text ; offset to di mov bx, di ; copy to bx mov al, 0 ; try to match zero cld ; clear DF (increment) string_end_loop: scasb ; search for zero jne string_end_loop dec di ; one too many , so decrement sub di, bx ; finish - start = length mov text_file_length, di ; length of text_file big_loop: ; get a word for the word search mov ax, offset entry_message call print_string mov ax, offset input_buffer call get_string ; find the end of string mov al, 0 ; compare with 0 mov bx, offset input_buffer mov cx, 0 ; letter count letter_count_loop: cmp al, [bx] ; compare to 0 je end_of_count_loop inc cx ; increment count inc bx ; increment pointer jmp letter_count_loop end_of_count_loop: cmp cx, 0 ; if 0, string is empty The PC Assembler Tutor 260 ______________________ je big_loop ; so start again mov letter_count, cx ; look for word match. In this program, the text string ; is referenced by si and the search string is referenced ; by di. mov si, offset tale_text mov cx, text_file_length ; length of file sub cx, letter_count ; last possible match inc cx ; +1 for boundary condition ; set up translation table ( it is in DATASTUFF ) mov bx, offset wordsearch_table word_search_loop: push si ; save a copy push cx ; save a copy mov di, offset input_buffer mov cx, letter_count letter_loop: mov al, es:[si] ; text to al cmp al, [di] ; same as search string? je next_letter xlat ; if not, translate cmp al, [di] ; allowable substitute? jne new_start ; if not, start at new place next_letter: inc di ; move to next letter inc si loop letter_loop ; we fell through, so we found a complete match jmp found_it ; no match. are we finished? new_start: pop cx pop si inc si ; move to next character loop word_search_loop ; we fell through. finished, but no match mov ax, offset no_match_message call print_string jmp big_loop found_it: pop cx ; take cx off the stack pop si ; start of the match ; move 25 characters to buffer for printing mov di, offset input_buffer mov cx, 25 Chapter 23 - Xlat 261 _________________ character_move: mov al, es:[si] mov [di], al inc si ; increment pointers inc di loop character_move mov BYTE PTR [di], 0 ; end of string mov ax, offset input_buffer call print_string jmp big_loop ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE You need to: link prog2+twotale+srchtbl+\asmhelp ; to get asmhelp and the two data files in the program. This program is very similar to the search program in the chapter on strings. However, because of where the files are, the pointers have been changed around. Therefore, it is safer if you simply cut out the program with a word processor and paste it into the template file rather than try to modify the prevoius search program.{2} It is assumed that you did the string match program. The logic is the same and will not be covered again. First we input a search string. Then starting at the beginning of the text to be search we check till we find the first match. If we find a match, we print out 25 characters starting with the first character of the match. If no match is found, a message to that effect is printed. The character match is a two step process. The character from the text is put in AL. It is compared with the search character for an EXACT match. If they match, we are done. If not, we use XLAT on AL (the character from the text) which will translate to its allowable substitute. In fact, all this is just: (1) all capital letters become small, (2) all punctuation becomes spaces, and (3) all whitespace becomes spaces. Once again, we compare AL with the search character. If we have a match, ok. If not, we start over. The text is in ES, the translation table is in DS, so it is inconvenient to use the string instructions in this program. Try to match a word at the beginning of the line, end of the line, with and without punctuation and with and without capitals. If you go across a line break, you need to substitute two blanks in the search string for CRLF (13,10). ____________________ 2. You should understand what is going on in the code before you run these programs. I didn't write the code for myself, I wrote it for you. If you run it but don't understand it, it won't help you a bit. The PC Assembler Tutor 262 ______________________ Suppose you are not interested in all 256 values of the translation table. Let's say that you only want to have a translation table for the numbers from 0 to 99. Can you still use this? Yes, but you need to put in some range checking to make sure that you have valid data. MAX_VALUE EQU 99 mov al, data_byte ; byte to al cmp al, MAX_VALUE ; too large? ja data_error ; report error xlat This insures that any data that is out of range is not translated. Therefore the translation table only needs to be 100 bytes long (0 - 99). If you want more than 256 elements in the translation table you need to use words, not bytes, and you cannot use XLAT. You can make your own code to do the same thing. MAX_VALUE EQU 999 my_translation_table dw 1000 dup (?) if you put the translation data into the table, you can then have the following code: mov bx, offset my_translation_table ; - - - - - translation block mov si, data_word ; word to si cmp si, MAX_VALUE ; too large? ja data_error shl si, 1 ; SI x 2 = number of BYTES into table mov ax, [bx+si] ; base + offset ; - - - - - end of translation block XLAT is about twice as fast as this last code, so when you have a choice always use XLAT. Chapter 23 - Xlat 263 _________________ SUMMARY XLAT BX holds the address of a 256 byte array called a translation table. AL holds the character to be translated. If x is the value in AL before XLAT, then after XLAT, AL=array[x]. Chapter 24 - Miscellaneous Instructions ======================================= 264 There are a few more assembler instructions which have not been covered. Some are seldom used and some are used in special circumstances. This chapter gives a brief explanation of them. XCHG XCHG, the exchange instruction, switches the contents of two registers or of a register and a memory location. xchg ax, bx xchg al, dl xchg variable1, si xchg ch, variable2 xchg [si], ax xchg bh, [di] It operates on either a word or a byte. It cannot switch two memory locations, and it does not operate on the segment registers, only on the 8 arithmetic registers. ESC The escape instruction is not a complete instruction, it is the beginning of an instruction. It signals the 8086 that the first two bytes of the instruction contain a co-processor instruction. The 8087 is a mathematics co-processor. It reads the instructions at the same time as the 8086, and when it sees an escape instruction meant for it, it performs that operation. The 8086 does nothing unless there is a memory address involved. In that case the 8086 calculates the address and gives the address to the 8087. fld DWORD PTR [si] fmul st, st(3) fisub WORD PTR [di] are all 8087 instructions that the assembler codes with the escape code. You will never code the escape instruction yourself. WAIT (FWAIT) The 8087 operates independently of the 8086. They can both perform operations at the same time, but it is possible for them to interfere with each other. If both the 8086 and 8087 are reading from or writing to memory, or if one is reading from while the other is writing to memory, the read/write will be corrupted. In order to stop this, whenever you access memory with the 8087, you need to put in a WAIT instruction. WAIT (or FWAIT ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson Chapter 24 - Miscellaneous Instructions 265 _______________________________________ which is the same thing), suspends operation of the 8086 until the co-processor is finished. It would look like this: fstp DWORD PTR [bx] fwait The 8086 will wait until the 8087 is finished storing into memory. There must also be a wait between 8087 instructions. This is to keep the 8087 from starting a second instruction before it is finished with the first one. The instruction execution is done by the 8087, but the timing is done by the 8086. If you had the folllowing code: fmul st, st(3) fadd st, st(2) fsub st, st(4) the 8086 would be past the third instruction before the 8087 had time to finish doing the first instruction. The proper coding is: fmul st, st(3) fwait fadd st, st(2) fwait fsub st, st(4) This should concern you only if you program the 8087. It is outside the realm of this book. Remember, WAIT and FWAIT are the same instruction. LOCK LOCK is not really of concern to a PC programmer. On some systems it is possible to have more than one 8086 that has access to the same memory. The problem then arises as it did with the 8087 that there is the possibility of two 8086s doing read/write operations to memory at the same time. This will result in corrupted data. LOCK allows an 8086 to lock out other 8086s. It will be the only one allowed to read to or write to memory during the next instruction. This is mostly of concern to people who write code for peripheral devices which have DMA (direct memory access). LOOPE/LOOPZ LOOPNE/LOOPNZ We have used the loop instruction, but these are special cases. Remember, the general loop instruction decrements CX by 1, and if the result is not zero, it jumps to the top of the loop. In special circumstances, you not only want to check the counter in CX, but you also want to check the result of some other operation. test ax, 5 loope outer_loop The PC Assembler Tutor 266 ______________________ will loop if cx is not zero AND ax = 5 test ax, 5 loopne outer_loop will loop if cx is not zero AND ax is not 5. HALT HALT actually halts the operation of the 8086. It can only be started again by a reset or an interrupt. If you write: cli ; clear interrupt flag halt Then normal interrupts can't happen and you have effectively shut down the system. One place this might be of use is if you have a copy protection scheme and you detect that someone has violated it. It halts the system and you need to reset to start again. Another place might be if you are writing a game - someone inadvertently enters the "Dungeon of Darkness" and you shut the system down. CMC CMC (complement the carry flag) toggles the carry flag. If it is off this turns it on, and if it is on, this turns it off. Why this instruction exists is a complete mystery to me. Why would you want it as part of the instruction set? LAHF LAHF (load AH from flags) stores half of the flags in AH. The register looks like this: 7 6 5 4 3 2 1 0 S Z A P C Where these are the Sign, Zero, Auxillary, Parity and Carry flags. This is a leftover from earlier Intel chips. All these flags are testable so you dont need to look at them, and if you want to save them, then: pushf does the trick. Notice that AH does not contain DF, the direction flag or IEF, the interrupt enable flag, which are things you CAN'T test with a jump instruction. To test for them you need to: pushf pop ax Then AX contains all the flags. Chapter 24 - Miscellaneous Instructions 267 _______________________________________ SAHF SAHF (store AH to flags) is the counterpart to the above instruction. It puts AH into the low half of the flags register. The comments about LAHF apply to this instruction also. NOP And finally, NOP does absolutely nothing. It is there in case you need to fill space, either because you have taken out an instruction or for reasons of alignment. It also allows a debugger to put the single byte INT (int 3) in the code and then replace it with NOP when the breakpoint is no longer desired. Chapter 25 - What Does it all mean? =================================== 268 What does it all mean? Not a whole lot, actually. We can now save memory space and we can save a lot of time. But with the incessant march of technology these things mean less and less. A few years ago, when 64k or 128k was a lot of memory and memory was expensive, having a 20k program instead of a 40k program was a significant advantage. Now it means almost nothing unless it is a memory resident program. What about disk space? Just a while back we were operating with two 360k floppy disks and a hard disk was too expensive. Nowdays everyone has a 20meg hard disk.{1} And speed? Those programs that were slow on an 8088 now seem o.k. on an 80386. Those programs that were unbearably slow on the 8088 died a quick death and are no longer around. Compilers are better and they have more subroutines available. They are also easier to program than going to the assembler level. What this chapter is about is when NOT to use the standard compiler functions and subroutines. First, you should understand that all compiler subroutines are general purpose subroutines. They need to be all things to all people. Imagine what a vehicle would be like if we gave the designer the following specification: We want to be able to drive to the store for groceries. It should be fuel efficient. In case we want to go into the mountains it should be an all terrain vehicle. We also want to be able to haul a roomful of furniture from coast to coast. Oh yes, and we want to be able to race it at Le Mans. Being universal requires a lot of code and it slows things down. Whether this extra code and time is too much is a question you need to decide for yourself. First, here are some examples of size. This is a C program that does almost nothing: #include <stdio.h> main() { int x ; x = 27 ; /* line 1 */ scanf ( "%d", &x ) ; /* line 2 */ printf ( "%d\n", x ) ; /* line 3 */ } ____________________ 1. Which has led to one of my pet peeves. All installation programs for compilers and word processors dump EVERYTHING on the hard disk. This gives us subdirectories that have 50 files in them, and we don't have the foggiest notion of what any of the files are for. If these installation programs would only prompt us by type of file to find out what we want to install and want to leave off the hard disk, we would all be better off. ______________________ The PC Assembler Tutor - Copyright (C) 1990 Chuck Nelson Chapter 25 - What Does It All Mean? 269 ___________________________________ I have made 3 programs from this. LINE1.C has line1 only. LINE2.C has lines 1 and 2. LINE3.C has lines 1, 2 and 3. For those non C people, scanf is an input function, printf is an output function. Guess how big each program is. Here's the directory listing. LINE1 EXE 3176 6-22-90 8:48a LINE2 EXE 7170 6-22-90 8:49a LINE3 EXE 9134 6-22-90 8:49a It takes 3000 bytes to start a C program (this is the startup module) 4000 bytes more to enter something and 2000 bytes extra to print something. That first 3000 bytes is unavoidable if you are writing in a high level language. If you are doing a lot of general purpose i/o, these extra amounts aren't too bad. There are two cases where you might want to use your own i/o routines. First, if you have something simple or secondly, if you have something special, you want to do your own i/o. If you don't need all that flexibility, you are better off doing your own i/o. Here are two files that write a text screen to a disk file. COPYSCRN EXE 10454 6-10-90 9:30a INTSCRN COM 445 6-12-90 7:40p They both do the same thing except that the .COM file is a little more sophisticated. Notice the difference in size. Speed really doesn't play a part here because what they do is so simple that it takes just a second in any case. The program was so simple that it only took an hour or two to write, so I didn't lose any time by writing it in assembler. The other case is when you have a specific idea of what the screen should look like. You want control of the whole screen all the time. This includes all word processors, databases, programming environments, etc. They all take charge of the screen because some DOS functions are too slow. If you remember from the ZOOM chapter, there is a radical difference between what you can do and what DOS can do. Even though these large programs are written in C, they all bypass the C i/o functions. That does not mean that they go down to the assembler level, however. INTERRUPTS You have done a few interrupts. They call the standard DOS or BIOS functions. Remember, they do this by going into low memory (the first 1k of memory) and getting the address of the subprogram that handles that particular interrupt. However, you do not need to call these interrupts from the assembler level. All modern compilers support interrupt calls in the language. If yours doesn't, you need a more recent compiler. Before going on with this chapter you need to read your compiler documentation about interrupts. TURBO Pascal has INTR, QuickBASIC and QuickC have INT86. Read the documentation now. The PC Assembler Tutor 270 ______________________ Have you read it? No cheating is allowed, because you won't understand the rest of this if you haven't read it. Though technically C is a structure and Pascal is a record, they are actually arrays where each array element has a specific name. The interrupt routine reads all these values into the corresponding register, calls the interrupt, then reads the register values back into the array. Some languages have one array for the input and another for the output. Int 21h is special so QuickC has a special function called INTDOS. It is the same as using Int 21h. The order of registers in the array is arbitrary and language dependent. For TurboPascal it is (AX, BX, CX, DX, BP, SI, DI, DS, ES, FLAGS). You enter values in the registers specified by the interrupt, and then call the interrupt. The routine does the rest. INTR ( int_no: byte, var the_regs: Registers) This will push the interrupt number, then the array address. On entry to the interrupt call and after initializing BP, we will have: int_no bp + 6 array_address bp + 4 old IP bp + 2 bp -> old BP bp What follows is not the exact code, but is similar to what the Pascal routine does: ; - - - - - - - - - - intr proc near push bp mov bp, sp push ax ; save all registers except SP, DS, SS, CS push bx push cx push dx push si push di push bp ; this is OUR bp push es ; insert the interrupt number in the interrupt mov al, [bp+6] ; AL now contains the interrupt number lea si, interrupt_spot ; where the interrupt is mov cs:[si+1], al ; insert it in the interrupt ; change all the registers mov si, [bp+4] ; array address is DS:SI mov ax, [si] mov bx, [si+2] mov cx, [si+4] mov dx, [si+6] mov bp, [si+8] Chapter 25 - What Does It All Mean? 271 ___________________________________ mov di, [si+12] mov es, [si+16] ; special manipulation for DS and SI push ds ; save ds push si ; save si push ax ; temp save of ax from array mov ax, [si+14] ; ds from array to ax mov si, [si+10] ; si from array to si mov ds, ax ; now move ax to ds pop ax ; restore ax ; call the interrupt interupt_spot: int 0 ; dummy number for the interrupt ; special needs for SI and DS ; our SI and DS are at the top of the stack ; save values of flags, si and ds from interrupt pushf ; value from interrupt push si ; value from interrupt push ds ; value from interrupt add sp, 6 ; get to our si and ds pop si ; our si pop ds ; our ds sub sp, 10 ; sp is where it was a moment ago. mov [si], ax ; DS:SI points to array mov [si+2], bx mov [si+4], cx mov [si+6], dx mov [si+8], bp mov [si+12], di mov [si+16], es pop [si+14] ; ds from the interrupt pop [si+10] ; si from the interrupt pop [si+18] ; flags from the interrupt add sp, 4 ; skip our DS and SI (already in regs) pop es pop bp ; this is OUR bp pop di pop si pop dx pop cx pop bx pop ax mov sp, bp pop bp ret (4) ; clear arguments off the stack ; - - - - - - - - - - - - - - - - - - - - This should test your insight into using code. DS and SI are needed for moving data, so we use some kludges to get it to work. There are two things here that you shouldn't normally do. First, The PC Assembler Tutor 272 ______________________ we are inserting the interrupt number directly in the machine code. Secondly, we are playing around with the value of SP. These are rare exceptions and shouldn't occur in your own code unless absolutely necessary. The first thing you are going to say is, "Gosh, that's a lot of code for one interrupt." True, especially when the interrupt is interrupt 12h. Here's int 12h inside of our template file: ; - - - - - START CODE HERE int 12h ; machine memory (return in ax) call print unsigned ; - - - - - END CODE HERE It finds out how much memory your computer has and returns the number of kbytes in AX. But how much extra time does using this Pascal interrupt routine take? About 700 clocks or about .0002 seconds (that's right, 2 ten thousandths) on the slowest machine. How many times will you call it during a program? Only one time. There is no point in going down to the assembler level to write a program that saves you .0002 seconds. In Pascal, you would write: INTR ( $12 , the_regs) ; and be done with it. No big loss of time and no trouble at all. In fact, as far as I can see, there is no reason for doing any interrupts from the assembler level. You may want to do a whole subprogram that contains interrupts, but if you just need one or two interrupts, it is easier to work from inside the high-level language. This includes the i/o we were talking about a minute ago. Yoy can write a screen program inside a high level language using arrays. Just think of a screen as a 80X25 array. If a two dimensional array is too slow you need to go to a one dimensional array. All interrupts that tell what kind of video card is in the computer, what mode the screen is in, etc. can be done from the high-level language. The most you need assembler for (depending on the language) is moving the text array into video memory. You want a bunch of help screens? Put all the help screens in a single file and use the interrupt for random access file read to read a screen when you need it.{2} Anything else? Yes, we still have the need for speed. There are certain types of operations like block moves of data, word searches and sorting of arrays that are characterized by large amounts of data and/or large amounts of computation. If you think you see a way to use registers effectively for one of these ____________________ 2. What you actually want to do is have the first block of data in the file tell you where each screen is and how long its data is. Then the first 2 bytes or words of the screen data should say the dimensions of the screen data ( 12 X 25, 17 X 3, etc.). This will allow you to store and use screens of any size. Chapter 25 - What Does It All Mean? 273 ___________________________________ things, you probably can beat a compiled version of the subprogram. Then the only question is whether or not it is worth the trouble. We have used the words "fast" and "slow" ambiguously so far, but now it is time to quantify them. Before you get the numbers, you need to know one thing about memory. People always talk about the "data bus". What is it? It is a group of wires connecting the 80x86 chip to memory. The 8088 has 8 wires, the 8086, 80286 and 80386/SX have 16 wires, and the 80386 has 32 wires. That means that the 8088 can transfer 8 bits of information at one time, the 8086 et. al. can transfer 16 bits at a time and the 80386 can transfer 32 bits at a time. This means one byte, two byte and four byte transfers respectively. This also means that the memory bytes are ordered a little differently. You will never notice it externally, but here is the different internal ordering. The 8088 has all bytes one after the other. All memory read/writes are done with the same 8 wires: 8088 MEMORY ADDRESSES 00005 00004 00003 00002 00001 00000 data lines |||||||| (8 bits) (All our examples will use absolute memory locations starting at 00000). The chips with a 16 bit data bus have all the even locations on the first 8 wires and the odd locations on the other 8 wires. They come in pairs - first even then odd: 8086 MEMORY ADDRESSES 00006 00007 00004 00005 00002 00003 00000 00001 data lines |||||||| |||||||| (16 bits) When one of these chips reads or writes, it can read/write either the left or the right byte or the whole word. What it cannot do is read the right byte from one pair along with the left byte from another pair. If you want to read the word at 00005:00006, the 8086 must: 1) read the 00005 byte. 2) read the 00006 byte. 3) join them together. This takes longer than just a single word read. The PC Assembler Tutor 274 ______________________ The true 80386 has a 32 bit data bus. This allows it to read 4 bytes at a time, and its physical memory structure looks like this: 80386 MEMORY ADDRESSES 00010 00011 00012 00013 0000C 0000D 0000E 0000F 00008 00009 0000A 0000B 00004 00005 00006 00007 00000 00001 00002 00003 data lines |||||||| |||||||| |||||||| |||||||| (32 bits) Instead of memory pairs, we now have memory quadruplets. As long as a word is totally inside of one quadruplet, the read/write time will be unaffected. If the read/write crosses the boundary (as we did above), the read/write time will be affected in the same way. The 80386 can also read 4 byte data quickly as long as the total data is inside of one memory quadruplet. In the 8086 family, data can always be read across these boundaries but it takes more time. (On the IBM 370, on the other hand, there are instructions that REQUIRE that data be aligned along 32 bit boundaries). This means you should order your data in the following way in the data segment: QWORD DATA DWORD DATA TBYTE DATA ; this is for the 8087 WORD DATA BYTE DATA ; all strings, etc. This insures that any read/write for that type of data will always be as fast as possible. If the segment definition has no alignment type, it will start on a paragraph boundary - i.e. every 16 bytes, and will work with anything. {3} In addition, if you ever subtract a number from SP to provide for a temporary data area, it should always be an even number. If SP is at an odd address instead of an even address, it takes longer for PUSHes and POPs. Also, when you define the size of the stack segment, it should be an even number of bytes. Having said that, it is now time for you to see the speeds of ____________________ 3. The alignment type is a word after the word SEGMENT which says how the segment should be aligned. The following: DATASTUFF SEGMENT BYTE PUBLIC 'DATA' says the segment can be aligned at any byte. The allowable forms are BYTE, WORD, DWORD, PARA, PAGE (256 bytes). If there is no explicit type, the default is PARA. Chapter 25 - What Does It All Mean? 275 ___________________________________ instructions. Read the introduction to APPENDIX III, then glance at the times to get the general idea of how fast times are. Come back to this chapter when you are comfortable with what the times look like. Have you read APPENDIX III? If not, do it before going on. The compiled languages all have one thing in common. They tell you that if you are writing a subroutine, you need to return from the subroutine with DS, BP, SS, and SP unchanged. They don't say a thing about any of the other registers. One thing this tells us is that they are doing everything from memory locations, not register locations. If you have taken a good look at the execution times, you will have noticed the phenomenal difference in time between a "memory, register" addition and a "register, register" addition. Now, if all you are going to do is move a number to a register, add it, and move it out again, a compiler can do it as fast as you can. But, if you run into a situation where you can use three or four registers at the same time, you can cut the execution time drastically. Compilers really can't use registers as efficiently as we can (yet). This is an ideal spot for using assembly language. The old adage that 10% of the code uses 90% of the computer time is appropriate here. You now know about assembler language, and you know what you want to do with it, so go out and enjoy. But before you do, try to slog your way through the next chapter on "simplified" segment definitions and linking to high level languages. Chapter 26 - Simplifying the Template ===================================== 276 By the time you have finished this chapter your assembler files will look cleaner. Unfortunately there is some heavy sledding before we get there. EXITING A PROGRAM Till now, we have exited most programs with CTRL-C; otherwise the program has done a return. A return to what? It has been returning to a section of code that does INT 20h, one of the ways of quitting a program when everything is in order. Notice the "everything is in order" in the last sentence. What happens if you have 2 files open, you are off in some subroutine, and you have things so hopelessly confused that you might as well give up? Can you call INT 20h? The answer is no for two reasons. First, you need CS to point to the PSP (program segment prefix). and you don't know where the PSP is. Secondly, you need to close files. Now, it is possible to make some code to do this, but why bother. We have a special interrupt for this: INT 21h function 4Ch AH = 4Ch AL = return code This will close all files, get you out of the program, and give a return code that is usable by the calling program. Here's a small program. Use template.asm and call this TEST4CH.ASM. ; - - - - - - - - - - - - - - - - - - - - - - - - - CODESTUFF SEGMENT PUBLIC 'CODE' ASSUME cs:CODESTUFF, ds:DATASTUFF EXTRN get_unsigned_byte:NEAR main proc far start: mov ax, DATASTUFF ; load ds mov ds,ax call get_unsigned_byte ; value is in al mov ah, 4Ch ; int 21h, function 4Ch int 21h main endp CODESTUFF ENDS ; - - - - - - - - - - - - - - - - - - - - - - - - - - We have revised the CODESTUFF segment so there is only one EXTRN statement and the beginning code: ______________________ The PC Assembler Tutor - Copyright (C) 1990 Chuck Nelson Chapter 26 - Simplifying The Template 277 _____________________________________ push ds sub ax, ax push ax is gone. Since we will never again do a return to the PSP, there is no need for this code anymore. The program gets a single byte to use as the exit code and then exits using int 21h function 4Ch. Get this assembled and linked. It should ask for a number and then exit. But where is that number? It is available through the operating system. Make the following batch file. It runs TEST4CH.EXE and then looks at the error code. Unfortunately ERRORLEVEL is not available as an exact number to a batch file, so we are checking to see if the return code was above a certain level. ----------------- DO4CH.BAT ----------------------- test4ch ECHO OFF IF ERRORLEVEL 1 ECHO The return code was over 0 IF ERRORLEVEL 51 ECHO The return code was over 50 IF ERRORLEVEL 101 ECHO The return code was over 100 IF ERRORLEVEL 151 ECHO The return code was over 150 IF ERRORLEVEL 201 ECHO The return code was over 200 ECHO ON ---------------------------------------------------- Here's one run of the batch file: >do4ch >int4ch The PC Assembler Helper Version 1.01 Copyright (C) 1989 Chuck Nelson All rights reserved. Enter a number from 0 to 255 172 >ECHO OFF The return code was over 0 The return code was over 50 The return code was over 100 The return code was over 150 This is what happens to DO4CH.BAT with a return code of 172. From now on, always use INT 21h function 4Ch to exit. SEGMENTS Our major simplification has to do with segment names. Before we go on with segment simplification, here are the rules the linker uses. If you don't remember them, you should review Chapter 10 before going on. During the link process, the linker will combine any segments which: The PC Assembler Tutor 278 ______________________ 1) have the same name. 2) are declared PUBLIC. 3) have the same class name (type). The linker processes object modules from left to right on the command line. The classes will be ordered in the ordering in which they were encountered (including the empty class type). Within each class, the segments will be ordered in the ordering in which they were encountered. If we have all these rules, how do high-level languages manage to combine their data and code correctly? The answer is that they use standardized segment definitions. Here are the basic ones for our data: ;--------------------------------------------------- _DATA SEGMENT WORD PUBLIC 'DATA' _DATA ENDS ;--------------------------------------------------- ;--------------------------------------------------- CONST SEGMENT WORD PUBLIC 'CONST' CONST ENDS ;--------------------------------------------------- ;--------------------------------------------------- _BSS SEGMENT WORD PUBLIC 'BSS' _BSS ENDS ;--------------------------------------------------- ;--------------------------------------------------- STACK SEGMENT PARA STACK 'STACK' STACK ENDS ;--------------------------------------------------- If all the code will fit in one segment we can use a single segment name: ;--------------------------------------------------- _TEXT SEGMENT WORD PUBLIC 'CODE' _TEXT ENDS ;--------------------------------------------------- otherwise we can make independent segments, each with an independant name: ;--------------------------------------------------- name_TEXT SEGMENT WORD PUBLIC 'CODE' name_TEXT ENDS ;--------------------------------------------------- where the "name" can be anything, but the "_TEXT" remains invariable. Any subroutine calls within the segment can be NEAR, while any calls to a different segment should be FAR. The "WORD" in these definitions says that when the linker combines segments into a larger segment, each subsegment must start at an even address (a word boundary). This has to do with the speed of word fetches from memory that we discussed in the last chapter. "WORD" is fine for 16 bit data busses, but for a Chapter 26 - Simplifying The Template 279 _____________________________________ 80386 you actually want "DWORD" so things are correctly aligned with a 32 bit data bus. "PARA" means paragraph and that means aligned with a segment starting address (every 16 bytes). Everything will work with "PARA". For reasons of convenience, compilers put different types of data in different segments. For you, there is no reason to use more than one segment, and that is: ;-------------------------------- _DATA SEGMENT WORD PUBLIC 'DATA' _DATA ENDS ;-------------------------------- Compilers use these different segments because they can. If they were constrained to use only one segment name, they could do it with no problem. What is in these different segments? _DATA standard initialized data CONST data constants _BSS uninitialized static data STACK room for the SS:SP stack So what do these things mean? _DATA The _DATA segment stores all initialized data which exists from the time the program starts till the time that the program ends. In C: static int x = 5; In Pascal: const my_salary : real = 52.77 These variables have a specific value at the start of the program, even before the first instruction is executed. This value may change during the program. The variable exists during the whole program. _BSS The _BSS segment stores all uninitialized data which exists from the time the program starts till the time that the program ends. In C: static int x ; In Pascal, any variable declared outside a procedure but without an initial value will be in _BSS. In compiled BASIC, everything except dynamic arrays is in the _BSS. These variables have an indeterminate value at the start of the program and exist during the whole program. The PC Assembler Tutor 280 ______________________ CONST CONST takes all constants which are longer than 2 bytes. If the compiler is on its toes, anything one or two bytes long will be coded into the machine instructions since this is much faster. What is a constant? It is anything that has a value but doesn't have a variable name: value = 275.29 ; printf ( "Mr. Yellow: 'Read my lips - no new taxis!'\n"); result = value / 27.619 ; file_ptr = fopen ( "stuff.doc", "r+") ; All the numbers and all the text strings need to be stored somewhere. They are stored in the CONST segment and given an internal name by the compiler so they can be used at the appropriate location. They are not available in other parts of the program.{1} These constants are sometimes called literals. STACK BASIC does not use the stack in the same way as Pascal and C. In BASIC it is used only for passing variables between subroutines. In C and Pascal, most variables are temporary. They come into existance at the beginning of the subroutine and they disappear upon leaving the subroutine. When you call the subroutine again, the values these variables have are indeterminate. These variables all exist on the stack relative to BP, the base pointer. This is why you can have recursion in C and Pascal but not in BASIC. As I said, you don't need to put your different types of data in different segments. It can all go into _DATA. GROUPS We now come to the bizarre. You will notice that when the linker links all these object modules together, it will have four distinct segments with each segment having a distinct class name. We will get: _DATA 'DATA' CONST 'CONST' _BSS 'BSS' STACK 'STACK' The problem here is that we want to set DS at the beginning of ____________________ 1. There is an exception. Some compilers check to make sure that there are no duplicates of the constant. These compilers give all duplicates the same address so there is only one copy of any one constant such as 0, 1, etc. Chapter 26 - Simplifying The Template 281 _____________________________________ the program so that it will reference all the data. How are we going to do this? The warped minds of electrical engineers and computer scientists spent hours and hours trying to find the most obscure way possible to unify data addressing and they came up with GROUPS. You can tell the linker that you want data from distinct segments to be referenced by the offset from the beginning of the lowest segment in memory that belongs to the group. Read this about five or ten times to get the hang of it. You tell the linker that a bunch of different segments belong to a group. It will find the segment which is lowest in memory and then whenever you ask for the GROUP offset, the linker will calculate the offset from the beginning of this first segment. The way you define a group is with a name, the word "GROUP", and then a list of those segments in the file which belong to the group: DGROUP GROUP _DATA, CONST, _BSS, STACK Note that it is the segment names, not the class names. DGROUP is the standard name for the data group. If the assembler gives the linker the correct information, the linker will adjust all offsets relative to the beginning of the group. The only limit on a group is that the distance from the first byte of the group to the last byte of the group must be 65535 bytes or less. This is because all the group segments must reside in one physical segment in memory. It is not even necessary for all the segments in a block of memory to belong to the group. Consider the following ordering of segments in memory. _DATA DATASTUFF CONST CODESTUFF _BSS EVENMORESTUFF STACK As long as the distance from one end of _DATA to the other end of STACK is 65535 bytes or less, the linker will adjust the offsets in _DATA, CONST, _BSS and STACK relative to the start of DGROUP and the linker will adjust the offsets of DATASTUFF, CODESTUFF and EVENMORESTUFF relative to their respective segment starting addresses. I didn't say that this was good programming, I only said that it was possible. Thoroughly confused? You're not alone. Just remember, in all compiled languages, we are going to combine these four types of segments into a single group where offsets are relative to the very beginning of the data. Before getting you even more confused, let's take a look at what we have so far. Make sure you actually do all of the following The PC Assembler Tutor 282 ______________________ examples. Use template.asm and at the very top, put in the following segments: ; - - - - - - - - - SEG1 SEGMENT 'STUFF' db 100 dup (?) seg1_data db ? db 899 dup (?) SEG1 ENDS ; - - - - - - - - - SEG3 SEGMENT 'STUFF' db 300 dup (?) seg3_data db ? db 699 dup (?) SEG3 ENDS ; - - - - - - - - - SEG5 SEGMENT 'STUFF' db 500 dup (?) seg5_data db ? db 499 dup (?) SEG5 ENDS ; - - - - - - - - - Call this program QGROUP1.ASM. These segments are 1000 bytes long, and the data names are 100, 300 and 500 bytes into their respective segments. Because these segments will be paragraph aligned, the second and third segments will start 1008 bytes (16 X 63) after the proceeding one. You need to tell the assembler that these are in a group and give the proper ASSUME statement. We'll call this QGROUP: QGROUP GROUP SEG1, SEG3, SEG5 ASSUME cs:CODESTUFF, ds:DATASTUFF, ds:QGROUP Here's some code: ; + + + + + + + + + + + + START CODE BELOW THIS LINE lea ax, seg1_data call print_unsigned lea ax, seg3_data call print_unsigned lea ax, seg5_data call print_unsigned ; + + + + + + + + + + + + END CODE ABOVE THIS LINE As you can see, all we are doing is putting the addresses into AX and then printing them as unsigned numbers. Here's the output: 00100 01308 02516 Remember, each segment is starting 1008 bytes after the start of the previous one. Here's the same program with a few extra segments thrown in. Call it QGROUP2.ASM.: ; - - - - - - - - - Chapter 26 - Simplifying The Template 283 _____________________________________ SEG1 SEGMENT 'STUFF' db 100 dup (?) seg1_data db ? db 899 dup (?) SEG1 ENDS ; - - - - - - - - - SEG2 SEGMENT 'STUFF' db 200 dup (?) seg2_data db ? db 799 dup (?) SEG2 ENDS ; - - - - - - - - - SEG3 SEGMENT 'STUFF' db 300 dup (?) seg3_data db ? db 699 dup (?) SEG3 ENDS ; - - - - - - - - - SEG4 SEGMENT 'STUFF' db 400 dup (?) seg4_data db ? db 599 dup (?) SEG4 ENDS ; - - - - - - - - - SEG5 SEGMENT 'STUFF' db 500 dup (?) seg5_data db ? db 499 dup (?) SEG5 ENDS ; - - - - - - - - - This is almost the same thing but we have added two more segments. We are NOT going to join these two segments into the group. Here's the GROUP and ASSUME statements: QGROUP GROUP SEG1, SEG3, SEG5 ASSUME ds:SEG2, ds:SEG4 ASSUME cs:CODESTUFF, ds:DATASTUFF, ds:QGROUP Make sure the ASSUME statements are in that order or things may get confused. We also add some code: ; + + + + + + + + + + + + START CODE BELOW THIS LINE lea ax, seg1_data call print_unsigned lea ax, seg2_data call print_unsigned lea ax, seg3_data call print_unsigned lea ax, seg4_data call print_unsigned lea ax, seg5_data call print_unsigned ; + + + + + + + + + + + + END CODE ABOVE THIS LINE This shows the addresses of all five variables. Here's the new output: The PC Assembler Tutor 284 ______________________ 00100 00200 02316 00400 04532 As you can see, the GROUPed segments have their offsets relative to the beginning of the group while the others have their offsets relative to the beginning of the segment. Make a copy of QGROUP1.ASM and call it QGROUP?.ASM. Leave the segment definitions, group definitions, ASSUME statements and code the same, but add six more lines of code at the end: ; compare the offsets mov ax, offset seg1_data call print_unsigned mov ax, offset seg3_data call print_unsigned mov ax, offset seg5_data call print_unsigned ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE After the three LEAs we now do 3 OFFSETS. Assemble and link this. Here's the output: 00100 01308 02516 00100 00300 00500 Wait a minute! Those last three numbers should be the same as the first three numbers. That's right, folks. This is a known error in the MASM assembler. In fact the Turbo Assembler copies this mistake when it is in "MASM" mode but does it right when it is in "IDEAL" mode. A86 does it right all the time. Here is the output from the same source file when assembled by A86: 00100 01308 02516 00100 01308 02516 You have told the assembler to calculate all offsets relative to the beginning of the group and MASM is ignoring you every time you use the OFFSET operator. The code fix for this is to use an override when you use OFFSET: ; compare the offsets mov ax, offset QGROUP:seg1_data call print_unsigned mov ax, offset QGROUP:seg3_data Chapter 26 - Simplifying The Template 285 _____________________________________ call print_unsigned mov ax, offset QGROUP:seg5_data call print_unsigned ; + + + + + + + + + + + + + + + END CODE ABOVE THIS LINE Better yet, use LEA whenever possible. If you do use OFFSET with groups, you need to go through the text file with a word search to make sure that all OFFSETs have a group override. This is a subtle error and it is very hard to find if you are not looking for it. This system is designed so that we can have 64k of data and stack, all of which is addressable with DS without changing DS's value. What happens if you have more data than that? One thing for sure is that you don't have more than 64k of individually named variables. Either that or you have some huge calluses on your typing fingers. What you do have is arrays. If you run into space problems, you move the least used or the biggest arrays into their own segments. You can have one segment per array if you want. The standardized high-level language names for these segments is: ; - - - - - - - - - - - - - - - - - - - - - FAR_DATA SEGMENT PARA 'FAR_DATA' FAR_DATA ENDP ; - - - - - - - - - - - - - - - - - - - - - FAR_BSS SEGMENT PARA 'FAR_BSS' FAR_BSS ENDP ; - - - - - - - - - - - - - - - - - - - - - Once again, the '_DATA' is for initialized data while the '_BSS' is for uninitialized data. Use only the 'FAR_DATA' kind.{2} You will notice that these segments are NOT PUBLIC. Although an assembler will unify all segments with the same definition that are in the same file, the linker will not unify segments from different files which are not PUBLIC. If we create 4 different .ASM files, each with one segment: ; FARDATA1.ASM PUBLIC data1 ; - - - - - - - - - - - - - - - - - - - - - FAR_DATA SEGMENT PARA 'FAR_DATA' data1 db 1A67h dup (0) FAR_DATA ENDS ; - - - - - - - - - - - - - - - - - - - - - ; FARDATA2.ASM PUBLIC data2 ____________________ 2. A high-level language has the right to set all the data of a 'BSS' segment to zero as part of its startup routine. Whether it does so or not depends on what it has told the linker. If you put initialized data into either a '_BSS' or a 'FAR_BSS' segment, it might easily wind up zero after startup. The PC Assembler Tutor 286 ______________________ ; - - - - - - - - - - - - - - - - - - - - - FAR_DATA SEGMENT PARA 'FAR_DATA' data2 db 0D4A8h dup (0) FAR_DATA ENDS ; - - - - - - - - - - - - - - - - - - - - - FARDATA3.ASM PUBLIC data3 ; - - - - - - - - - - - - - - - - - - - - - FAR_DATA SEGMENT PARA 'FAR_DATA' data3 db 200h dup (0) FAR_DATA ENDS ; - - - - - - - - - - - - - - - - - - - - - FARDATA4.ASM PUBLIC data4 ; - - - - - - - - - - - - - - - - - - - - - FAR_DATA SEGMENT PARA 'FAR_DATA' data4 db 8716h dup (0) FAR_DATA ENDS ; - - - - - - - - - - - - - - - - - - - - - and link these with TEMPLATE.OBJ and ASMHELP, we will get the following .MAP file: Start Stop Length Name Class 00000H 01A66H 01A67H FAR_DATA FAR_DATA 01A70H 0EF17H 0D4A8H FAR_DATA FAR_DATA 0EF20H 0F11FH 00200H FAR_DATA FAR_DATA 0F120H 17835H 08716H FAR_DATA FAR_DATA 17840H 1823FH 00A00H STACKSEG STACK 18240H 1875DH 0051EH DATASTUFF DATA 18760H 1A02FH 018D0H CODESTUFF CODE Program entry point at 1876:0000 The numbers in the segment definitions were in hex so you could read the .MAP file more easily. We have created four different FAR_DATAs - one for each variable. The idea here is to leave DS alone if possible and use ES:SI or ES:DI for your manipulation of the array. mov ax, seg data1 mov es, ax mov si, offset data1 Of course, if you using two different FAR_DATA arrays from two different segments at the same time, you will probably need to use DS temporarily. This is the kind of thing you need to plan before you start a program which contains large arrays. Chapter 26 - Simplifying The Template 287 _____________________________________ You have now seen all possible segments for any Microsoft language and for Turbo C.{3} These are: _DATA SEGMENT WORD PUBLIC 'DATA' CONST SEGMENT WORD PUBLIC 'CONST' _BSS SEGMENT WORD PUBLIC 'BSS' STACK SEGMENT PARA STACK 'STACK' _TEXT SEGMENT WORD PUBLIC 'CODE' name_TEXT SEGMENT WORD PUBLIC 'CODE' FAR_DATA SEGMENT PARA 'FAR_DATA' FAR_BSS SEGMENT PARA 'FAR_BSS' DGROUP GROUP _DATA, CONST, _BSS, STACK We have another problem on our road to simplification. We want DS to have the address of the start of DGROUP. How do we do it? Well, before we had: mov ax, DATASTUFF mov ds, ax DATASTUFF was a segment. We do the same thing for groups: mov ax, DGROUP mov ds, ax We use a group name instead of a segment name. This means that our ultimate code segment will look like this ; - - - - - - - - - - _TEXT SEGMENT WORD PUBLIC 'CODE' DGROUP GROUP _DATA, CONST, _BSS, STACK ASSUME cs:_TEXT, ds:DGROUP start: mov ax, DGROUP mov ds, ax ; - - - - - - - - - - ; the program goes here ; - - - - - - - - - - mov ah, 4Ch ____________________ 3. If you are using Turbo PASCAL, then there are only two segments possible. They are: DATA SEGMENT WORD PUBLIC CODE SEGMENT BYTE PUBLIC There is no class name. You can substitute DSEG for DATA and CSEG for CODE if you want. Turbo Pascal has no DGROUP. The PC Assembler Tutor 288 ______________________ mov al, ? ; replace ? with error code int 21h _TEXT ENDS ; - - - - - - - - - - Say, if all this stuff is standardized text, why are we forced to type all this drivel over and over again. The answer is that we aren't. All the segment information has a shorthand. Here's how it works. Every shorthand symbol starts with a dot. The assembler will then generate the desired text.{4} This is from MASM 5.0 on, so if you have an earlier assembler you'll have to write the full text. To start out, use the two starting directives DOSSEG (with no dot) and .MODEL. MODEL will be explained later.{5} DOSSEG .MODEL Medium For now, 'medium' is what we want. From that point, if you want a data segment, you just write .DATA, if you want code, you write .CODE. Every time that the assembler sees a segment directive it will close any segment that is open and start the segment indicated by the directive. (You can always reopen a segment). Here is what replaces the directives: DIRECTIVE REPLACEMENT TEXT .DATA _DATA SEGMENT WORD PUBLIC 'DATA' .CONST CONST SEGMENT WORD PUBLIC 'CONST' .DATA? _BSS SEGMENT WORD PUBLIC 'BSS' .STACK [size] STACK SEGMENT PARA STACK 'STACK' .CODE _TEXT SEGMENT WORD PUBLIC 'CODE' .CODE [name] name_TEXT SEGMENT WORD PUBLIC 'CODE' .FARDATA [name] FAR_DATA SEGMENT PARA 'FAR_DATA' .FARDATA? [name] FAR_BSS SEGMENT PARA 'FAR_BSS' The [name] in brackets will be explained in a minute. The [size] after the stack declaration allows you to customize the size of the stack. Without any size, the declaration .STACK will allocate 1k of memory for the stack. A size allocates a ____________________ 4. It really generates no text. It is just that the assembler will generate the same machine code as if that text had been generated. 5. DOSSEG tells the assembler to tell the linker that the .EXE file should have the standard segment order. It is not necessary but it doesn't hurt. Chapter 26 - Simplifying The Template 289 _____________________________________ specific number of bytes: .STACK 2000h You can make it anything you want, but make sure it is an even number and remember that the limit for all four parts of DGROUP is 64k. To see how the names work, we need some text files. Here is a complete main file: ; FARDATA.ASM - driver module DOSSEG .MODEL medium EXTRN data2_routine:FAR, data3_routine:FAR .STACK 200h .FARDATA data1 db 0100h dup (0) .CODE main: mov ax, DGROUP mov ds, ax call data2_routine call data3_routine mov ax, 4C00h int 21h END main It has some data and some code though it doesn't really do anything. We will use this along with two other files for the examples. Here is FARDATA2: ; FARDATA2.ASM DOSSEG .MODEL medium PUBLIC data2_routine .FARDATA data2 db 0200h dup (0) .CODE data2_routine proc ret data2_routine endp END Notice that data2_routine doesn't have a FAR or NEAR. That's being taken care of by the memory model. Data2_routine's type does need to be declared EXTRN in the main module. The third routine has similar code. Here is the .MAP file when they are combined: Start Stop Length Name Class 00000H 00013H 00014H FARDATA_TEXT CODE 00014H 00014H 00001H FARDATA2_TEXT CODE 00016H 00016H 00001H FARDATA3_TEXT CODE 00020H 0011FH 00100H FAR_DATA FAR_DATA 00120H 0031FH 00200H FAR_DATA FAR_DATA The PC Assembler Tutor 290 ______________________ 00320H 0061FH 00300H FAR_DATA FAR_DATA 00620H 00620H 00000H _DATA DATA 00620H 0081FH 00200H STACK STACK You can see the FAR_DATAs there, but where did the FARDATA3_TXT come from? The assembler decided that we wanted independent code segments and gave each one the name of the assembler file it came from. Since all the object files in a program must have unique names, these segment names should also be unique. If we change the .MODEL from MEDIUM to COMPACT without touching anything else, then we get: Start Stop Length Name Class 00000H 00022H 00023H _TEXT CODE 00030H 0012FH 00100H FAR_DATA FAR_DATA 00130H 0032FH 00200H FAR_DATA FAR_DATA 00330H 0062FH 00300H FAR_DATA FAR_DATA 00630H 00630H 00000H _DATA DATA 00630H 0082FH 00200H STACK STACK If we now put a name after the .FARDATA directive, it will give the segment a unique name. Putting: .FARDATA jake_the_snake in FARDATA2.ASM, along with name changes in the other modules results in the following .MAP file: Start Stop Length Name Class 00000H 00022H 00023H _TEXT CODE 00030H 0012FH 00100H HACKSAW FAR_DATA 00130H 0032FH 00200H JAKE_THE_SNAKE FAR_DATA 00330H 0062FH 00300H HULKSTER FAR_DATA 00630H 00630H 00000H _DATA DATA 00630H 0082FH 00200H STACK STACK We are doing a number of interrelated things here, so let's try to unify what is going on. You have seen both NEAR and FAR routines in the Tutor. A NEAR routine alters IP and restores IP on the return. A FAR routine alters both CS and IP and restores them on the return. When we passed addresses of data, we have almost always passed just the offset of the data. That is because the data has almost always been in the DATASTUFF SEGMENT, and the value of DS has been known. In Chapter 19 we did "move_pascal_string" which was a subroutine where we passed both the segment and offset of the data. These are our two choices for passing addresses: OFFSET 1 word SEGMENT:OFFSET 2 words Chapter 26 - Simplifying The Template 291 _____________________________________ This gives us four basic possiblilities for program structure: SUBROUTINE CALL DATA ADDRESSES PASSED AS NEAR OFFSET FAR OFFSET NEAR SEGMENT:OFFSET FAR SEGMENT:OFFSET Each of these structural possibilities has a name called a MODEL name. They are: SUBROUTINE CALL ADDRESSES PASSED AS MODEL NAME NEAR OFFSET SMALL FAR OFFSET MEDIUM NEAR SEGMENT:OFFSET COMPACT FAR SEGMENT:OFFSET LARGE You tell the assembler which model you are working with by using the .MODEL directive: .MODEL medium The assembler will then make either NEAR or FAR the default type. This can be overridden if you have explicitly given a NEAR or FAR: my_proc procedure will generate the correct subroutine calls and returns for that model, while: my_proc1 procedure near my_proc2 procedure far will remain unaltered. At the assembler level, you need to code the address passing yourself, but if you have a MODEL and you are connected to a high-level language (with the same .MODEL type), the high-level language will pass all addresses as stated above. The advantage of this system is that using the .MODEL directive and appropriate EQU statements and MACROS (which we have not covered), it is possible to write a single subroutine which can then be assembled in all four model configurations. Coding this is non-trivial, but when you have done more programming you will see how to deal with the stack using EQUs and MACROs. For now, you want to stay with data addresses which are passed by offset only. This is much easier. These are the SMALL and MEDIUM models. Whether you choose NEAR or FAR procedures doesn't affect much except where parameters are on the stack (because of that extra CS). The PC Assembler Tutor 292 ______________________ Are these .MODELS important? They are nice, but not particularly vital. What happens is that in a manual you see a sample program like this: DOSSEG .MODEL medium .STACK .DATA variable1 dw 25 .CODE sample proc mov ax, variable1 ret sample endp ENDS END and you start comparing the size of this to the size it would be if you used the standard segment definitions. I have news for you. This is not a legitimate program. A legitimate program is a page or two long.{6} Also, at least to my way of thinking, you want visual separation between segments. The above is a disordered presentation of segments. We want order in our programs and the segment headers provide a visual structure. In the text file for ASMHELP, (which is about 3600 lines long), the SEGMENT declarations occupy about 20 lines. This is about 0.5% of the total length of the file. If you are going to assemble a file in multiple models, then it is worthwhile to use the .MODEL directives, otherwise it is optional depending more on your concept of what looks clear than any major difference. ____________________ 6. Perhaps you want to scan \COMMENTS\MISHMASH.DOC which contains some real subroutines. They are all long. Chapter 26 - Simplifying The Template 293 _____________________________________ SUMMARY To exit a program, use INT 21h Function 4Ch mov ah, 4Ch ; exit program mov al, ? ; replace ? with error code int 21h A GROUP is a group of segments whose data will be referenced by the offset from the beginning of the group. You declare a group with: DGROUP GROUP _DATA, CONST, _BSS, STACK MASM calculates OFFSETS incorectly with groups, so you should either use LEA or the DGROUP override: lea ax, variable1 mov ax, offset DGROUP:variable1 To get the address of DGROUP in DS you need: ASSUME ds:DGROUP and: mov ax, DGROUP mov ds, ax The standardized segment definitions, along with their simplified directives are: DIRECTIVE REPLACEMENT TEXT .DATA _DATA SEGMENT WORD PUBLIC 'DATA' .CONST CONST SEGMENT WORD PUBLIC 'CONST' .DATA? _BSS SEGMENT WORD PUBLIC 'BSS' .STACK [size] STACK SEGMENT PARA STACK 'STACK' .CODE _TEXT SEGMENT WORD PUBLIC 'CODE' .CODE [name] name_TEXT SEGMENT WORD PUBLIC 'CODE' .FARDATA [name] FAR_DATA SEGMENT PARA 'FAR_DATA' .FARDATA? [name] FAR_BSS SEGMENT PARA 'FAR_BSS' In addition you have the different model names: SUBROUTINE CALL ADDRESSES PASSED AS .MODEL NAME NEAR OFFSET SMALL FAR OFFSET MEDIUM NEAR SEGMENT:OFFSET COMPACT FAR SEGMENT:OFFSET LARGE Using BASIC - 1 =============== mmi So you like using BASIC. I can't blame you. BASIC has several very nice features. (1) Its graphics interface is easy to use and is very powerful. If you want to draw lines, circles, patterns or other graphics images, you can do it. Your only limitations are the aspect ratio of the screen and the possible screen colors. These limitations have to do with hardware, not with BASIC. (2) Its string handling capabilities are unparalleled. If you want to have an array of strings like the following: DIM STRING.ARRAY$ (50, 25) ' 1250 elements and some of the elements are only a couple of characters long but some of the strings are several hundred characters long, BASIC is the ONLY language that can handle it correctly. In fact, if you wanted one of the strings to be 1000 characters (it's allowed in the most recent BASIC), then you would need 1.25 megabytes for this array in C or in PASCAL. This would probably be too big for the computer. BASIC, on the other hand, implements this without wasting any space in memory. BASIC has a few things that require a little more work than with other languages. It has strings, integers, and floating point numbers but it is missing characters and unsigned numbers - these can be implemented by using integers (if you are careful). This is not all that limiting. If BASIC is so nice, why do I think that you should have some experience with a structured language before doing assembler? It all has to do with data INTEGRITY. The question is, are you sure that you haven't inadvertantly screwed up your data? There are two ways that this might happen. You need to know how BASIC works to understand the problem. A variable name in BASIC consists of a name followed by a type identifier. The identifiers are '%', '!', '$', '#' (double precision) and possibly '&' (long integer). The name: CURRENT.TEMPERATURE is NOT a variable inside of BASIC while: CURRENT.TEMPERATURE% CURRENT.TEMPERATURE! CURRENT.TEMPERATURE$ ______________________ The PC Assembler Tutor - Copyright (C) 1990 Chuck Nelson The PC Assembler Tutor mmii ______________________ are all valid variables. In fact, all three can be in the same program, as we shall see shortly. I just said that that first example is not a valid variable inside of BASIC, but that is exactly what you have been writing since the first time that you wrote a BASIC program. What's going on here? Well, when the BASIC interpreter or compiler reads the text file, it reads the name and the identifier. If the type identifier is missing, it tacks the default identifier on to the end of the name. What's the default type identifier? The answer is '!'. BASIC maintains a default list. The legal first characters in a name are A-Z. For each legal first letter, there is an entry in the list which says what the default type for that letter is. When you start the program, the type list looks like this: !!!!!!!!!!!!!!!!!!!!!!!!!! When the interpreter sees a name without an identifier, it takes the first character of the name and goes to the default-type list, gets the appropraiate type identifier for that first character, and adds the identifier on to the end of the name. Externally (in your text file) there may be names without type identifiers, but internally (inside the interpreter/compiler) ALL variables have the type identifier after the name. Though the list starts out with all '!'s, you can change this list with a DEF statement. Here are some DEF statements and how they change the list. I am assuming that we are always starting at the beginning of the program (with all '!'s). (1) DEFINT A-Z ' This is probably your favorite %%%%%%%%%%%%%%%%%%%%%%%%%% (2) DEFINT A-E %%%%%!!!!!!!!!!!!!!!!!!!!! (3) DEFINT A-E DEFSTR F-K DEFDBL L-R %%%%%$$$$$$#######!!!!!!!! This last one is a recipe for disaster, but it's legal. Have you understood everything so far? Then let's go on. This ability to NOT use the type identifier leads to laziness. The usual thing is to always identify strings with a '$' but to be rather lax about the distinction between integers and floating point numbers. Now we get to the problem. Look at the following code: BASIC I - Using Basic mmiii _____________________ 10 DEFINT A-Z 20 FOR I = 1 to 5 30 LARGE.NUMBER! = 300.0 * I 40 SMALL.NUMBER! = LARGE.NUMBER / 7 50 PRINT LARGE.NUMBER!, SMALL.NUMBER! 60 NEXT I If you can't see the error, I forgot to put the '!' after LARGE.NUMBER in line 40. The problem here is that BASIC is not going to complain. When it sees LARGE.NUMBER without a type identifier, it will go to the default list (because of the DEFINT statement, all defaults are integer) and put a % after it. BASIC has created a SECOND variable with the same initial name, but a different type. This is not because you wanted it to, but rather because you forgot a '!'. Lets do a BASIC program. Here's some code: ******************* NINJA.BAS ******************* 10 ' place for defint statement 20 ' 30 ' enter all the data 40 NINJA% = 20 50 NINJA! = 4.7134E+13 60 NINJA$ = "Hey dude!" 70 ' 80 NINJA$ (0) = "howdy! " 90 FOR I% = 1 TO 5 100 NINJA$ (I%) = NINJA$ ( I% - 1 ) + "doody! " 110 NINJA! (I%) = I% * 25.31 120 NINJA% (I%) = I% * 3 130 NEXT I% 140 ' 150 ' print all the data 160 PRINT NINJA%, NINJA!, NINJA$ 170 FOR I% = 1 TO 5 180 PRINT NINJA% (I%), NINJA! (I%), NINJA$ (I%) 190 NEXT I% 200 PRINT NINJA , NINJA (3) 210 ' 220 END ********************************************************* I actually want you to run this through your BASIC. I have defined SIX different NINJAs. They are: 1) NINJA% 2) NINJA% ( ) 3) NINJA! 4) NINJA! ( ) 5) NINJA$ 6) NINJA$ ( ) where the parentheses indicate an array. If I had gotten carried away, I could have included 4 more NINJAs for double precision, The PC Assembler Tutor mmiv ______________________ double precision array, long integer and long integer array. Running this through my computer I get: 20 4.7134E+13 Hey dude! 3 25.31 howdy! doody! 6 50.62 howdy! doody! doody! 9 75.93 howdy! doody! doody! doody! 12 101.24 howdy! doody! doody! doody! doody! 15 126.55 howdy! doody! doody! doody! doody! doody! 4.7134E+13 75.93 Notice that I have 3 different array types and 3 different variable types. As you can see from the bottom line (which prints the defaults), the default here is '!'. If we now put DEFINT A-Z on line 10: 10 DEFINT A-Z everything will stay the same except the bottom line which now is: 20 9 which are the integers. Finally, if we have DEFSTR A-Z 10 DEFSTR A-Z we get: Hey dude! howdy! doody! doody! doody! as the bottom line of output. If you didn't know all of this before, make sure you have assimilated this before going on. PASCAL and C Both PASCAL and C require you to account for every variable. You need to specifically state what kind of variable it is and you need to enter it in a list of variables used in that block of code: int ch, variable1, variable2, variable3[60] ; char long_array[4027] ; A name can be used in only one way in any block of code. If you try to use it differently, the compiler will generate an error. Though people moan and groan about having to specifically list all the variables used in a block of code, over time they get used to it. It offers two advantages. (1) No variable can be inadvertently used in an incorrect way (used as a string when you wanted an integer, used as an integer when you wanted a floating point number, etc.). (2) There can be no variables which are misspelled, are left BASIC I - Using Basic mmv _____________________ over from a previous version of the code or appear by mistake, without both the compiler and you knowing about them. Slightly modifying names is one of my major problems. I might use 'long_array' and 'longarray' in different places in the C code, though I want the same thing. C will catch this, BASIC won't. Every time BASIC encounters a slightly different name it simply creates a new variable (and normally sets it to 0). In programs larger than a few pages, this situation is almost impossible to detect or control. THE RANGE OF A VARIABLE When talking about PASCAL and C I kept using the word BLOCK. Both languages have a modular (or block) design. The philosophy behind this is that when you write a program you want to divide a task into a number of distinct subtasks, and you don't want these subtasks to interfere with each other. You put each of these in a subprogram. In BASIC you can put sections of a program in subprograms too. This is good programming practice. There is a difference, however. In C and PASCAL, a variable is valid ONLY in the block in which it is declared (unless you take specific steps to make it universally valid). If you try to use this variable in a different block of code the compiler will generate an error - it literally has no information about where this variable is. In BASIC, all names are valid everywhere in a file. If on page 1 you have: 210 CURRENT.PRICE! = 427.11 you may want to use this as a constant number throughout the program. But if in a subprogram on page 23 you have: 12060 CURRENT.PRICE! = 0 you have changed the value. You're saying to yourself "But why would I change a variable like "CURRENT.PRICE!"? You probably wouldn't. But you will have dozens and dozens of variables around with names like I, J, K, COUNT, START, STARTING.NUMBER, STARTING.VALUE, TOTAL, RESULT and LAST.VALUE (whatever their type). It is exactly these that will be corrupted in BASIC but will be reliable in C or PASCAL. ********************************************** 1920 MINIMUM! = RATE! * 5.0 1930 MAXIMUM! = RATE! * 7.32 1940 COUNT% = (MAXIMUM! - MINIMUM!) / 3.0 1950 CALL CHECK.ANSWER 1960 PRINT COUNT% ********************************************** In this section of code, did 'CALL CHECK.ANSWER' change any of the four variables MINIMUM!, MAXIMUM!, RATE! or COUNT%? In BASIC, we hope so, but we can't be sure. In PASCAL or C we KNOW that it The PC Assembler Tutor mmvi ______________________ didn't change any code. This is the difference between a structural engineer HOPING that a bridge won't fall down and KNOWING that a bridge won't fall down. In fact, the PASCAL and C programmers will probably be annoyed when they start using assembler because in assembler, ALL names are valid everywhere. More importantly, however, these programmers will make great efforts to insure that all constants stay constants and that each variable is used for one job only. Finally, some BASIC programmers still do most of their programming using GOTO statements instead of CALLs. This is the antethesis of modular programming and is called spaghetti programming. The code winds up as an intertwined mess. If you are one of these people, once you start writing assembler code it will be impossible to figure out what the code is doing once it is longer than a page or two. All that being said, if you feel that your programming style is clear and orderly, have a fun time learning assembler. When you are done with all the chapters, read the second BASIC appendix on the problems associated with linking a non-BASIC subprogram to a BASIC program. BASIC II - Interfacing BASIC with ASSEMBLER =========================================== mmvii Have you finished reading all the chapters? If not, go back and do them, then come back to this when you are done. This chapter assumes you know about segments, subroutines, and the general information about linking subroutines to high-level languages. In order to do this appendix I had to dust off my old QuickBASIC 3.0. If you have QuickBASIC 4.x, some things will have been updated. If you have TurboBASIC, the subroutine conventions are different. However, the structure will be the same. You will have to consult your manual for exact details. If you are trying to this with the interpreter that came with DOS, I have a simple comment -> Forget it! I won't go into the details, but the BASIC interpreter is so much slower (about 10 times slower) and so much more difficult to use with assembler that I won't even cover it. There is no reason not to use one of the compiled BASICs if BASIC is your language of choice. This material only covers how to deal with COMPILED BASIC. In BASIC, all individual numeric data, strings, "static" arrays and the stack must fit into one 64k segment. The word 'segment' here has the same meaning as in assembler. Both the DS register and the SS register are set to this segment, and must stay set to this segment whenever BASIC has control of the program. "Dynamic" arrays can be located somewhere else in memory. You allocate a "static" array with a constant number as a dimension: DIM array1! (277), array2% (346), array3$ (500) and you allocate a "dynamic" array by using a variable to dimension the array: length1% = 277 length2% = 346 length3$ = 500 DIM array1!(length1%), array2%(length2%), array3$(length3%) Even though the first and second dimension statements produce the same size and type arrays, the first ones must be located inside DS and the second ones can be located outside of DS. "Static" means that once the array is defined, its length and number of dimensions cannot be changed for the rest of the program. It will occupy a specific amount of space for the rest of the program. "Dynamic" means that you can change the length of the array whenever you want to. You do this with: REDIM array1! (495) ______________________ The PC Assembler Tutor - Copyright (C) 1990 Chuck Nelson The PC Assembler Tutor mmviii ______________________ BASIC does this by deallocating space for the old array and then reallocating space for the new array. All the old information is lost. There are certain restrictions. You cannot change the number of dimensions in an array (if it starts out with 2 dimensions like DIM A!(47,63), it must always have 2 dimensions). In order to understand BASIC's memory strategy, we need to look at strings, the reason for it all.{1} The limit for a single string is 32,767 bytes. If the total amount of data you can have in the DS segment is only 65536 bytes, how does BASIC allocate memory so you can have long strings without runnung out of space? It uses only as much space as it needs. Let's define 3 strings (the dots will indicate a space): mystring$ = "You.say.either" yourstring$ = "And.I.say.either" ourstring$ = "Let's.call.it.off" After defining these three strings one after the other, memory will look like this: 17150 |You.say.eitherAnd.I.say.eitherLet's.call.it.off| (For clarity, the memory image will be between the '|'s and each row will be 50 bytes long. The next row down would start at 17200). For our example we will assume that this data starts at memory location 17150. There is no empty space. How does BASIC know where and how long mystring$ is? It has something called a string descriptor. This is a two word (4 byte) block, also in DS, which says exactly where and how long the string is. The first word is the length and the second word is the location (offset) in DS. From BASIC's view, we have: STRING DESCRIPTOR length:location mystring$ 14:17150 -> |You.say.either| yourstring$ 16:17164 -> |And.I.say.either| ourstring$ 17:17180 -> |Let's.call.it.off| Now let's change one of the strings: yourstring$ = "But.oh!,.If.we.call.the.whole.thing.off" We now have a problem. The current "yourstring$" is only 16 bytes long, but the new one is 39 bytes long. What does BASIC do? It (1) deallocates the space for the old "yourstring", (2) allocates new space for the new string and (3) updates the string ____________________ 1. This is an outline of what BASIC does, but it will not include the parts of memory management that you will never see. BASIC II - Interfacing BASIC With Assembler mmix ___________________________________________ descriptor. Memory will now look like this: 17150 |You.say.either Let's.call.it.offBut| |.oh!,.If.we.call.the.whole.thing.off| and the descriptors will now look like this: STRING DESCRIPTOR length:location mystring$ 14:17150 -> |You.say.either| yourstring$ 39:17197 -> |But.oh!,.if.we... ourstring$ 17:17180 -> |Let's.call.it.off| BASIC is aware that there is an empty block of space and has a strategy for dealing with empty spaces, though each BASIC has its own strategy. We don't know exactly WHEN it will take action, but we do know WHAT action it will take. At some point BASIC will decide that it has too many empty spaces in memory and will REORGANIZE the segment. This is known as GARBAGE COLLECTION. Exactly how this is done is up to the person who wrote the BASIC compiler/interpreter. After reorganization, the addresses of ALMOST ALL strings and MANY dynamic arrays will have changed. The string locations themselves will have changed, but the string descriptors will still be in the same place in DS, and they will have been updated. Here is the new memory: 12724 |You.say.eitherLet's.call.i| |t.offBut.oh!,.If.we.call.the.whole.thing.off| and here are the updated descriptors: STRING DESCRIPTOR length:location mystring$ 14:12724 -> |You.say.either| yourstring$ 39:12755 -> |But.oh!,.if.we... ourstring$ 17:12738 -> |Let's.call.it.off| The strings have been moved several thousand bytes from where they were just a second ago. The information that was in the string descriptors a second ago is no longer valid. Old information about dynamic arrays is also unreliable. This means that if you have a subroutine written in assembler, you must get any address information at the time the subroutine is called. We'll come back to this later. Let's go on to data input and output. When you first started The PC Assembler Tutor mmx ______________________ doing BASIC, you did i/o using only: WRITE #1, my.data! Perhaps you you do it differently now, perhaps not. In any case, you need to know about i/o speed and how different file i/o works. Here's the simplest file output: *********************************** DIM large.array! (10000) FOR i% = 1 to 10000 large.array! (i%) = 2.1 NEXT i% OPEN "2-1.doc" for output as # 1 PRINT time$ FOR i% = 1 to 10000 WRITE #1, large.array! (i%) NEXT i% PRINT time$ CLOSE #1 *********************************** Of course, to make it a challenge we are going to write an array of 10,000 numbers. How long does it take?{2} For this output it took 38 seconds. The same program, inputting the same data with: INPUT #1, large.array(i%) took 49 seconds. These are fairly large amounts of time. But wait, it gets worse. Let's change one line of the above program: large.array! (i%) = 2.1678319E+19 This is a different constant which is put into each element of the array. How long does output take now? 59 seconds. And input? a whopping 79 seconds! What's going on here? When you do i/o with INPUT #, WRITE # or PRINT #, it is exactly like doing i/o to the screen. For output, BASIC converts the binary numbers into TEXT and then writes the TEXT to the disk. When it does input, it reads the TEXT from the disk and converts the TEXT into a binary number. Here is the beginning of the output file from the first example: 2.1 2.1 2.1 2.1 ____________________ 2. All times from now on are with a slower PC with a slower hard disk, but an 8087. Since these are floating point numbers, your results should be slower if you don't have an 80x87, while if you have an 80386 with an 80387 and a fast hard disk, your times will be much faster. BASIC II - Interfacing BASIC With Assembler mmxi ___________________________________________ 2.1 Each data item has been converted into "2.1" + CHR$(13) + CHR$(10). These last two things are a carriage return on the IBM PC. That's (5 bytes X 10000 items) plus 1 byte for the end of file marker, or 50001 bytes: 2-1 DOC 50001 6-29-90 12:39p Here's the beginning of the output file from the second example: 2.167832E+19 2.167832E+19 2.167832E+19 2.167832E+19 2.167832E+19 Each data item has been converted into "2.167832E+19" + CHR$(13) + CHR$(10). That's (14 bytes * 10000 items) plus 1 byte for the end of file marker, or 140001 bytes: 2-1E19 DOC 140001 6-29-90 12:47p These files are unnecessarily large, and i/o is slow: if you don't have an 8087 and you are doing floating-point i/o, it can be slower still. Can we do it faster? Yes. Using GET and PUT, we get a certain number of bytes from the disk, then transfer them to the array. Some of you have never used random access i/o, so this is a brief summary. When you open a file as text (as we did in the above examples), BASIC divides the text by looking for carriage returns. When you open a file as a random access file, you are telling BASIC that you want to divide the file into distinct blocks of information. It may be text or it may be something else - BASIC doesn't care. If you say nothing, BASIC assumes that you want the blocks to be 128 bytes long, but the length can be anything. In the example that we will do, we will use 1024 byte blocks because that is exactly 2 disk sectors long, so the disk can read information easily and efficiently. If we had a block length of 4 bytes, the disk would have to do 10000 disk writes; that would be very slow and be hard on the disk. Here's how we open the file: OPEN "packed.doc" for RANDOM as #1 LEN = 1024 This will be a random access file and the block length will be 1024 bytes. When you tell it to read or write, it will do it 1024 bytes at a time. That is getting faster. Where is the block of data that it is going to write to disk? Here life starts getting complicated, so I hope you have understood everything that we have done so far. When you open a file, BASIC assigns it a BUFFER. The buffer has a fixed length (either 128 bytes or the length you have designated), and is The PC Assembler Tutor mmxii ______________________ located somewhere in the DS data segment along with the numbers and strings. Like a string, it is relocatable. We need a way to pin it down. The easy and nice way would be if it were an array and we cound address it like an array: buffer#1 (45) = 20 We are not that lucky. The only thing you can do is overlay a template on the buffer, and work from the template. This template MUST be made up of strings. We make up the template with a FIELD statement. FIELD #1, 1024 AS out.string$ The FIELD statement starts out with the file # followed by a list of strings and the length of each string. FIELD #1, 100 AS string1$, 200 AS string2$, 300 AS string3$ The total length of the strings may be shorter than the buffer, but may not be longer than the buffer. What does the FIELD statement do? The first thing that it does is set the string descriptor for all of these strings. Let's say that at the moment file #1 buffer is at 46217: STRING DESCRIPTOR length:location string1$ 100:46217 string2$ 200:46317 string3$ 300:46517 The first string starts at the first byte of the buffer. The second string starts right where the first string ends and the third string starts right where the second string ends. This is true for any FIELD statement, no matter how many strings are defined. Because of the way BASIC does memory management, if it moves the buffer, it will also update these string descriptors to point to the same relative places in the buffer. These string descriptors are on auto pilot. Suppose now that we have the following string: "Let's get physical" and we want to write it to disk as string1$. All we need to do is: string1$ = "Let's get physical" Right? No, that's very, very, very wrong. What you have just done is alter the string descriptor of string1$ to point to an entirely different place in memory. The string descriptors are now: string1$ 18:58902 string2$ 200:46317 BASIC II - Interfacing BASIC With Assembler mmxiii ___________________________________________ string3$ 300:46517 BASIC deallocated the space for string1, reallocated it somewhere else in memory, and changed the file descriptor. Not only is string1 in a different place in memory, but BASIC may think that part of the file #1 buffer is actually empty space, and the next time it reorganizes memory, who knows what is going to happen. From the moment you define strings in a FIELD statement until the time you close the corresponding file, you can NEVER have them on the left side of an equal sign. Having them on the left side is sure to change the file descriptor. How are we going to transfer data to these strings? There are three special operators in BASIC - LSET, MID$ and RSET. Their job is to put something into a string without altering the string length or location (i.e. without altering the string descriptor). LSET string1$ = "Let's get physical" MID$ (string1$,17) = "Let's get physical" RSET string1$ = "Let's get physical" LSET will insert the string at the very left of string1, RSET will insert the string at the very right of string1, and MID$ will insert the string starting at the 17th byte of string1. This is the strategy for all random access i/o in BASIC. We: 1) open a file as RANDOM and declare a block size. 2) define some "fixed length" strings inside the buffer with a FIELD statement. 3) insert data in the strings using LSET, RSET or MID$. This is true whether the data is strings or numbers. There's only one problem left. For LSET, RSET and MID$, the thing on the RIGHT side of the equal sign must be a string. You can't have: LSET string1$ = number! It's illegal. To counter this, BASIC has some pseudo-functions. Let's take integers as an example: a.string$ = MKI$ (number%) number% = CVI (a.string$) MKI$ doesn't actually DO anything. It just tells BASIC that it is o.k. to move two bytes from "number%" to "a.string$". The bytes are binary data and are moved unaltered. Similarly, CVI tells BASIC that it is alright to move two bytes of binary data from "a.string$" to "number%". We are tricking BASIC into moving binary data from one data type to another. This is simply data movement, and there is no data conversion. The forms are: NUMERIC DATA MOVE TO STRING FROM STRING integer <-> string MKI$ CVI long integer <->string MKL$ CVL The PC Assembler Tutor mmxiv ______________________ single precision <-> string MKS$ CVS double precision <-> string MKD$ CVD In contrast, the functions STR$ and VAL convert text representations to binary representations and binary representations to text representations. This is the same as what happens with PRINT and INPUT. Here's a program: ********************************************** number! = 2.1678319E+19 binary.string$ = MKS$ (number!) text.string$ = STR$ (number!) PRINT LEN(text.string$), LEN(binary.string$) PRINT text.string$, binary.string$ ********************************************** and here's the output: 13 4 2.167832E+19 nl _ You probably won't be able to see all of that last output on your printer because it is four bytes long and the number is: 6E6C965F hex or 110, 108, 150, 95 decimal The third byte is outside of ASCII 33-127, the standard ASCII characters. STR$ gives us the text representation of the number, while MKS$ stuffs the binary representation of a number into a string. In the opposite direction, VAL gives us the numeric value of a text string (if it has a numeric representation), while CVS stuffs 4 binary bytes from a string into a single precision number. STR$ from binary value to text representation VAL from text representation to binary value Note that STR$ can convert ANY type of number to a text string and VAL can convert a text string to ANY type of number, while CVI, CVL, CVS, CVD, MKI$, MKL$, MKS$, and MKD$ can only stuff a specific type of number into a string or a string into a specific type of number. We want our output program to stuff the binary value from a single precision number to selected bytes of a string. To stuff a floating-point number into string1$ above, all we need to do is: LSET string1 = MKS$ ( number!) The following program has a single string which is the size of the entire buffer, and we are going to stuff the single precision numbers in one at a time with MID$. ************************************************************ number% = 10240 DIM large.array! (number%) BASIC II - Interfacing BASIC With Assembler mmxv ___________________________________________ FOR i% = 1 to 10240 large.array! (i%) = 2.1678319e+19 NEXT i% OPEN "packed.doc" for RANDOM as #1 LEN = 1024 FIELD #1 , 1024 AS out.string$ PRINT time$ k% = 0 record.count% = 0 FOR i% = 1 to 40 record.count% = record.count% + 1 spot% = 1 FOR j% = 1 to 256 k% = k% + 1 MID$ (out.string$,spot%,4) = MKS$ (large.array!(k%)) spot% = spot% + 4 NEXT j% PUT #1, record.count% NEXT i% PRINT time$ CLOSE #1 *********************************************************** The array length has been increased slightly so that we have an exact number of blocks. We use MID$ to make sure that the string descriptor for out.string$ does not get changed. Each file write will be (256 numbers * 4 bytes/number) 1024 bytes long. We start with the first record and increase the record number by 1 each time we write. Does this increase the speed any? Well, this takes 11 seconds. TYPE OUTPUT INPUT num <-> text 38 - 59 sec 49 - 79 secs num <-> bin. string 11 sec 11 sec I didn't show you the equivalent input routines but here are the times they took. Note that the complexity of the single precision number has no effect on the last (the binary) routine. Also, the last routine does not suffer if there is no 8087. If you are running an 80286 with a fast hard disk, this last routine should only take a second or two. Here are the file sizes: 2-1 DOC 50001 6-29-90 12:39p 2-1E19 DOC 140001 6-29-90 12:47p PACKED DOC 40960 6-29-90 1:08p The first two are the different sizes depending on whether the constant was 2.1 or 2.1678319E+19. The last one is for our last routine. Notice that it is more compact. Can we do any better than 11 seconds? Yes, but we need to take over disk i/o and we need to know a few more things before we do that. The PC Assembler Tutor mmxvi ______________________ LOCATION OF DATA BASIC is designed to pass subroutines the location of the data, not the data itself. This is called passing by reference. Though it is possible to pass the data itself, there are certain problems with the stack if you do.{3} We will always pass the addresses. All single numeric variables are in the DS segment. BASIC passes the offset address of these variables in DS (1 word). All strings are in the DS segment. Their string descriptors are also in the DS segment. BASIC always passes the offset address of the STRING DESCRIPTOR. This, in fact, is what we want. We need to know both where the string is and how long it is. If we write past the end of the string we may destroy BASIC's memory management system. Static arrays are in the DS segment but dynamic arrays can be anywhere. If we want to write a general purpose routine with arrays, we need to handle them no matter where they are. BASIC has a special function called VARPTR that tells you where a variable is in memory. Here's a program that uses it for a couple of variables: *********************************************************** ' check out the use of varptr n% = 5000 p% = 50 DIM b!(800),a!(900) DIM d!(n%), c!(p%) mystring$ = "What's up, doc?" addressA! = varptr (n%) addressB! = varptr (p%) address1! = varptr (a!(0)) address2! = varptr (b!(0)) address3! = varptr (c!(0)) address4! = varptr (d!(0)) address5! = varptr (mystring$) PRINT addressA!, addressB! PRINT address1!, address2!, address3!, address4!, address5! *********************************************************** It gives us the addresses of all sorts of things. a!() and b!() are static arrays, so they should be in the DS segment. c!() and d!() are dynamic arrays, so they might be anywhere. Remember, the DS segment is from offset 0 to offset 65535. Let's see where they ____________________ 3. If you make a mistake and pass a single precision number instead of an integer, you will pass 4 bytes instead of 2. From that moment on the stack will have 2 extra bytes on it and you won't know where they came from. BASIC II - Interfacing BASIC With Assembler mmxvii ___________________________________________ are: 6230 6232 9438 6234 87616 67584 13062 The individual numbers are in DS, and the two static arrays are in DS, but c!() and d!() are outside of DS. These numbers tell us the address relative to the start of DS, but we don't know where DS is at the moment. Where exactly are these arrays? It would be tedious to pass the subroutine these numbers because they are floating-point numbers and would be very difficult to deal with. QuickBASIC has a function called PTR86. It is in an external object file called INT86.OBJ.{4} This object file has the routines that you need if you want to do interrupts from BASIC itself. We'll come back to that. PTR86's job is to take the floating-point number which we got from VARPTR, add the starting address of the DS segment to get an absolute address in memory, and then calculate both a segment and an offset for that address in memory. The segment will always be the highest segment that contains the first byte of the variable or array and the offset will always be a number from 0 to 15. In order to use an object file from inside of QuickBASIC you need to put it in a library file and then load the library file when starting QuickBASIC. Building the library file is quite easy. QuickBASIC comes with a program called BUILDLIB.EXE which builds the library for you. For now, you need only INT86.OBJ and PREFIX.OBJ in your library.{5} Put these two things in every library that you build from now on. PREFIX.OBJ insures proper segment ordering in the executable file. >buildlib int86+prefix This will create a library with the default name USERLIB.EXE. To load a library with this default file name, just put '/l' on the command line: >qb /l If you have given the library a different name like XQRTYF.EXE, then put that name after the '/l': >qb /lXQRTYF.EXE These object files will now be loaded and their subroutines will be usable from inside BASIC. ____________________ 4. PTR86 has been replaced by VARSEG in QuickBASIC 4.0. 5. Both of these object files come with your QuickBASIC. The PC Assembler Tutor mmxviii ______________________ If you now load the user library along with BASIC, you can look at the segments and offsets of the arrays. Here is a program with a couple of arrays: ********************************************** k% = 12000 DIM array1!(k%), array2!(k%), array3!(12000) value1! = VARPTR (array1!(0)) value2! = VARPTR (array2!(0)) value3! = VARPTR (array3!(0)) PRINT value1!, value2!, value3! ********************************************** Each array is 12001 * 4 (bytes) or 48004 bytes long. The first two have been defined as dynamic, so they can be anywhere in memory as long as they're after the start of DS. The third one is static, so it must be entirely in DS. Here's the output: 68032 116064 6252 Remember, these offsets are not relative to the start of memory, they are relative to the start of the DS segment. The DS segment only goes up to offset 65535 so the first two are outside of DS while the third one is totally inside (6252 + 48004 = 54006 which is less than 65536). PTR86 is going to give us a SEGMENT:OFFSET pair relative to the start of memory.{1} The form for the call is: CALL PTR86 (segment%, offset%, value!) where value! is the number returned by VARPTR. We'll add the following code to the bottom of the above program: *************************************** CALL PTR86 ( seg1% , off1%, value1! ) CALL PTR86 ( seg2% , off2%, value2! ) CALL PTR86 ( seg3% , off3%, value3! ) PRINT seg1%, off1% PRINT seg2%, off2% PRINT seg3%, off3% *************************************** What does the program print now? 68032 116064 6252 ____________________ 1. If you are using QuickBasic 4.0 or later this has become simplified. You can just use VARSEG and VARPTR. These will give you a segment offset pair which is usable in a subroutine call. ______________________ The PC Assembler Tutor - Copyright (C) 1990 Chuck Nelson BASIC II - Interfacing BASIC With Assembler mmxix ___________________________________________ 19125 0 22127 0 15263 12 PTR86 has calculated the segments. The first two offsets are 0 and the third one is 12. Even though the third array is in the DS segment, PTR86 has recalculated the segment to find the highest segment which contains the first byte of the data. If you want to do a little calculation, you can figure out that while this program was running, DS was set to segment 14873. VARPTR and PTR86 can be used to do this calculation for any array element, not just array(0). Here's VARPTR: value1! = VARPTR (array1!(0)) value2! = VARPTR (array1!(196)) value3! = VARPTR (array1!(2781)) PRINT value1!, value2!, value3! And here's the output: 68032 68816 79156 From now on, we will have VARPTR inside of the PTR86 call: CALL PTR86 ( seg1% , off1%, VARPTR (array1!(0)) ) CALL PTR86 ( seg2% , off2%, VARPTR (array2!(0)) ) CALL PTR86 ( seg3% , off3%, VARPTR (array3!(0)) ) It is clearer and uses less space. Of course, we can use this for any element in the array, not just the beginning of the array: CALL PTR86 ( seg1% , off1%, VARPTR (array1!(0)) ) CALL PTR86 ( seg2% , off2%, VARPTR (array1!(5076)) ) CALL PTR86 ( seg3% , off3%, VARPTR (array1!(1983)) ) We now have all the ammunition we need to do a quicker disk write. We can pass the offset address of any numeric data in the DS segment, we can pass the string descriptor address of any string, and we can pass the SEGMENT:OFFSET pair of any array anywhere. Before doing our disk write program, I need to say something about subroutine calls. You notice that when I show a subroutine call I am always showing what numeric type I am passing. There is a reason for this. Let's step back from assembler for a minute and do a BASIC program with a subroutine. Here's the program: ********************************************************** floatA! = 27.925 floatB! = 16.96 integerA% = 300 integerB% = 140 The PC Assembler Tutor mmxx ______________________ CALL CheckTheNumbers (integerA%, integerB%, floatA!, floatB!) PRINT floatA!, floatB!, integerA%, integerB% END SUB CheckTheNumbers ( int1%, int2%, flt1!, flt2!) STATIC int1% = int1% + 7 int2% = int2% * 45 flt1! = flt1! + 19.0 flt2! = flt2! * 43.0 END SUB *********************************************************** This gives the following output: 46.925 729.28 307 6300 This is nothing earthshaking. Now let's change one line in the program: CALL CheckTheNumbers (floatA!, floatB!, integerA%, integerB%) In the call statement I have put single precision numbers where the integers were and integers where the single precision numbers were. Now let's look at the output: Overflow. ENTER to debug, SPACEBAR to edit We had an overflow. Here's where the debugger said the error was: int2% = int2% * 45 But int2% received the floating point number for floatB!, and that is 16.96. Since 16.96 * 45 = 763.2, where was the overflow? What the subroutine saw was not 16.96 but the first two binary bytes of floatB! (since we passed the address of floatB!). It thought these were an integer, performed a multiplication and got an error because the result was too big for an integer. Now, change the value in floatB! to: floatB! = 1.696E-29 and run the program again. Your results should be: 27.92501 1.693756E-29 0 16792 These numbers have no relation to either what we started out with or what we did. Why? Because the subroutine mixed binary information from single precision numbers with binary information from integers and came up with unmitigated garbage. It was not only doing that, it was also writing 4 bytes of information into a 2 byte integer. Both flt1! and flt2! were writing past the end of the data and overwriting something else. There is NEVER any checking between a subroutine and the calling program to see that the correct numeric types are being passed (integers, long integers, single precision, double precision). In BASIC II - Interfacing BASIC With Assembler mmxxi ___________________________________________ C and Pascal this checking is done by the compiler at compile time and the compiler will howl if you try to do something like this. In BASIC (my BASIC at least), this checking is not being done. Therefore, it is IMPERATIVE that you make sure that you pass the correct data types. Having given that warning, we are going to build an assembler subprogram that opens a file for writing, then writes a block of data from memory to disk. The form of the call will be: CALL BlockToDisk ("filename$"+CHR$(0), seg%, offset%, # of bytes) Notice that there MUST be a 0 after the filename. When you use this function, you must always have: filename$ = "my.file.name" + CHR$(0) The disk interrupt that is used in this subroutine expects a 'C' string (terminated by a number 0, not an ASCII character '0'). If you don't do it, you will almost certainly get an error. This is followed by the segment of the first byte of data, the offset of the first byte of data, and the number of BYTES (not array elements) to write. Which way does BASIC load the arguments to a subroutine? From left to right, just like PASCAL. Also, in BASIC, ALL subroutine calls are far calls. Therefore, BASIC will do the following when it calls BlockToDisk: PUSH address of file_descriptor PUSH address of block segment PUSH address of block offset PUSH address of length CALL FAR PTR BlockToDisk There are two things to notice here. First, these are all ADDRESSES of the data, not the data items themselves. Upon entry to the subroutine and initialization of BP, the stack will look like this: address of file_descriptor bp+12 address of block segment bp+10 address of block offset bp+8 address of length bp+6 old CS bp+4 old IP bp+2 BP-> old BP bp Secondly, the name of the subroutine does not have any periods. BASIC allows periods '.' but does not allow underscores '_' while assembler allows underscores but doesn't allow periods. (Periods have a special meaning in assembler; they are used in structures). In order to go on from here, you need a book about interrupts. This information is from "DOS Programmer's Reference" by Terry The PC Assembler Tutor mmxxii ______________________ Dettmann, but if you have "The Peter Norton Programmer's Guide to The IBM PC", that's fine too. I'm going to give only partial information about these interrupts and you should have complete information. Here's the program. The explaination will come afterwards. ******************************************************* ; BASOUT.ASM include \pushregs.mac PUBLIC BLOCKTODISK DGROUP GROUP _DATA ; - - - - - - - - - - - - - - - - - - - - - _DATA SEGMENT PUBLIC 'DATA' file_handle dw ? error_message db "Disk i/o error #" error_byte db " ", 13, 10, "$" _DATA ENDS ; - - - - - - - - - - - - - - - - - - - - - _TEXT SEGMENT 'CODE' ASSUME cs:_TEXT, ds:DGROUP ; - - - - - - - - - - - - - - - - - - - - - print_error proc far mov ah, 9 ; print error message mov dx, offset DGROUP:error_message int 21h ret print_error endp ; - - - - - - - - - - - - - - - - - - - - - - - - - ; BlockToDisk ( filename , array SEG, array OFF, # of bytes) ; this is for BASIC DESCRIPTOR_LOCATION EQU [bp + 12] SEGMENT_LOCATION EQU [bp + 10] OFFSET_LOCATION EQU [bp + 8] LENGTH_LOCATION EQU [bp + 6] BLOCKTODISK proc far push bp mov bp, sp PUSHREGS ax, bx, cx, dx, si, ds ; open a new file or truncate an old one mov si, DESCRIPTOR_LOCATION mov ah, 3Ch ; open new or truncate old mov cx, 0 ; normal file attribute mov dx, [si+2] ; [si] =length, [si+2] =location int 21h jnc write_the_file ; ok if CF=0, error if CF=1 mov error_byte, '1' ; cannot open call print_error jmp exit write_the_file: mov file_handle, ax ; store handle for later use mov bx, ax ; file_handle to bx BASIC II - Interfacing BASIC With Assembler mmxxiii ___________________________________________ mov ah, 40h ; int 21h ah = 40h, write block mov si, LENGTH_LOCATION mov cx, [si] ; # of bytes into CX push ds ; save BASIC's DS mov si, OFFSET_LOCATION mov dx, [si] ; offset to DX mov si, SEGMENT_LOCATION mov ds, [si] ; segment to DS int 21h pop ds ; restore BASIC's DS jnc normal_exit ; ok if CF=0, error if CF=1 mov error_byte, '2' ; bad file write call print_error normal_exit: mov ah, 3Eh ; close the file mov bx, file_handle int 21h exit: POPREGS ax, bx, cx, dx, si, ds mov sp, bp pop bp ret (8) ; pop 4 words (8 bytes off the stack) BLOCKTODISK endp ; - - - - - - - - - - - - - - - - - - - - - - - - - _TEXT ENDS END ************************************************************** This is using the standardized segment names so the data will be in the DS segment. Notice the DGROUP GROUP declaration. Also notice that in the 'print_error' subroutine, we have done an offset override with 'offset DGROUP:error_message'. You need to do this every time to avoid errors with the offset addressing whenever you are using the DGROUP GROUP directive. Go back to the discussion of simplified segment directives if you don't remember this. The main program has 3 interrupts. The first one opens a new file or truncates an old one to zero length. The file will be usable for reading and/or writing: Int 21h function 3Ch AH = 3Ch CX = 0 0 indicates a normal file DS:DX address of ASCII filename (terminated by 0) Returns: AX = file handle if CF = 0 or: AX = error code if CF = 1 This filename can be any legitimate pathname specification, and must be terminated by a zero. Like all the disk interrupts we will see in this chapter, if there is an error, the interrupt will set CF = 1. Otherwise it will clear CF = 0. If CF = 0 you The PC Assembler Tutor mmxxiv ______________________ can go on; if CF = 1, there was an error and you need to terminate the subroutine and do some error reporting. The file handle is a number from 0 to 65535 which the operating system gives your program to uniquely identify that open file. There is no other open file in the system which has that number. Guard it carefully because it is your ONLY access to the file. The second interrupt writes a block of data to disk. Int 21h function 40h AH = 40h BX = file handle CX = number of bytes to write DS:DX = address of first byte of data Returns: AX = actual number of bytes written if CF = 0 AX = error code if CF = 1 This too sets the carry flag if there was an error and clears it if there wasn't. It is limited to writing 65535 bytes at a time, but the largest array we can have is 65535 bytes (actually 65534 since all data types have an even number of bytes), so this is no problem. The third interrupt closes the file. Int 21h function 3Eh AH = 3Eh BX = File handle Also, the print-error subroutine has an interrupt Int 21h function 09h AH = 9 DS:DX = first byte of string. This string must be terminated by a dollar sign '$' (of all things). The message is on two lines so we can insert an error number into the middle of the message. This is a quick and dirty interrupt for string printing. All interrupt numbers and function numbers are hex. This is standard for interrupts. If things go wierd, always check first to make sure that you have a hex number and not a decimal number. The data has an 'ASSUME ds:DGROUP' statement. Like Pascal, BASIC requires that the CALLED subroutine pop the arguments off the stack, so we pop 4 extra words (8 extra bytes) with: ret (8) Assemble this program and put the object file in a library with the other object files by using BUILDLIB.EXE. Now all we need is a BASIC program to use this. Here it is: BASIC II - Interfacing BASIC With Assembler mmxxv ___________________________________________ ************************************************************ DIM large.array! (10000) FOR i% = 1 to 10000 large.array! (i%) = 2.167832E+19 NEXT i% filename$ = "blocktxt.doc" + CHR$ (0) length% = 40000 - 65536 PRINT time$ CALL PTR86 (segment%, offset%, VARPTR (large.array!(1)) ) CALL BlockToDisk ( filename$ , segment% , offset%, length% ) PRINT PRINT time$ ************************************************************ There is an extra PRINT statement there which I will explain later. We are starting at large.array!(1) because that is where we started with the other programs. why are we subtracting 65536? Because BASIC has a limit of -32768 to +32767, so we store 40000 as its modular equivalent (mod 65536). How long does the disk write take? From 2 to 3 seconds, and much of that time was spent opening and truncating the file. This is significantly better than the other ways of doing i/o. In fact, the limits of this routine are the limits of your system. It is literally impossible to do disk i/o any faster than this. Try using a filename that doesn't have a CHR$(0) at the end. You should get an error message. In my BASIC, here is the output: 19:50:23 Disk i/o error #1 19:50:23 Now remove that lone PRINT statement (the next to the last line). Here's my output: 19:50:00 D9:50:00 error #1 1 For the QuickBASIC 3.0 environment, BASIC thinks that it has complete control of screen i/o, so it is not doing its i/o in a standard way and is overwriting the error message. If you are going to do any screen i/o from assembler, you will have to think of a way to live in harmony with BASIC.{2} We simply trick BASIC into writing an empty line where the message was. This may not always work correctly, especially if the window is scrolling up. ____________________ 2. The easiest way to do this is to save the whole screen image and cursor location, do what you want using the whole screen, and then restore the screen and the cursor before returning. The PC Assembler Tutor mmxxvi ______________________ This whole program only involved interrupts. There is nothing intrinsically assembler-like in its capabilities. In fact, we'll do its disk read counterpart entirely in BASIC. Let's do something that requires assembler language. The BASIC program: ******************************************* FOR i% = 1 to 10000 to.array!(i%) = from.array!(i%) NEXT i% ******************************************* get's the job done, but is isn't all that fast. It requires about 5.5 seconds. This is a natural for assembler. Dive down into the assembler level, move the string, and come back up for air. Our BASIC program will be: ******************************************* n% = 10000 DIM from.array! (n%), to.array! (n%) PRINT time$ FOR i% = 1 to 10000 to.array!(i%) = from.array!(i%) NEXT i% PRINT time$ FOR j% = 1 to 50 cnt% = 40000 - 65536 'count CALL PTR86 (from.seg%, from.off%, VARPTR( from.array!(1)) ) CALL PTR86 (to.seg%, to.off%, VARPTR ( to.array!(1)) ) CALL BlockMove(from.seg%, from.off%, to.seg%, to.off%, cnt%) NEXT j% PRINT time$ ******************************************** We are doing 50 repeats of the bottom section of code so you will be able to average the time. Here's the assembler program: ; - - - - - - - - - - include /pushregs.mac _TEXT SEGMENT PUBLIC 'CODE' ASSUME cs:_TEXT PUBLIC BlockMove ; - - - - - - - - - - ; BlockMove ( from.seg, from.off, to.seg, to.off, byte.count) ; for BASIC ; MOVSW is from DS:[SI] to ES:[DI] FROM_SEG_ADDRESS EQU [bp+14] FROM_OFFSET_ADDRESS EQU [bp+12] TO_SEG_ADDRESS EQU [bp+10] TO_OFFSET_ADDRESS EQU [bp+8] BYTE_COUNT_ADDRESS EQU [bp+6] ; - - - - - - - - - - BlockMove proc far push bp mov bp, sp BASIC II - Interfacing BASIC With Assembler mmxxvii ___________________________________________ PUSHREGS ax, bx, cx, dx, si, di, es, ds mov si, TO_SEG_ADDRESS mov es, [si] ; to_seg to ES mov si, TO_OFFSET_ADDRESS mov di, [si] ; to_offset to DI mov si, BYTE_COUNT_ADDRESS mov cx, [si] ; byte count to CX mov si, FROM_SEG_ADDRESS mov ax, [si] ; temporary storage for new DS mov si, FROM_OFFSET_ADDRESS mov si, [si] ; from_offset to SI mov ds, ax ; now move from_seg to DS sub bx, bx ; clear BX shr cx, 1 ; divide by 2, remainder in CF rcl bx, 1 ; move CF to low bit of BX cld ; clear DF (go up) rep movsw ; the block move (count in CX) and bx, bx ; one extra byte? jz exit movsb ; move one last byte exit: POPREGS ax, bx, cx, dx, si, di, es, ds mov sp, bp pop bp ret (10) BlockMove endp ; - - - - - - - - - - _TEXT ENDS END ; - - - - - - - - - - This is a string block move using MOVSW. The count is the number of BYTES, not the number of array elements. CX contains the byte count. It is divided by 2 so we can move words, and if there is a remainder (i.e. if the number was odd), BX is set to 1. We move words instead of bytes, and afterwards we check BX to see if we need to move 1 byte more. This routine takes about 1/8 second instead of 5.5 seconds. This is a considerable savings in time. There is a small problem, however. If the FROM block and the TO block overlap (e.g. move 400 bytes from array!(11) to array!(26)), then the data may be compromised. To be exact, if the start of the FROM data is below the start of the TO data, the data will be screwed up. The general solution of this for BASIC is in BLKMOVE.ASM, which is in a file called MISHMASH.DOC which is in \XTRAFILE. Finally, here's the disk read done entirely in BASIC. Once again you need your DOS interrupt book. ******************************************************** ' READBLK.BAS ' reads a block from the disk into memory DIM in.regs%(9), out.regs%(9) The PC Assembler Tutor mmxxviii ______________________ DIM big.array! (10000) AX% = 0 BX% = 1 CX% = 2 DX% = 3 BP% = 4 SI% = 5 DI% = 6 FLGS% = 7 DS% = 8 ES% = 9 filename$ = "blocktxt.doc" + CHR$(0) PRINT time$ ' open an existing file for reading in.regs%(AX%) = &H3D00 in.regs%(DX%) = SADD (filename$) CALL INT86 (&H21,VARPTR (in.regs%(0)),VARPTR (out.regs%(0))) IF (out.regs%(FLGS%) AND &H0001) <> 0 THEN PRINT "Can't open the file." GOTO ExitProgram END IF ' set the i/o pointer to 0 file.handle% = out.regs%(AX%) in.regs%(AX%) = &H4200 in.regs%(BX%) = file.handle% in.regs%(CX%) = 0 in.regs%(DX%) = 0 CALL INT86 (&H21, VARPTR(in.regs%(0)), VARPTR(out.regs%(0))) IF (out.regs%(FLGS%) AND &H0001) <> 0 THEN PRINT "File pointer error" GOTO CloseFile END IF in.regs%(AX%) = &H3F00 in.regs%(BX%) = file.handle% in.regs%(CX%) = 40000 - 65536 CALL PTR86 ( segment%, offset%, VARPTR (big.array!(1) )) in.regs%(DX%) = offset% in.regs%(DS%) = segment% CALL INT86X (&H21,VARPTR(in.regs%(0)),VARPTR(out.regs%(0))) IF (out.regs%(FLGS%) AND &H0001) <> 0 THEN PRINT "Disk read error" END IF CloseFile: in.regs%(AX%) = &H3E00 in.regs%(BX%) = file.handle% CALL INT86 (&H21, VARPTR(in.regs%(0)), VARPTR(out.regs%(0))) PRINT time$ BASIC II - Interfacing BASIC With Assembler mmxxix ___________________________________________ ExitProgram: END ****************************************************** This shows the use of INT86 in BASIC. We have two 10 element integer arrays (0 - 9). One is used for putting the data into the registers before the interrupt call and the other is used for getting the data out of the registers after the interrupt. At the top we have substituted variable names for the array elements they represent. This is the only way to make sense of things in BASIC. What is the difference between INT86 and INT86X? INT86X also changes ES and DS. We need a different file opening call because the last one TRUNCATED the file. This one just opens a pre-existing file and we put 00 in AL to signal a file read: INT 21h Function 3Dh AH = 3dh ; open a pre-existing file AL = 0 ; file read DS:DX ; pointer to a 00h terminated string Returns AX = file handle if CF = 0 AX = error code if CF = 1 We use SADD to get the filename offset in DS because this is exactly what DOS wants. SADD is a function that gives the offset of a string (the string itself) relative to the DS segment. It gives no length information. At every step along the way we check CF to make sure it is 0 and not 1. We need to make sure that the file pointer is at the beginning of the file. This is: INT 21h Function 42h AH = 42h ; move file pointer AL = 0 ; count from beginning of the file CX:DX = 0 ; 4 byte offset from beginning of file Returns DX:AX = new file-pointer location if CF = 0 AX = error code if CF = 1 Then we do the block read: INT 21h Function 3Fh AH = 3Fh ; read a block from disk BX = file handle CX = byte count DS:DX = pointer to first byte of block Returns AX = # of bytes read (can be less than CX) if CF = 0 AX = error code if CF = 1 The PC Assembler Tutor mmxxx ______________________ Finally, we close the file: INT 21h Function 3Eh AH = 3Eh ; close a file BX = file handle Returns nothing if CF = 0 (close was successful) AX = error code if CF = 1 All you need to know about CF is that when the flags are represented as a word (2 bytes) CF = &H0001. Is this faster than our other access methods? Our worst case before took 79 seconds and this one takes 1 second. This is certainly worth using for large disk reads. We don't need to go down to the assembler level, either. What's the difference between this and bload? Bload requires that the file has already been stored from memory. When you use BSAVE, the binary information is written to disk, but the first seven bytes is BLOAD information. The first byte (0FDh) is a signature byte. The next six bytes are three words. (1) the segment where the data came from, (2) the offset where the data came from, and (3) the length of the data. Of course, having these seven bytes in the front makes the file incompatible with everything else in the world. It even makes it difficult to load the information into BASIC the first time, since this seven byte header is missing in the original data unless the data came from a BASIC file. So what sorts of things are candidates for assembler subroutines? Things that are cumbersome in BASIC. If you want the top byte of a number, that's difficult. If you want to rotate the bits of a number, that's extremely hard. Shifting bits is hard. Practically everything involving unsigned numbers is problematic. For every assembler instruction, if you can't do it easily in BASIC you should make a subroutine that does it in assembler. How about one that does unsigned division? Another subroutine that you might want to make is one that returns both the quotient and the remainder from signed division. This cuts the work in half if you need both of them. Well, use BASIC any way you want, but most of all, have fun! BASIC II - Interfacing BASIC With Assembler mmxxxi ___________________________________________ SUMMARY BASIC strings are defined by STRING DESCRIPTORS. A string descriptor is a 4 byte block that contains the LENGTH of the string and its LOCATION in the DS segment. Though you may modify individual bytes of a string from the assembler level, you may not alter the length without interfering with BASIC's memory management system. BASIC passes all arguments by reference. That is it sends the offset address of the data instead of the data itself. The ruiles are: 1) If it is a single piece of numeric data, the offset is relative to the DS segment. 2) If it is a string, the address is the address of the STRING DESCRIPTOR which contains both the length and location of the string. 3) To reference an array, use VARPTR (array(0)) to get the offset and then PTR86 to convert this to a SEGMENT:OFFSET pair which is usable by the assembler subroutine. 4) If you pass a single array element array(x) instead of using VARPTR, BASIC will pass the location of that element, but the element might be separated from the rest of the array, so only pass an individual element if you want the element itself and not the array.{3} SADD (stringname$) SADD [Microsoft's string address function] is used to pass the offset address of stringname$ relative to BASIC's DS segment. It should only be used with 00h terminated strings since this gives no length information. It can, however, be used in conjunction with LEN (stringname$). number! = VARPTR (variable) In older BASICs, VARPTR gives the offset address of "variable" relative to the first byte of the DS segment. This variable can be anywhere in memory from DS:0000 to the end of memory, and the number returned will be a single precision number in the range 0 to 1,048,576. VARSEG and VARPTR In more recent BASICs, using a combination of VARSEG and ____________________ 3. The rule here is that if the array itself is outside of the DS segment (if it is a dynamic array), BASIC will make a copy of the element inside of DS before the CALL, give you the address of the COPY, and return the copy to its appropriate place in the array after the CALL. This copy can be hundreds of thousands of bytes away from the actual array. If you want the element itself this works properly, but if you want the array, the address will be the wrong address. The PC Assembler Tutor mmxxxii ______________________ VARPTR has supplanted the use of PTR86. PTR86 ( segment%, offset%, VARPTR (variable) ) PTR86 [Microsoft's segmentation scheme] takes the result provided by VARPTR and adds it to DS to come up with a total address. It then converts this absolute address into a SEGMENT:OFFSET pair where the segment is the highest segment that contains the first byte of the variable from VARPTR and the offset is a number from 0 to 15 which is the offset of this variable in this segment. RET (x) When executing a return, all called subroutines must pop the arguments passed to them by BASIC. The number of BYTES popped is twice the number of arguments (as long as you are passing addresses and not actual data). CALL MY_ROUTINE (arg1, arg2, arg3, etc) Arguments are always PUSHed on the stack from left to right, so this call will: PUSH address of arg1 PUSH address of arg2 PUSH address of arg3 etc. in that order. INT86 ( interrupt.number%, in.reg.array%(9), out.reg.array%(9) ) INT86 executes a DOS interrupt (interrupt.number%). The integers in in.reg.array% are put into the arithmetic registers before the call and the arithmetic registers are put into out.reg.array after the call. INT86X does the same thing but also changes the DS and ES segment registers. Consult your BASIC manual for the proper ordering of the registers in the array. Neither of these changes CS, SS or SP. The PC ASSEMBLER Helper ======================= i "The PC Assembler Helper" is an object module (ASMHELP.OBJ) which is designed as a companion for "The PC Assembler Tutor". It can also be used as an aid in the development of assembler programs and subprograms. It allows for input to and output from the assembler level, as well as displaying the data in all the 8086 registers. Part 1 will give a discription of all callable subroutines and part 2 will explain how to correctly link a program to ASMHELP.OBJ. THE SUBROUTINES There are a number of routines for displaying data and for inputting data. They all follow a standard format. (1) All subroutines which display output on the monitor at the current cursor position start with the word "print_". On the screen, all hex output is followed by an 'H'. All signed output has a + or a -. Unsigned output has no sign. All binary output is distinguishable because it is either 8 digits or 16 digits long. ASCII output is followed by a single asterisk '*' under normal conditions. If one or more of the ASCII characters cannot be printed (i.e. if it is less than 33d or it is 127d or 255d){1} then it will be displayed as a two digit hex number instead of a single character. A double asterisk '**' is then used to signal the presence of a hex number in an ASCII format. The only time there might be confusion is if one of the characters is also an asterisk. (2) All subroutines which get input from the keyboard start with the word "get_". They all display prompts to tell you what kind of input is desired. (3) One byte input or output is passed through register AL. One word (two byte) input or output is passed through register AX. Any input or output that is longer than two bytes is passed by reference. The offset address of the data is put into AX before calling the subroutine. Each subroutine returns with all 8086 registers (including the flags register) unchanged from when the subroutine was called. As an example, if "variable3" is the name of a variable (in memory) which we want to print as a signed number, the proper way to set up for the calls is: ____________________ 1 In this chapter, as in all others, 'd' stands for decimal, 'h' for hex. ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor ii ______________________ ONE BYTE: mov al, variable3 call print_signed_byte ONE WORD: mov ax, variable3 call print_signed FOUR BYTES: lea ax, variable3 call print_signed_4byte EIGHT BYTES: lea ax, variable3 call print_signed_8byte For a byte or a word we use the data itself, while for data larger than a word we pass the data by reference. The same applies for getting input. Here are the subroutines: get_num Enters a number 5 digits or less, either signed or unsigned. It does no checking to see if the number is too positive or too negative. Returns a two byte value in AX. print_num Prints a word value (from AX) in its signed, unsigned, hex, ASCII, and binary representation. All the other input routines do strict error checking. If an illegal character is entered or if the input is out of range, the subroutine will ask for new input until it receives valid input. get_string Enters a string of 79 characters or less, and puts a 00h at the end of the string (a C string with a maximum total length of 80 bytes). The address of the string must be in AX before making the call. print_string Displays a 00h terminated string (a C string) on the monitor. The address of the first byte of the string must be in AX before making the call. get_ascii_byte Enters a single ascii character and returns it in AL. print_ascii_byte Displays one byte (from AL) as an ascii character. get_ascii Enters one or two characters and returns it (them) in AX. Appendix I - The PC Assembler Helper iii ____________________________________ print_ascii Displays the value in AX as two characters. get_bcd Enters a one to eighteen digit signed number as a 10 byte bcd number. Commas are allowed. The address of the bcd variable must be in AX before calling the routine. print_bcd Displays a 10 byte bcd number as a one to eighteen digit signed number with commas. The address of the bcd number must be in AX before calling the subroutine. get_binary_byte Enters a one to eight digit binary number and returns it in AL. print_binary_byte Displays a one byte number (from AL) as an eight digit binary number. get_binary Enters a one to sixteen digit binary number and returns it in AX. print_binary Displays a one word number (from AX) as a sixteen digit binary number. get_hex_byte Enters a one or two digit hex number and returns it in AL. print_hex_byte Displays a one byte number (in AL) as two hex digits. get_hex Enters a one to four digit hex number and returns it in AX. print_hex Displays a one word number (from AX) as a four digit hex number. get_signed_byte Enters a one byte signed number (-128 to +127) and returns it in AL. print signed_byte Displays a one byte number (from AL) as a signed number The PC Assembler Tutor iv ______________________ (-128 to +127). get_signed Enters a one word signed number (-32768 to +32767) and returns it in AX. print_signed Displays a one word number (from AX) as a signed number (-32768 to +32767). get_signed_4byte Enters a 4 byte signed number (-2,147,483,648 to +2,147,483,647). Commas are allowed. The address of the 4 byte number must be in AX before calling the subroutine. print_signed_4byte Displays a 4 byte signed number (-2,147,483,648 to +2,147,483,647) with commas. The address of the 4 byte number must be in AX before calling the subroutine. get_signed_8byte Enters an 8 byte signed number (-9,223,372,036,854,775,808 to +9,223,372,036,854,775,807). Commas are allowed. The address of the 8 byte number must be in AX before calling the subroutine. The screen prompt will show the last 3 negative digits as -807 instead of -808, but this is because of lack of screen space. print_signed_8byte Displays an 8 byte signed number (-9,223,372,036,854,775,808 to +9,223,372,036,854,775,807) with commas. The address of the 8 byte number must be in AX before calling the subroutine. get_unsigned_byte Enters a one byte unsigned number (0 to 255) and returns it in AL. print_unsigned_byte Displays a one byte number (from AL) as an unsigned number (0 to 255). get_unsigned Enters a one word unsigned number (0 to 65535) and returns it in AX. print_unsigned Displays a one word number (from AX) as an unsigned number (0 to 65535). get_unsigned_4byte Enters a 4 byte unsigned number (0 to 4,294,967,295). Commas are allowed. The address of the 4 byte number must be in AX before calling the subroutine. Appendix I - The PC Assembler Helper v ____________________________________ print_unsigned_4byte Displays a 4 byte unsigned number (0 to 4,294,967,295) with commas. The address of the 4 byte number must be in AX before calling the subroutine. get_unsigned_8byte Enters an 8 byte unsigned number (0 to 18,446,744,073,709,551,615). Commas are allowed. The address of the 8 byte number must be in AX before calling the subroutine. print_unsigned_8byte Displays an 8 byte unsigned number (0 to 18,446,744,073,709,551,615) with commas. The address of the 8 byte number must be in AX before calling the subroutine. Some of the above routines allow commas to be used for data entry. These routines strip the commas before looking at the number, so the following numbers all give the equivalent input: 2134875 2,134,875 21,,,34875 2,1,3,4,8,7,5 21,34,87,5 show_regs Displays the 8086 registers on the top of the screen. Each call to show_regs increments a resettable counter so you can know where you are in the program. The counter starts with an initial value of 0. It also scrolls the screen up if the cursor is past line 19. show_regs_and_wait The same as show_regs except that it waits for you to press ENTER before continuing. set_count Sets the counter in show_regs to the value in AX. The new counter value (incremented by 1) will appear the next time show_regs is called. set_blue If you have a color monitor which is displaying color text, it sets the background to blue. get_continue Waits for you to press ENTER before continuing the program. It enters no data. See also set_timer below. set_timer In order to allow use with a debugger, it is possible to set a timer so the print functions keep control of the screen for a specific amount of time. To use set_timer, you put a number from 1 to 5 in AL, and call set_timer. This will cause a 1 to 5 second delay every time a print function or The PC Assembler Tutor vi ______________________ show_regs is called. A 0 in AL will reset the timer to 0. A number larger than 5 in AL will cause ASMHELP to wait for you to press the ENTER key every time a print function or show_regs is called. kill_timer Resets the timer to 0. set_reg_style Sets the display style of the individual registers. The correct order of the definition is: AX, BX, CX, DX, SI, DI, BP, SP The style values are: FULL REGISTER OR RIGHT HALF REGISTER signed = 1d 1h unsigned = 2d 2h binary = 3d 3h hex = 4d 4h ascii = 5d 5h plus (if applicable) LEFT HALF REGISTER AND HALF REG. BIT signed = 144d 90h unsigned = 160d A0h binary = 176d B0h hex = 192d C0h ascii = 208d D0h set_reg_style makes a COPY of the information, which is used the next (and subsequent) times that show_regs is called. The next several paragraphs explain how this works. AX must contain the address of the 8 byte style definition array before calling set_reg_style. REGISTER DISPLAY STYLE The 8086 contains a number of different registers. Some of these always contain addresses and are always displayed in hex. These are CS, DS, ES, SS and IP. They are always in hex and cannot be changed. Also, each flag has an unchangeable style which will be explained later. This still leaves us with AX, BX, CX, DX, SI, DI, BP and SP. Any of these registers can be displayed in any style desired. In addition, AX, BX, CX and DX can each be broken into two half registers, and each half register has an independent style. The default style is full register, unsigned. If you call show_regs without setting a style, the eight abovementioned registers will all display full, unsigned numbers. Appendix I - The PC Assembler Helper vii ____________________________________ In order to change the style, you need to set up an 8 byte style definition array in your data segment. The correct definition for this is the following: ax_byte db 2 bx_byte db 2 cx_byte db 2 dx_byte db 2 si_byte db 2 di_byte db 2 bp_byte db 2 sp_byte db 2 This is the order that ASMHELP.OBJ expects. It is also the order that the registers appear on the screen. The number 2 is the default value for each byte. Once you have defined the 8 bytes, you may alter any of them that you want to. The possible styles are signed, unsigned, binary, hex, and ascii. If you look at a byte: 76543210 HLLL0RRR the leftmost bit (80h or 128d) signals a half or a full register. If it is 1, you get a half register, if it is 0, you get a full register. bits 4,5 and 6 represent the left half register (if appropriate), while bits 0,1 and 2 represent either the full register or the right half register. For SI, DI, BP and SP, only bits 0, 1 and 2 are significant. For AX, BX, CX and DX, all bits except bit 3 are significant. The code for bits 0, 1 and 2 is the following: signed = 1d 1h unsigned = 2d 2h binary = 3d 3h hex = 4d 4h ascii = 5d 5h This is the correct code for either a full register or the right half register. If you want half registers, you must add 80h (128d) and the code for the left half register. We have for the left half register: DECIMAL HEX signed = 16d + 128d 10h + 80h unsigned = 32d + 128d 20h + 80h binary = 48d + 128d 30h + 80h hex = 64d + 128d 40h + 80h ascii = 80d + 128d 50h + 80h which is the same as the above codes but shifted left 4 bits. Since the 128d and the left half register code always appear in The PC Assembler Tutor viii ______________________ tandem, we may simply add them together. This gives us: DECIMAL HEX signed = 144d 90h unsigned = 160d A0h binary = 176d B0h hex = 192d C0h ascii = 208d D0h Here are some examples: signed left + hex right = 144 + 4 = 148 ascii left + unsigned right = 208 + 2 = 210 binary left + ascii right = 176 + 5 = 181 Simply move the constant to the appropriate byte: mov cx_byte, 210 ; ascii left, unsigned right mov ax_byte, 5 ; full register, ascii mov si_byte, 1 ; full register signed mov dx_byte, 148 ; signed left, hex right mov bx_byte, 4 ; full register, hex After you have changed all the styles you want to, put the address of ax_byte (the first byte of the array) in ax and call set_reg_style: lea ax, ax_byte call set_reg_style ASMHELP.OBJ uses the address in AX and stores a COPY of those 8 bytes. The next time you call show_regs, it will use this copy to define the styles. If you make a mistake, it will default to unsigned. Registers are displayed the same way as with the 'print_' subroutines. Unsigned numbers have no sign. Signed numbers have a + or -, hex has an 'H', binary is either 8 or 16 digits, and ascii has a single asterisk '*' unless one of the characters is not printable ( 00d to 32d, 127d and 255d) in which case the non- printable character will be written as a two digit hex number, and a double asterisk will signal the event. Finally, the flags are displayed as follows: OF, IEF, TF, ZF, AF, and CF are blank if they are not set and have an X if they are set. DF has a + or a - to indicate increment or decrement. SF has a + or a - to indicate the sign. PF has E (for even) or O (for odd). Appendix I - The PC Assembler Helper ix ____________________________________ LINKING In order to use ASMHELP.OBJ you must have some standard segments in your program. The standard code segment is defined as: CODESTUFF SEGMENT PUBLIC 'CODE' and the standard data segment is: DATASTUFF SEGMENT PUBLIC 'DATA' In addition you need a stack segment: STACKSEG SEGMENT STACK 'STACK' which should have at least 100 bytes for ASMHELP to use. ASMHELP.OBJ expects that when you call one of its routines the following conditions are met: (1) It is a near call, and CS is set to the CODESTUFF segment. (2) DS is set to the DATASTUFF segment. Any data which is longer than one word long and is being transferred must be in the DATASTUFF segment and its offset address must be in AX. For one byte or one word data, the data itself is in AX/AL. (3) There is available stack space. In order to link properly, any subroutine which is called must have an EXTRN statement. EXTRN show_regs:NEAR If the above conditions hold, then simply link your object file with asmhelp.obj: C> link myfile.obj+asmhelp.obj and the subroutines will be usable. ALIASES The subroutine names were chosen so their functions would be completely clear. However, they tend to be long so if you use them with some frequency it would be easier to use aliases. Make a redefinition macro file and then include it at the beginning of the program. For instance, if you have: call print_unsigned_8byte The PC Assembler Tutor x ______________________ you might want the EQU statement prt_u8 EQU print_unsigned_8byte and then rewrite the call: call prt_u8 Include redefinitions which are reasonable to you and put them in a file REDEF.MAC. prt_s8 EQU print_signed_8byte get_u4 EQU get_unsigned_4byte show_rw EQU show_regs_and_wait Then all you need on the first line of your program is: include REDEF.MAC and the assembler will do the work for you. MEMORY RESIDENT SHOW_REGS If you are not using a debugger, it is possible to have the show_regs portion of The Assembler Helper resident in memory. This means that once it is installed, it will stay there till you turn the machine off or reset the machine. The name of the program is HELPMEM.COM and it is in \XTRAFILE. It operates exactly the same way as show_regs except you get to it by using INT 3. At that point you can set TF to do single stepping. You load it into memory by typing: >helpmem and it will wait for you to send interrupts. The first time you do INT 3, you will see the normal show_regs screen. The count is reset to 0 each time you have an INT 3. There will also be two menu lines: ---------- 0=clear TF ; X=set TF ; A=regs from ax ; B=set count ; C=continue 1=ax ; 2=bx ; 3=cx ; 4=dx ; 5=si ; 6=di ; 7=bp ; 8=sp ---------- This is pretty clear. 0 clears TF and X sets TF. If you press B you will be prompted for a new count number. C lets you continue. You can set the registers individually. Each register has a number; ax is 1, bx is 2, etc. When you press the number, you will be prompted for a HEX style code. This is the same style code you have been using all the time. Finally, if you want to transfer the style information from your program, do the following: Appendix I - The PC Assembler Helper xi ____________________________________ mov ax, offset ax_byte int 3 Load the address of ax_byte in AX before the interrupt, and then press selection A. The 8 bytes located at the address in AX will be transfered to HELPMEM. The advantage of this is that you can run ASMHELP concurrently, and they will both show the same register styles. When you move into single step mode, you get a different prompt line: ---------- 0 = clear TF and continue ; 2 = menu ; other keys = continue ---------- 0 clears TF, 2 sends you to the above menu, and any other key will continue the program with TF set. As is true with all debuggers, you should not single step through subroutine calls since the subroutine might have hundreds or thousands of steps. In addition, the subroutines in ASMHELP store the flags. This means that they may store the flags with the trap flag set. If you clear the trap flag while inside ASMHELP, when the code exits ASMHELP it will POP the flags with the trap flag set. If this happens, keep pressing 0 until you get out of single step mode. The memory resident version works fine for single stepping as long as there are no interrupts or subroutine calls. The show_regs in ASMHELP works fine inbetween subroutine calls but requires a lot of coding for single stepping. Therefore, a strategy for using these two programs in conjunction is: 1) use the same register style definition array for both programs. 2) when there is only 8086 code with no interrupts or subroutine calls, use INT 3 and single step. 3) When there are interrupts and subroutines interspersed with the 8086 instructions, switch to ASMHELP. The screens should look the same except the count will be different. I/O All data i/o is done to the current screen at the current cursor position. As long as you are in a text mode, ASMHELP should write to whatever is current. The PC Assembler Tutor xii ______________________ SHOW_REGS is a different matter. The only allowed modes are 2, 3 and 7. If you enter in modes 2, 3 or 7, it will keep the same mode. If it is in any other mode, the video card will be forced to mode 3. The memory resident version does almost the same thing. Every time it is called, it checks for modes 2, 3 or 7, and if the mode is not one of these, changes the mode to mode 3. If the mode is 3, then HELPMEM looks at the first character on page 0 and uses the attribute of that character for all its display operations. Whatever display attribute is in the upper left hand corner of page 0 will be the display attribute for everything. The register display occupies the first 10 lines of page 0 of whatever mode you are in. This is written directly to memory and cannot be changed. Both versions change the page to page 0 upon entry. If you are on another page, after a call to show_regs you will be on page 0 at its current cursor position. If you use ASMHELP in a programming environment or with a debugger there may be conflicts as to who controls the screen display. HELPMEM.COM should not be used in a programming environment and cannot be used with a debugger, since both HELPMEM and the debugger try to use the same interrupts. The 8086 Instruction Set ======================== xiii Before I go through the instructions I need to say something about the structure of the instructions. For the majority of instructions, two bits in the first instruction byte and the whole second byte determine whether it is a byte or word instruction and whether it is: 1. register, register 2. register, memory 3. memory, register These refer to the 8 arithmetic registers, not the segment registers. "Memory" is any allowable addressing mode. Therefore, it is simpler to abbreviate this as: reg/mem, reg/mem This will always mean: 1. either a byte or word operation is allowed 2. any of the above 3 possibilities is allowed unless specifically stated otherwise. Another possibility is the operations with constants: add ax, 7 xor sign_flag, 80h These follow the same form. There are two possibilities: 1. register, constant 2. memory, constant where this may be a byte or a word, "register" is any arithmetic register, and "memory" is any allowable addressing mode. These will be abbreviated: reg/mem, constant If we are taliking about a segment register, they will be abbreviated: segreg If anything does not follow this pattern, it will be explicitly stated. DESTINATION,SOURCE The standard way of writing instructions is with the destination on the left and the source on the right: ______________________ The PC Assembler Tutor - Copyright (C) 1989 Chuck Nelson The PC Assembler Tutor xiv ______________________ add destination, source The register or memory location on the left ALWAYS gets the result of the operation. The thing on the right is the second operand (if any). ADDRESSING MODES These are the natural (default) segments of all addressing modes: (1) DS variable + (constant) [bx] + (constant) [si] + (constant) [di] + (constant) [bx+si] + (constant) [bx+di] + (constant) (2) SS [bp] + (constant) [bp+si] + (constant) [bp+di] + (constant) Where the constant is optional. Segment overrides may be used. The segment overrides are: SEGMENT OVERRIDE MACHINE CODE (hex) CS: 2E DS: 3E ES: 26 SS: 36 These default segments apply to all instructions except the string instructions, which will be explained individually. THE INSTRUCTION SET AAA (ascii adjust for addition) adjusts AL, assuming that it contains the result of a legitimate unpacked BCD addition. If the lower half-byte has generated a result 10 or over, it subtracts 10, carries 1 to AH, and sets the carry flag. If the result is 9 or less, it clears CF. In either case it zeroes the upper half-byte of AL. AAD (ascii adjust for division) PREPARES AL and AH for division. It assumes that AH contains the 10's digit and AL contains the 1's digit of a two byte unpacked BCD number. It multiplies AH by 10 and adds it to AL, thus making a single integer between 0 and 99. It zeroes AH in preparation for unsigned division. Appendix II - The 8086 Instruction Set xv ______________________________________ AAM (ascii adjust for multiplication) adjusts AL, assuming that it contains the result of a legitimate BCD multiplication. It divides the result by 10, putting the quotient in AH and the remainder in AL. If AL contains 73 before AAM, then afterwards AH will contain 7 and AL will contain 3 AAS (ascii adjust for subtraction) adjusts AL, assuming that it contains the result of a legitimate unpacked BCD subtraction. If the lower half-byte has generated a result -1 or less, it borrows 1 from AH, adds 10 to AL, and sets the carry flag. If the result is 0 or more, it clears CF. In either case it zeroes the upper half-byte of AL. ADC (add with carry) adds two integers, either signed or unsigned, and adds in the carry from the previous arithmetic operation. It is used for multiple word (byte) addition. adc reg/mem, reg/mem adc reg/mem, constant ADD adds two integers, either signed or unsigned. add reg/mem, reg/mem add reg/mem, constant AND ands two bytes or words. A bit remains set only if both its respective source and destination bits were set. and reg/mem, reg/mem and reg/mem, constant CALL calls a procedure. The two forms we have used are NEAR and FAR calls. call set_regs call FAR PTR calculate_array There are two other forms where the addresses of the subprograms are in memory and changable. You should only use these forms if you are writing an operating system or a compiler, since they are an invitation to disaster. CBW sign extends the signed byte in AL to AH:AL in preparation for signed division. It is used alone without any register specification. cbw The PC Assembler Tutor xvi ______________________ CLC clears the carry flag (CF = 0). CLD clears the direction flag (increments string pointers). CLI clears the interrupt flag (disables interrupts). All maskable interrupts must wait till the flag is set again before they will be processed. CMC changes the value of the carry flag. If CF =1, then CF = 0; if CF = 0, then CF = 1. CMP compares two values. It is the same as SUB except it does not alter the "destination" variable. Its job is to set the flags as if a subtraction had been performed, in preparation for a conditional jump instruction. cmp reg/mem, reg/mem cmp reg/mem, constant CMPS compares the byte (or word) at DS:[si] with the one at ES:[di], (calculating [si] - [di]) and sets the flags accordingly. It increments (or decrements) SI and DI depending on the setting of DF, the direction flag (by 1 for bytes and by 2 for words). You may use CS:[si], SS:[si] or ES:[si], but you MAY NOT OVERRIDE ES:[di]. Notice that SI and DI are reversed from their position in the other string instructions. cmpsb cmpsw cmps BYTE PTR SS:[si], ES:[di] ;or CS, DS, ES:[si] cmps WORD PTR SS:[si], ES:[di] ;or CS, DS, ES:[si] CWD sign extends the signed integer in AX to DX:AX in preparation for signed word division. No register specification is used. cwd DAA (decimal adjust for addition) adjusts AL, assuming that it contains the result of a legitimate packed BCD addition. It treats AL as two independent half-bytes. If the result of the lower half-byte is 10 or over, it subtracts 10 from the lower half-byte and adds the carry to the upper half byte. It then looks at the upper half byte. If its result is 10 or over, DAA subtracts 10 from the upper half byte and sets the carry flag. Appendix II - The 8086 Instruction Set xvii ______________________________________ Otherwise the carry flag is cleared. DAS (decimal adjust for subtraction) adjusts AL, assuming that it contains the result of a legitimate packed BCD subtraction.It treats AL as two independent half-bytes. If the result of the lower half-byte is -1 or less, it adds 10 to the lower half-byte and borrows 1 from the upper half byte. It then looks at the upper half byte. If its result is -1 or less, DAA adds 10 to the upper half byte and sets the carry flag to indicate a borrow. Otherwise the carry flag is cleared. DEC decrements the byte or word by 1. It does not alter CF. dec reg/mem DIV performs unsigned division. If byte division, the unsigned double byte must be in AH:AL. Afterwards, AL has the quotient and AH has the remainder. If word division, the unsigned word must be in DX:AX. Afterwards, AX has the quotient, and DX the remainder. Division by a constant is not allowed. The DIV instruction does not mention DX:AX or AH:AL. They are understood. div reg/mem ESC (escape) signals to the 8086 that the bytes starting with the ESC byte are a coprocessor instruction. The 8086 will calculate a memory address (if appropriate) and give it to the coprocessor. The 8086 will then go on to the next instruction. HLT (halt) halts the machine. It can be restarted with a reset, a non-maskable interrupt, or (if IEF is set) with a maskable interrupt. IDIV performs signed integer division. For byte division, the value in AL must be sign extended to AH (with CBW), giving the double signed byte AH:AL, after division, AL contains the quotient and AH contains the remainder. For word division, the value in AX must be sign extended to DX (with CWD), giving the double signed word DX:AX, after division, AX contains the quotient and DX contains the remainder. Division by a constant is not allowed. IDIV cannot perform multiple word division. To do that, you need to make the numbers positive and use DIV, adjusting the signs afterwards. IDIV does not mention the AL or AX register, it is understood. The PC Assembler Tutor xviii ______________________ idiv reg/mem IMUL performs signed integer multiplication. The multiplicand must be in AX (or AL for bytes). After the multiplication, the result is in DX:AX for words and AH:AL for bytes. If DX (or AH for bytes) contains significant information, then CF is set (=1). If all the information is in AX (or AL for bytes) then CF = 0. Multiplication by a constant is not allowed. The IMUL instruction does not mention AX (or AL); it is understood. imul mem/reg IN allows you to move data from a port address to AX (for a word) or AL (for a byte). There are two forms: IN al, port_number IN al, dx Where 'port_number' is a CONSTANT that is hard coded into the machine instruction and is from 0 - 255. The port address in DX may be from 0-65535. It must be the DX register. INC increments a register or a variable by 1. It does not effect CF. inc reg/mem INT (interrupt) sends the program to a subprogram whose address is located in the interrupt vector table in low memory. For any interrupt number I, the 8086 goes to segment 0000, offset (I X 4). It loads IP with the first two bytes, then loads CS from the next two bytes. The next instruction executed will be at the new CS:IP. Before loading the new CS and IP, it pushes (1) the flags, (2) the old CS and (3) the old IP in that order. int 3 int 21h INTO checks OF. If OF = 1, INTO generates interrupt 4. Otherwise it does nothing. It is used for error handling of signed numbers. into IRET (interrupt return) is a special return instruction for interrupt routines. It pops (1) the old IP, (2) the old CS, and (3) the old flags in that order. Appendix II - The 8086 Instruction Set xix ______________________________________ iret JUMPS --------------------------------------------------------- There are two kinds of jumps. JMP is unconditional and can jump anywhere in the segment. All other jumps are conditional and are limited jumps from -128 to +127 bytes from the END of the jump instruction.{1} THESE ARE THE JUMP INSTRUCTIONS FOR SIGNED NUMBERS jg ; jump if greater jnle ; jump if not (less or equal) jl ; jump if less jnge ; jump if not (greater or equal) je ; jump if equal jz ; jump if zero jge ; jump if greater or equal jnl ; jump if not less jle ; jump if less or equal jng ; jump if not greater jne ; jump if not equal jnz ; jump if not zero THESE ARE THE JUMP INSTRUCTIONS FOR UNSIGNED NUMBERS ja ; jump if above jnbe ; jump if not (below or equal) jb ; jump if below jnae ; jump if not (above or equal) je ; jump if equal jz ; jump if zero jae ; jump if above or equal jnb ; jump if not below jbe ; jump if below or equal jna ; jump if not above ____________________ 1 There is also a long jump which can jump anywhere in memory (from 00000 to FFFFF). Except under special circumstances, using this is truly lousy programming practice. The long jump is for the Olympics, not for quality programming. The PC Assembler Tutor xx ______________________ jne ; jump if not equal jnz ; jump if not zero THESE JUMPS CHECK A SINGLE FLAG These come in opposite pairs jc ; jump if the carry flag is set jnc ; jump if the carry flag is not set jo ; jump if the overflow flag is set jno ; jump if the overflow flag is not set jp or jpe ; jump if parity flag is set (parity is even) jnp or jpo ;jump if parity flag is not set (parity is odd) js ; jump if the sign flag is set (negative ) jns ; jump if the sign flag is not set (positive or 0) THIS CHECKS THE CX REGISTER jcxz ; jump if cx is zero This is used before entry to a loop to make sure the loop counter is not set to 0. INDIRECT JUMPS are jumps where the information for the jump is stored in memory (or in a register for an in-segment jump). These forms are dangerous and should be used only when writing things like multi-tasking operating systems. Have you written one lately? END OF JUMPS - - - - - - - - - - - - - - - - - - - - - - - LAHF (load AH from flags) loads the lower half of the flags register into AH. LDS (load DS) loads the first two bytes of the memory address into the named arithmetic register and the next two bytes into DS. The register may be any arithmetic register, but this is designed for the 4 addressing registers: BX, SI, DI and BP. lds reg, memory_address LEA calculates an offset address and puts it in the named register. Appendix II - The 8086 Instruction Set xxi ______________________________________ lea ax, [bp+si]+145 LES (load ES) loads the first two bytes into the named arithmetic register and the next two bytes into ES. The register may be any arithmetic register, but this is designed for the 4 addressing registers: BX, SI, DI and BP. les reg, memory_address LOCK is for use if there is more than one 8086 operating in a computer. LOCK allows one 8086 to be the only one which can read from or write to memory during the following instruction. LODS moves a byte or word from DS:[si] to AL or AX, and increments (or decrements) SI depending on the setting of DF, the direction flag (by 1 for bytes and by 2 for words). You may use CS:[si], SS:[si] or ES:[si]. lodsb lodsw lods BYTE PTR SS:[si] ; or CS, DS, ES:[si] lods WORD PTR SS:[si] ; or CS, DS, ES:[si] LOOP/LOOPE/LOOPNE are conditional jumps (-128 to + 127 bytes). They will jump back to the start of the loop (which is identified by a label), under the following conditions: loop decrement cx ; jump back if cx is not zero loope decrement cx ; jump back if cx not zero AND zf = 1 loopz decrement cx ; jump back if cx not zero AND zf = 1 loopne decrement cx ; jump back if cx not zero AND zf = 0 loopnz decrement cx ; jump back if cx not zero AND zf = 0 Here, 'e' stands for equal, 'z' is zero and 'n' is not. loop some_label MOV is the general instruction for moving bytes or words on the 8086. The forms are: reg/mem, reg/mem reg/mem, constant segreg, reg/mem reg/mem, segreg Notice that you may not (1) move a constant to a segment register, (2) move a segment register to another segment The PC Assembler Tutor xxii ______________________ register, or (3) move from memory to memory. MOVS moves a byte (or a word) from DS:[si] to ES:[di], and increments (or decrements) SI and DI depending on the setting of DF, the direction flag (by 1 for bytes and by 2 for words). You may use CS:[si], SS:[si] or ES:[si], but you MAY NOT OVERRIDE ES:[di]. movsb movsw movs BYTE PTR ES:[di], SS:[si] ;or CS, DS, ES:[si] movs WORD PTR ES:[di], SS:[si] ;or CS, DS, ES:[si] MUL performs unsigned integer multiplication. The multiplicand must be in AX (or AL for bytes). After the multiplication, the result is in DX:AX (or AH:AX for bytes). If DX (or AH for bytes) contains significant information, then CF = 1. If all the significant information is in AX (or AL for bytes) then CF = 0. Multiplication by a constant is not allowed. The MUL instruction does not mention AX (or AL). It is understood. mul reg/mem NEG performs '0 - number' on a number and sets the flags accordingly. neg reg/mem NOP (no operation) does absolutely nothing. It can fill up space which was previously occupied by a different instruction. NOT performs bitwise logical NOT on a word or a byte. not reg/mem OR performs bitwise logical OR on a byte or word or reg/mem, reg/mem or reg/mem, constant OUT moves a byte (or word) from AL (or AX) to the named port. There are two forms: Appendix II - The 8086 Instruction Set xxiii ______________________________________ out port_number, ax out dx, ax where 'port_number' is a CONSTANT which is coded into the machine code. The constant may be from 0-255. DX (and it must be DX) may hold a port number from 0-65535. POP pops a WORD off the stack. Technically, POP takes the word at SS:SP and puts it into the named register or memory location, then ADDS 2 to SP. pop mem/reg pop segreg POPF pops a WORD off the top of the stack into the flags register. Its technical operation is the same as for POP. PUSH pushes a WORD on the stack from the named register or memory location. Technically, PUSH subtracts 2 from SP, then moves the word to SS:SP. push reg/mem push segreg PUSHF pushes the flags register on the stack. Its technical operation is the same as for PUSH. RCR (rotate through carry right) and RCL (rotate through carry left) rotate the bits of a register or of memory data right and left respectively. The bit which is shoved off the register (or data) is placed in CF and CF is placed on the other side of the register (or data). There are two fixed forms. (1) rotate 1 bit and (2) rotate by the number in CL. rcr reg/mem, 1 rcl reg/mem, cl REP. The string instructions may be prefixed by REP/REPE/REPNE which will repeat the instructions according to the following conditions: rep decrement cx ; repeat if cx is not zero repe decrement cx ; repeat if cx not zero AND zf = 1 repz decrement cx ; repeat if cx not zero AND zf = 1 repne decrement cx ; repeat if cx not zero AND zf = 0 The PC Assembler Tutor xxiv ______________________ repnz decrement cx ; repeat if cx not zero AND zf = 0 Here, 'e' stands for equal, 'z' is zero and 'n' is not. These repeat instructions should NEVER be used with a segment override, since the 8086 will forget the override if a hardware interrupt occurs in the middle of the REP loop. rep movsb repe scasb repne cmpsw RET returns from the subroutine to the calling program. The return will be either NEAR or FAR depending on the type of procedure it is in. It may take the parameters off the stack (for Pascal) or leave them on the stack (for C). ret (6) ; Pascal - take 6 bytes off the stack ret ; C - do nothing ROR (rotate right) and ROL (rotate left) rotate the bits of a register or memory data right and left respectively. The bit which is shoved off one end is moved to the other end. CF indicates whether the last bit moved from one end to the other was a 1 or a 0. There are two fixed forms. (1) rotate 1 bit and (2) rotate by the number in CL. ror reg/mem, 1 rol reg/mem, cl SAHF (store AH into flags) puts AH into the low byte of the flags register. SAL (shift arithmetic left) and SHL (shift logical left) are exactly the same instruction. They move bits left. 0s are placed in the low bit. Bits are shoved off the register or memory data on the left side, and CF indicates whether the last bit shoved off was a 1 or a 0. It is used for multiplying an unsigned number by powers of 2. There are two fixed forms. (1) shift 1 bit and (2) shift by the number in CL. shl reg/mem, 1 sal reg/mem, cl SAR (shift arithmetic right) shifts bits right. The high (sign) bit stays the same throughout the operation. Bits are shoved off the register or memory location on the right side. CF indicates whether the last bit shoved off was a 1 or a 0. It is used (with Appendix II - The 8086 Instruction Set xxv ______________________________________ difficulty) for dividing a signed number by powers of 2. There are two fixed forms. (1) shift 1 bit and (2) shift by the number in CL. sar reg/mem, 1 sar reg/mem, cl SBB (subtract with borrow) subtracts one integer from another and then subtracts 1 more if CF is set. It works with both signed and unsigned numbers and is used for multiple word arithmetic. sbb reg/mem, reg/mem sbb reg/mem, constant SCAS compares AL (or AX) to the byte (or word) pointed to by ES:[di], and increments (or decrements) DI depending on the setting of DF, the direction flag (by 1 for bytes and by 2 for words). NO OVERRIDES ARE ALLOWED. scasb scasw scas BYTE PTR ES:[di] ; no override allowed scas WORD PTR ES:[di] ; no override allowed SHR (shift logical right) does the same thing as SHL but in the opposite direction. Bits are shifted right. 0s are placed in the high bit. Bits are shoved off the register or memory location on the right side and CF indicates whether the last bit shoved off was a 0 or a 1. It is used for dividing an unsigned number by powers of 2. There are two fixed forms. (1) shift 1 bit and (2) shift by the number in CL. shr reg/mem, 1 shr reg/mem, cl STC sets the carry flag (CF = 1). STD sets the direction flag (DF = 1, which decrements). STI sets the interrupt flag (IEF = 1; interrupts enabled). STOS (store to string) moves a byte (or a word) from AL (or AX) to ES:[di], and increments (or decrements) DI depending on the setting of DF, the direction flag (by 1 for bytes and by 2 for words). NO SEGMENT OVERRIDES ARE ALLOWED. stosb The PC Assembler Tutor xxvi ______________________ stosw stos BYTE PTR ES:[di] ; no override allowed stos WORD PTR ES:[di] ; no override allowed SUB subtracts one integer from another. It works for both signed and unsigned numbers. sub reg/mem, reg/mem sub reg/mem, constant TEST performs logical AND, but does not alter the value of the "destination" variable. Its purpose is to set the flags for conditional jumps. test reg/mem, reg/mem test reg/mem, constant WAIT forces the 8086 to wait until the coprocessor is finished with its instruction before the 8086 can go on to the next instruction. XCHG exchanges the values in two registers or memory locations. xchg reg/mem, reg/mem XLAT. If BX contains the address of a 256 byte translation table, then XLAT goes to (BX + AL), takes the byte there and puts it into AL. DS is the normal segment, but segment overrides are allowed. XOR performs logical exclusive OR on two numbers. xor reg/mem, reg/mem xor reg/mem, constant APPENDIX III - Instruction Speed and Flags ========================================== xxvii This appendix contains a list of all the 8086 instructions along with their relatives speed and the flags they affect. These speed numbers are from INTEL and are in clock ticks.{1} If you have a 25 mhz clock, then it is doing 25 million ticks per second. If you have a 4.77 mhz clock, then it is doing 4.77 million ticks per second. Instead of calling them ticks, I'll be calling them clocks. In order to understand these numbers, you need a little extra information. In order to do an instruction, ANY microprocessor must: (1) calculate any memory addresses. (2) fetch the two operands (or the single operand if it is an instruction like NOT). (3) process the instruction and (4) put the result in the destination. While (3) will require the same amount of time whether the operands are in memory or in registers, (1), (2) and (4) are all different depending on where the operands are. Some machines allow both operands to be in memory, but the 8086 does not. Therefore, (1) is either NO memory addresses (if everything is in registers) or ONE memory address. Calculating a memory address involves (a) calculating the offset from the beginning of the segment and (b) adding it to the segment starting address. Part (a) is different depending on the addressing mode. In terms of the speed of calculating (a) and (b), the order is: (i) one pointer (ii) named variable (+ constant) (iii) two pointers (iv) one pointer + constant (v) two pointers + constant The reason that (ii) contains both "variable" and "variable + constant" is that the addition (variable + constant) is done by the assembler, not the 8086. By the time the 8086 sees it, it is a single number. The constants in both (iv) and (v) need to be added by the 8086, and this takes extra time. According to INTEL, here is the time required for calculating all possible memory addresses. Note that some pointers are marginally faster than other pointers (this is trivial - don't worry about it). ____________________ 1 All the speed numbers are from "Programmer's Pocket Reference Guide", (c)1980, 1982 Intel Corporation. ______________________ The PC Assembler Tutor - Copyright (C) 1990 Chuck Nelson The PC Assembler Tutor xxviii ______________________ ADDRESSING MODE CLOCKS (EA) (i) [bx], [si], [di], [bp] 5 (ii) variable (+ constant) 6 (iii) [bp+di] or [bx+si] 7 [bp+si] or [bx+di] 8 (iv) ([bx] or [si] or [di] or [bp]) + constant 9 (v) ([bp+di] or [bx+si]) + constant 11 ([bp+si] or [bx+di]) + constant 12 The most complicated memory address takes 2.4 times longer to calculate than the simplest address. These calculation times will be noted by EA (calculate Effective Address). Remember, if both operands are in registers, this calculation does not have to be done. In order for you to see how all of this works, we'll use ADD as an example. Don't start using the table till you understand this example. On the 8086, we normaly have the following possibilities for source and destination: register, register register, memory memory, register register, constant memory, constant In Appendix II, we simply combined them as: reg/mem, reg/mem reg/mem, constant We can't do this here because they all have different times. For ADD they are: ADD CLOCKS register, register 3 register, memory 9 + EA memory, register 16 + EA register, constant 4 memory, constant 17 + EA Notice how much faster using a register is. The EA stands for "calculate Effective Address" and is the number from the list above. If you have: add ax, bx that is "register, register", and it will take 3 clocks to execute. If you have: add [bx+di+9], 17 that is "memory, constant" and will take 17 + EA. What is EA here? According to the above list, [BX+DI+CONSTANT] takes 12 Appendix III - Speeds and Flag Settings xxix _______________________________________ cycles, so 17 + EA is 17 + 12 is 29, so this will take 29 clocks. That's right. The one instruction is almost 10 times slower than the other. If you can move things into some registers, do a number of calculations, and then move them back to memory, you can save a lot of time. Let's do a few examples to make sure that you see all of them: INSTRUCTION TYPE TIME add variable1, bl memory, register 16 + EA = 22 add bl, variable1 register, memory 9 + EA = 15 add [si], di memory, register 16 + EA = 21 add di, [si] register, memory 9 + EA = 14 Examples 1 and 2 are the same except that source and destination have been switched. The same applies to examples 3 and 4. Notice that when the 8086 has to fetch the variable and then put the result back in memory, it is significantly slower than when it just gets the variable from memory and puts the result in a register. INSTRUCTION TYPE TIME add ax, cx register, register 3 add di, 1876 register, constant 4 add variable1, 199 memory, constant 17 + EA = 23 To show you how the different types of EA effect the time, let's do all 3 types of "source, destination" that involve memory. First, "memory, register": INSTRUCTION TIME add [bx], ax 16 + EA = 21 add variable1, ax 16 + EA = 22 add [bp+di], ax 16 + EA = 23 add [bp+si], ax 16 + EA = 24 add [bx+9], ax 16 + EA = 25 add [bp+di+294], ax 16 + EA = 27 add [bx+di+294], ax 16 + EA = 28 Now let's do the same things but go "register, memory": INSTRUCTION TIME add ax, [bx] 9 + EA = 14 add ax, variable1 9 + EA = 15 add ax, [bp+di] 9 + EA = 16 add ax, [bp+si] 9 + EA = 17 add ax, [bx+9] 9 + EA = 18 add ax, [bp+di+294] 9 + EA = 20 add ax, [bx+di+294] 9 + EA = 21 The PC Assembler Tutor xxx ______________________ And finally we have "memory, constant": INSTRUCTION TIME add [bx], 177 17 + EA = 22 add variable1, 177 17 + EA = 23 add [bp+di], 177 17 + EA = 24 add [bp+si], 177 17 + EA = 25 add [bx+9], 177 17 + EA = 26 add [bp+di+294], 177 17 + EA = 28 add [bx+di+294], 177 17 + EA = 29 Is this everything you need to know before looking at the list? Not quite. Most of the 8086 family has a 16 bit data bus. That means that there are 16 wires connecting the processor to memory, and the processor reads 1 word (2 bytes) at a time. These memory reads ALWAYS start at an even location 1472d, 88026d, 198752d, etc. If you are reading one byte, it makes no difference whether it is at an even or odd location. If you are reading a word at an even location, then everything is normal. If you are reading a word at an ODD location, however, the processor must: 1) start reading at the first even location that contains the variable. 2) read the next even location (which contains the last part of the variable). 3) join the parts together. As an example, let's take a word at address 21957 (i.e. 21957- 21958). The processor will: 1) read the high byte from the word at 21956 2) read the low byte from the word at 21958 3) join them together. It now has 21957-21958. The processor can do this, but it takes extra time (4 extra clock ticks), so our speed listing will also contain the following notice: WORDS WHICH ARE AT ODD ADDRESSES NEED 4 EXTRA CLOCKS Thus for: add ax, variable1 (9 + EA = 15) if the address is an even location, the instruction will require 15 clocks. If it is an odd location, the instruction will require 19 clocks (4 extra ticks). It is worth your while to keep words at even locations if at all possible. Appendix III - Speeds and Flag Settings xxxi _______________________________________ INSTRUCTION SPEEDS AND FLAGS REGISTER: One of the arithmetic registers - either word (AX, BX, CX, DX, SI, DI, BP, SP for word operations) or byte (AH, AL, BH, BL, CH, CL, DH or DL for byte operations). AX or AL: AX and AL are considered special on the 8086. Sometimes there is a special AX/AL form of the instruction, (such as for ADD). This form will be shorter. It will only be noted if it is FASTER (such as for MOV) or if it is the only form allowed. It will either say (AX/AL) or (AX only) depending on whether both words and bytes are allowed or only words are allowed. Though both multiplication and division require the use of the (AX/AL) register, the AX/AL is understood, and is not mentioned in the instruction. SEGREG: One of the 4 segment registers - CS, DS, ES or SS. MEMORY: Either a byte or word in memory. It may be addressed with any possible addressing mode, and the extra time needed to calculate the address in memory (EA - calculate Effective Address) is the following: ADDRESSING MODE CLOCKS (EA) (i) [bx], [si], [di], [bp] 5 (ii) variable (+ constant) 6 (iii) [bp+di] or [bx+si] 7 [bp+si] or [bx+di] 8 (iv) ([bx] or [si] or [di] or [bp]) + constant 9 (v) ([bp+di] or [bx+si]) + constant 11 ([bp+si] or [bx+di]) + constant 12 EXTRA TIME: 1) WORDS WHICH ARE AT ODD ADDRESSES NEED 4 EXTRA CLOCKS. 2) A segment override adds 2 clocks to the instruction. FLAGS Following the instruction mnemonic is information in square brackets that indicates how the instruction effects the flags register. There are three possibilities 1) The instruction may alter the value of a flag in a particular way depending on the result. For AND, the sign, zero and parity flags [SZP] will be set according to the result. The PC Assembler Tutor xxxii ______________________ 2) The instruction may set a flag to a specific number (either 1 or 0). For AND, the overflow flag and the carry flag are cleared [(OC=0)]. 3) The instruction may unreliably alter a flag. That means that the instruction might change the flag, but that it gives you no information. This kind of flag cannot be trusted after this operation. For AND, the auxillary flag is unreliable [?A?]. The information will always be displayed in this order. The flags which are reliably set according to the result will be listed first [SZP]. Next, any flags which are either set or cleared will be put inside parentheses followed by an equal sign followed by the value of the flag (all inside of the parentheses) [SZP,(OC=0)]. Finally, any unrelaible flags will be put between question marks [SZP,(OC=0),?A?]. Each part will be separated by commas. If no flags are changed by the instruction, the brackets will have [none] written between them. Each flags will be indicated by a single letter. The letters are: O overflow flag 0 = no overflow, 1 = overflow D direction flag direction of movement for string instructions 0 = upwards, 1 = downwards I interrupt enable 0 = no ints, 1 = ints o.k. T trap flag trap next instruction? 0 = no trap, 1 = trap S sign flag 0 = positive, 1 = negative Z zero flag 0 = non-zero, 1 = zero A auxillary flag carry out of bottom half register? 0 = no, 1 = yes P parity flag 0 = odd, 1 = even C carry flag 0 = no carry, 1 = carry ***************** THE INSTRUCTIONS ********************* INSTRUCTION TIMING AAA [AC,?OSZP?] 4 clocks AAD [SZP,?OAC?] 60 clocks AAM [SZP,?OAC?] 83 clocks AAS [AC,?OSZP?] 4 clocks Appendix III - Speeds and Flag Settings xxxiii _______________________________________ ADC [OSZAPC] see ADD ADD [OSZAPC] register, register 3 register, memory 9 + EA memory, register 16 + EA register, constant 4 memory, constant 17 + EA AND [SZP,(OC=0),?A?] see ADD CALL [none] near call 19 far call 28 near call (reg-ind) 16 {2} near call (mem-ind) 21 + EA far call (mem-ind) 37 + EA CBW [none] 2 clocks CLC [(C=0)] 2 clocks CLD [(D=0)] 2 clocks CLI [(I=0)] 2 clocks CMC [C] 2 clocks CMP [OSZAPC] register, register 3 register, memory 9 + EA memory, register 9 + EA register, constant 4 memory, constant 10 + EA CMPS [OSZAPC] 22 clocks CWD [none] 5 clocks DAA [SZAPC,?O?] 4 clocks DAS [SZAPC,?O?] 4 clocks DEC [OSZAP] word register 2 byte register 3 word/byte memory 15 + EA DIV [?OSZAPC?] byte register 80 - 90 {3} word register 144 - 162 ____________________ 2 These three last ones are indirect calls. They get the address of the subroutine from a register (reg-ind) or from memory (mem-ind). 3 The smaller the numbers, the faster this operation can be accomplished. This applies for signed and unsigned multiplication and division. They all show a range of values rather than a specific value. The PC Assembler Tutor xxxiv ______________________ byte memory (86 - 96) + EA word memory (150 - 168) + EA ESC [none] memory 8 + EA coproc. register 2 This only makes sense if you know about coprocessors. HLT [none] 2 clocks IDIV [none] byte register 101 - 112 word register 165 - 184 byte memory (107 - 118) + EA word memory (171 - 190) + EA IMUL [OC,?SZAP?] byte register 80 - 98 word register 128 - 154 byte memory (86 - 104) + EA word memory (134 - 160) + EA IN [none] (AX/AL), port# 10 (AX/AL), dx 8 INC [OSCAP] word register 2 byte register 3 word/byte memory 15 + EA INT [(IT=0) {4} ] 51 clocks {5} INTO [ {6} ] overflow 54 no overflow 4 IRET [ {7} ] 24 clocks J(condition) [none] This includes all conditional jumps (JAE, JZ, JNO, JLE, JP, etc.) with the exception of JCXZ. jump 16 no jump 4 ____________________ 4 Although this sets the trap flag and interrupt flag to 0, it doesn't do it to YOUR flags, it does it to the flags that the interrupt will see. Your flags are safely stored on the stack and will return unaltered at the end of the interrupt. 5 There is one exception. INT 3, when coded as the single byte trap interrupt, is 52 clocks. 6 If there is overflow, it pushes your flags and sets (IT=0) for the interrupt. If there was no overflow, it does nothing. In either case your own flags will remain unaffected. 7 This puts the copy of your old flags back in the flags register. The flags will be the same as they were when you called the interrupt. Appendix III - Speeds and Flag Settings xxxv _______________________________________ JCXZ [none] jump 18 no jump 6 JMP [none] same segment 15 different segment 15 near (reg-ind) 11 {8} near (mem-ind) 18 + EA far (mem-ind) 24 + EA LAHF [none] 4 clocks LDS [none] 16 + EA LES [none] 16 + EA LEA [none] 2 + EA LOCK [none] 2 clocks LODS [none] 12 clocks LOOP [none] jump 17 no jump 5 LOOPE/LOOPZ [none] jump 18 no jump 6 LOOPNE/LOOPNZ [none] jump 19 no jump 5 MOV [none] register, register 2 register, memory 8 + EA memory, register 9 + EA register, constant 4 memory, constant 10 + EA (AX/AL) <-> memory 10 {9} ____________________ 8 These last three are indirect jumps. The information about where to jump to is coming from a register (reg-ind) or from memory (mem-ind). 9 This is a special instruction which moves a directly addressed variable to or from AX (or AL for bytes). Pointers are not allowed, only the forms: mov ax, variable1 mov variable1, ax This takes 10 clocks instead of 14 or 15 for the other form. Whether this form gets used is up to the assembler, not you. Fortunately, MASM, TurboAssembler and A86 all use this form when appropriate. The PC Assembler Tutor xxxvi ______________________ segreg <-> register 2 {10} segreg, memory 8 + EA memory, segreg 9 + EA MOVS [none] 11 clocks MUL [OC,?SZAP?] byte register 70 - 77 word register 118 - 133 byte memory (76 - 83) + EA word memory (124 - 139) + EA NEG [OSZAPC] {11} register 3 memory 16 + EA NOP [none] 3 clocks NOT [none] register 3 memory 16 + EA OR [SZP,(OC=0),?A?] see ADD OUT [none] (AX/AL), port# 10 (AX/AL), dx 8 POP [none] register 8 segreg 8 memory 17 + EA POPF [ {12} ] 8 clocks PUSH [none] register 11 segreg 10 memory 16 + EA PUSHF [none] 10 clocks RCL [OC] register by 1 bit 2 memory by 1 bit 15 + EA register by # in CL 8 + (4 * #) {13} memory by # in CL 20