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Frequently Asked Questions (FAQS);faqs.221 Besides listing frequently-asked questions, this article also summarizes frequently-posted answers. Even if you know all the answers, it's worth skimming through this list once in a while, so that when you see one of its questions unwittingly posted, you won't have to waste time answering. This article is always being improved. Your input is welcomed. Send your comments to scs@adam.mit.edu, scs%adam.mit.edu@mit.edu, and/or mit-eddie!adam.mit.edu!scs; this article's From: line may be unusable. The questions answered here are divided into several categories: 1. Null Pointers 2. Arrays and Pointers 3. Memory Allocation 4. Expressions 5. ANSI C 6. C Preprocessor 7. Variable-Length Argument Lists 8. Boolean Expressions and Variables 9. Structs, Enums, and Unions 10. Declarations 11. Stdio 12. Library Subroutines 13. Lint 14. Style 15. Floating Point 16. System Dependencies 17. Miscellaneous (Fortran to C converters, YACC grammars, etc.) Herewith, some frequently-asked questions and their answers: Section 1. Null Pointers 1.1: What is this infamous null pointer, anyway? A: The language definition states that for each pointer type, there is a special value -- the "null pointer" -- which is distinguishable from all other pointer values and which is not the address of any object. That is, the address-of operator & will never yield a null pointer, nor will a successful call to malloc. (malloc returns a null pointer when it fails, and this is a typical use of null pointers: as a "special" pointer value with some other meaning, usually "not allocated" or "not pointing anywhere yet.") A null pointer is conceptually different from an uninitialized pointer. A null pointer is known not to point to any object; an uninitialized pointer might point anywhere. See also questions 3.1, 3.9, and 17.1. As mentioned in the definition above, there is a null pointer for each pointer type, and the internal values of null pointers for different types may be different. Although programmers need not know the internal values, the compiler must always be informed which type of null pointer is required, so it can make the distinction if necessary (see below). References: K&R I Sec. 5.4 pp. 97-8; K&R II Sec. 5.4 p. 102; H&S Sec. 5.3 p. 91; ANSI Sec. 3.2.2.3 p. 38. 1.2: How do I "get" a null pointer in my programs? A: According to the language definition, a constant 0 in a pointer context is converted into a null pointer at compile time. That is, in an initialization, assignment, or comparison when one side is a variable or expression of pointer type, the compiler can tell that a constant 0 on the other side requests a null pointer, and generate the correctly-typed null pointer value. Therefore, the following fragments are perfectly legal: char *p = 0; if(p != 0) However, an argument being passed to a function is not necessarily recognizable as a pointer context, and the compiler may not be able to tell that an unadorned 0 "means" a null pointer. For instance, the Unix system call "execl" takes a variable-length, null-pointer-terminated list of character pointer arguments. To generate a null pointer in a function call context, an explicit cast is typically required, to force the 0 to be in a pointer context: execl("/bin/sh", "sh", "-c", "ls", (char *)0); If the (char *) cast were omitted, the compiler would not know to pass a null pointer, and would pass an integer 0 instead. (Note that many Unix manuals get this example wrong.) When function prototypes are in scope, argument passing becomes an "assignment context," and most casts may safely be omitted, since the prototype tells the compiler that a pointer is required, and of which type, enabling it to correctly convert unadorned 0's. Function prototypes cannot provide the types for variable arguments in variable-length argument lists, however, so explicit casts are still required for those arguments. It is safest always to cast null pointer function arguments, to guard against varargs functions or those without prototypes, to allow interim use of non-ANSI compilers, and to demonstrate that you know what you are doing. (Incidentally, it's also a simpler rule to remember.) Summary: Unadorned 0 okay: Explicit cast required: initialization function call, no prototype in scope assignment variable argument in comparison varargs function call function call, prototype in scope, fixed argument References: K&R I Sec. A7.7 p. 190, Sec. A7.14 p. 192; K&R II Sec. A7.10 p. 207, Sec. A7.17 p. 209; H&S Sec. 4.6.3 p. 72; ANSI Sec. 3.2.2.3 . 1.3: What is NULL and how is it #defined? A: As a matter of style, many people prefer not to have unadorned 0's scattered throughout their programs. For this reason, the preprocessor macro NULL is #defined (by <stdio.h> or <stddef.h>), with value 0 (or (void *)0, about which more later). A programmer who wishes to make explicit the distinction between 0 the integer and 0 the null pointer can then use NULL whenever a null pointer is required. This is a stylistic convention only; the preprocessor turns NULL back to 0 which is then recognized by the compiler (in pointer contexts) as before. In particular, a cast may still be necessary before NULL (as before 0) in a function call argument. (The table under question 1.2 above applies for NULL as well as 0.) NULL should _only_ be used for pointers; see question 1.8. References: K&R I Sec. 5.4 pp. 97-8; K&R II Sec. 5.4 p. 102; H&S Sec. 13.1 p. 283; ANSI Sec. 4.1.5 p. 99, Sec. 3.2.2.3 p. 38, Rationale Sec. 4.1.5 p. 74. 1.4: How should NULL be #defined on a machine which uses a nonzero bit pattern as the internal representation of a null pointer? A: Programmers should never need to know the internal representation(s) of null pointers, because they are normally taken care of by the compiler. If a machine uses a nonzero bit pattern for null pointers, it is the compiler's responsibility to generate it when the programmer requests, by writing "0" or "NULL," a null pointer. Therefore, #defining NULL as 0 on a machine for which internal null pointers are nonzero is as valid as on any other, because the compiler must (and can) still generate the machine's correct null pointers in response to unadorned 0's seen in pointer contexts. 1.5: If NULL were defined as follows: #define NULL (char *)0 wouldn't that make function calls which pass an uncast NULL work? A: Not in general. The problem is that there are machines which use different internal representations for pointers to different types of data. The suggested #definition would make uncast NULL arguments to functions expecting pointers to characters to work correctly, but pointer arguments to other types would still be problematical, and legal constructions such as FILE *fp = NULL; could fail. Nevertheless, ANSI C allows the alternate #define NULL ((void *)0) definition for NULL. Besides helping incorrect programs to work (but only on machines with homogeneous pointers, thus questionably valid assistance) this definition may catch programs which use NULL incorrectly (e.g. when the ASCII NUL character was really intended; see question 1.8). 1.6: I use the preprocessor macro #define Nullptr(type) (type *)0 to help me build null pointers of the correct type. A: This trick, though popular in some circles, does not buy much. It is not needed in assignments and comparisons; see question 1.2. It does not even save keystrokes. Its use suggests to the reader that the author is shaky on the subject of null pointers, and requires the reader to check the #definition of the macro, its invocations, and _all_ other pointer usages much more carefully. See also question 8.1. 1.7: Is the abbreviated pointer comparison "if(p)" to test for non- null pointers valid? What if the internal representation for null pointers is nonzero? A: When C requires the boolean value of an expression (in the if, while, for, and do statements, and with the &&, ||, !, and ?: operators), a false value is produced when the expression compares equal to zero, and a true value otherwise. That is, whenever one writes if(expr) where "expr" is any expression at all, the compiler essentially acts as if it had been written as if(expr != 0) Substituting the trivial pointer expression "p" for "expr," we have if(p) is equivalent to if(p != 0) and this is a comparison context, so the compiler can tell that the (implicit) 0 is a null pointer, and use the correct value. There is no trickery involved here; compilers do work this way, and generate identical code for both statements. The internal representation of a pointer does _not_ matter. The boolean negation operator, !, can be described as follows: !expr is essentially equivalent to expr?0:1 It is left as an exercise for the reader to show that if(!p) is equivalent to if(p == 0) "Abbreviations" such as if(p), though perfectly legal, are considered by some to be bad style. See also question 8.2. References: K&R II Sec. A7.4.7 p. 204; H&S Sec. 5.3 p. 91; ANSI Secs. 3.3.3.3, 3.3.9, 3.3.13, 3.3.14, 3.3.15, 3.6.4.1, and 3.6.5 . 1.8: If "NULL" and "0" are equivalent, which should I use? A: Many programmers believe that "NULL" should be used in all pointer contexts, as a reminder that the value is to be thought of as a pointer. Others feel that the confusion surrounding "NULL" and "0" is only compounded by hiding "0" behind a #definition, and prefer to use unadorned "0" instead. There is no one right answer. C programmers must understand that "NULL" and "0" are interchangeable and that an uncast "0" is perfectly acceptable in initialization, assignment, and comparison contexts. Any usage of "NULL" (as opposed to "0") should be considered a gentle reminder that a pointer is involved; programmers should not depend on it (either for their own understanding or the compiler's) for distinguishing pointer 0's from integer 0's. NULL should _not_ be used when another kind of 0 is required, even though it might work, because doing so sends the wrong stylistic message. (ANSI allows the #definition of NULL to be (void *)0, which will not work in non-pointer contexts.) In particular, do not use NULL when the ASCII null character (NUL) is desired. Provide your own definition #define NUL '\0' if you must. Reference: K&R II Sec. 5.4 p. 102. 1.9: But wouldn't it be better to use NULL (rather than 0) in case the value of NULL changes, perhaps on a machine with nonzero null pointers? A: No. Although symbolic constants are often used in place of numbers because the numbers might change, this is _not_ the reason that NULL is used in place of 0. Once again, the language guarantees that source-code 0's (in pointer contexts) generate null pointers. NULL is used only as a stylistic convention. 1.10: I'm confused. NULL is guaranteed to be 0, but the null pointer is not? A: When the term "null" or "NULL" is casually used, one of several things may be meant: 1. The conceptual null pointer, the abstract language concept defined in question 1.1. It is implemented with... 2. The internal (or run-time) representation of a null pointer, which may or may not be all-bits-0 and which may be different for different pointer types. The actual values should be of concern only to compiler writers. Authors of C programs never see them, since they use... 3. The source code syntax for null pointers, which is the single character "0". It is often hidden behind... 4. The NULL macro, which is #defined to be "0" or "(void *)0". Finally, as red herrings, we have... 5. The ASCII null character (NUL), which does have all bits zero, but has no relation to the null pointer except in name; and... 6. The "null string," which is another name for an empty string (""). The term "null string" can be confusing in C (and should perhaps be avoided), because it involves a null ('\0') character, but not a null pointer, which brings us full circle... This article always uses the phrase "null pointer" (in lower case) for sense 1, the character "0" for sense 3, and the capitalized word "NULL" for sense 4. 1.11: Why is there so much confusion surrounding null pointers? Why do these questions come up so often? A: C programmers traditionally like to know more than they need to about the underlying machine implementation. The fact that null pointers are represented both in source code, and internally to most machines, as zero invites unwarranted assumptions. The use of a preprocessor macro (NULL) suggests that the value might change later, or on some weird machine. The construct "if(p == 0)" is easily misread as calling for conversion of p to an integral type, rather than 0 to a pointer type, before the comparison. Finally, the distinction between the several uses of the term "null" (listed above) is often overlooked. One good way to wade out of the confusion is to imagine that C had a keyword (perhaps "nil", like Pascal) with which null pointers were requested. The compiler could either turn "nil" into the correct type of null pointer, when it could determine the type from the source code, or complain when it could not. Now, in fact, in C the keyword for a null pointer is not "nil" but "0", which works almost as well, except that an uncast "0" in a non-pointer context generates an integer zero instead of an error message, and if that uncast 0 was supposed to be a null pointer, the code may not work. 1.12: I'm still confused. I just can't understand all this null pointer stuff. A: Follow these two simple rules: 1. When you want to refer to a null pointer in source code, use "0" or "NULL". 2. If the usage of "0" or "NULL" is an argument in a function call, cast it to the pointer type expected by the function being called. The rest of the discussion has to do with other people's misunderstandings, or with the internal representation of null pointers (which you shouldn't need to know), or with ANSI C refinements. Understand questions 1.1, 1.2, and 1.3, and consider 1.8 and 1.11, and you'll do fine. 1.13: Given all the confusion surrounding null pointers, wouldn't it be easier simply to require them to be represented internally by zeroes? A: If for no other reason, doing so would be ill-advised because it would unnecessarily constrain implementations which would otherwise naturally represent null pointers by special, nonzero bit patterns, particularly when those values would trigger automatic hardware traps for invalid accesses. Besides, what would this requirement really accomplish? Proper understanding of null pointers does not require knowledge of the internal representation, whether zero or nonzero. Assuming that null pointers are internally zero does not make any code easier to write (except for a certain ill-advised usage of calloc; see question 3.9). Known-zero internal pointers would not obviate casts in function calls, because the _size_ of the pointer might still be different from that of an int. (If "nil" were used to request null pointers rather than "0," as mentioned in question 1.11, the urge to assume an internal zero representation would not even arise.) 1.14: Seriously, have any actual machines really used nonzero null pointers, or different representations for pointers to different types? A: The Prime 50 series used segment 07777, offset 0 for the null pointer, at least for PL/I. Later models used segment 0, offset 0 for null pointers in C, necessitating new instructions such as TCNP (Test C Null Pointer), evidently as a sop to all the extant poorly-written C code which made incorrect assumptions. Older, word-addressed Prime machines were also notorious for requiring larger byte pointers (char *'s) than word pointers (int *'s). Some Honeywell-Bull mainframes use the bit pattern 06000 for (internal) null pointers. The Symbolics Lisp Machine, a tagged architecture, does not even have conventional numeric pointers; it uses the pair <NIL, 0> (basically a nonexistent <object, offset> handle) as a C null pointer. Depending on the "memory model" in use, 80*86 processors (PC's) may use 16 bit data pointers and 32 bit function pointers, or vice versa. Section 2. Arrays and Pointers 2.1: I had the definition char a[6] in one source file, and in another I declared extern char *a. Why didn't it work? A: The declaration extern char *a simply does not match the actual definition. The type "pointer-to-type-T" is not the same as "array-of-type-T." Use extern char a[]. References: CT&P Sec. 3.3 pp. 33-4, Sec. 4.5 pp. 64-5. 2.2: But I heard that char a[] was identical to char *a. A: Not at all. (What you heard has to do with formal parameters to functions; see question 2.4.) Arrays are not pointers. The array declaration "char a[6];" requests that space for six characters be set aside, to be known by the name "a." That is, there is a location named "a" at which six characters can sit. The pointer declaration "char *p;" on the other hand, requests a place which holds a pointer. The pointer is to be known by the name "p," and can point to any char (or contiguous array of chars) anywhere. As usual, a picture is worth a thousand words. The statements char a[] = "hello"; char *p = "world"; would result in data structures which could be represented like this: +---+---+---+---+---+---+ a: | h | e | l | l | o |\0 | +---+---+---+---+---+---+ +-----+ +---+---+---+---+---+---+ p: | *======> | w | o | r | l | d |\0 | +-----+ +---+---+---+---+---+---+ It is important to remember that a reference like x[3] generates different code depending on whether x is an array or a pointer. Given the declarations above, when the compiler sees the expression a[3], it emits code to start at the location "a," move three past it, and fetch the character there. When it sees the expression p[3], it emits code to start at the location "p," fetch the pointer value there, add three to the pointer, and finally fetch the character pointed to. In the example above, both a[3] and p[3] happen to be the character 'l', but the compiler gets there differently. (See also question 17.14.) 2.3: So what is meant by the "equivalence of pointers and arrays" in C? A: Much of the confusion surrounding pointers in C can be traced to a misunderstanding of this statement. Saying that arrays and pointers are "equivalent" does not by any means imply that they are interchangeable. "Equivalence" refers to the following key definition: An lvalue of type array-of-T which appears in an expression decays (with three exceptions) into a pointer to its first element; the type of the resultant pointer is pointer-to-T. (The exceptions are when the array is the operand of a sizeof or & operator, or is a literal string initializer for a character array.) As a consequence of this definition, there is not really any difference in the behavior of the "array subscripting" operator [] as it applies to arrays and pointers. In an expression of the form a[i], the array reference "a" decays into a pointer, following the rule above, and is then subscripted exactly as would be a pointer variable in the expression p[i]. In either case, the expression x[i] (where x is an array or a pointer) is, by definition, exactly equivalent to *((x)+(i)). References: K&R I Sec. 5.3 pp. 93-6; K&R II Sec. 5.3 p. 99; H&S Sec. 5.4.1 p. 93; ANSI Sec. 3.2.2.1, Sec. 3.3.2.1, Sec. 3.3.6 . 2.4: Then why are array and pointer declarations interchangeable as function formal parameters? A: Since arrays decay immediately into pointers, an array is never actually passed to a function. As a convenience, any parameter declarations which "look like" arrays, e.g. f(a) char a[]; are treated by the compiler as if they were pointers, since that is what the function will receive if an array is passed: f(a) char *a; This conversion holds only within function formal parameter declarations, nowhere else. If this conversion bothers you, avoid it; many people have concluded that the confusion it causes outweighs the small advantage of having the declaration "look like" the call and/or the uses within the function. References: K&R I Sec. 5.3 p. 95, Sec. A10.1 p. 205; K&R II Sec. 5.3 p. 100, Sec. A8.6.3 p. 218, Sec. A10.1 p. 226; H&S Sec. 5.4.3 p. 96; ANSI Sec. 3.5.4.3, Sec. 3.7.1, CT&P Sec. 3.3 pp. 33-4. 2.5: Why doesn't sizeof properly report the size of an array which is a parameter to a function? A: The sizeof operator reports the size of the pointer parameter which the function actually receives (see question 2.4). 2.6: Someone explained to me that arrays were really just constant pointers. A: This is a bit of an oversimplification. An array name is "constant" in that it cannot be assigned to, but an array is _not_ a pointer, as the discussion and pictures in question 2.2 should make clear. 2.7: I came across some "joke" code containing the "expression" 5["abcdef"] . How can this be legal C? A: Yes, Virginia, array subscripting is commutative in C. This curious fact follows from the pointer definition of array subscripting, namely that a[e] is exactly equivalent to *((a)+(e)), for _any_ expression e and primary expression a, as long as one of them is a pointer expression and one is integral. This unsuspected commutativity is often mentioned in C texts as if it were something to be proud of, but it finds no useful application outside of the Obfuscated C Contest (see question 17.9). References: ANSI Rationale Sec. 3.3.2.1 p. 41. 2.8: My compiler complained when I passed a two-dimensional array to a routine expecting a pointer to a pointer. A: The rule by which arrays decay into pointers is not applied recursively. An array of arrays (i.e. a two-dimensional array in C) decays into a pointer to an array, not a pointer to a pointer. Pointers to arrays can be confusing, and must be treated carefully. (The confusion is heightened by the existence of incorrect compilers, including some versions of pcc and pcc-derived lint's, which improperly accept assignments of multi-dimensional arrays to multi-level pointers.) If you are passing a two-dimensional array to a function: int array[YSIZE][XSIZE]; f(array); the function's declaration should match: f(int a[][XSIZE]) {...} or f(int (*ap)[XSIZE]) {...} /* ap is a pointer to an array */ In the first declaration, the compiler performs the usual implicit parameter rewriting of "array of array" to "pointer to array;" in the second form the pointer declaration is explicit. Since the called function does not allocate space for the array, it does not need to know the overall size, so the number of "rows," YSIZE, can be omitted. The "shape" of the array is still important, so the "column" dimension XSIZE (and, for 3- or more dimensional arrays, the intervening ones) must be included. If a function is already declared as accepting a pointer to a pointer, it is probably incorrect to pass a two-dimensional array directly to it. References: K&R I Sec. 5.10 p. 110; K&R II Sec. 5.9 p. 113. 2.9: How do I declare a pointer to an array? A: Usually, you don't want to. Consider using a pointer to one of the array's elements instead. Arrays of type T decay into pointers to type T (see question 2.3), which is convenient; subscripting or incrementing the resultant pointer accesses the individual members of the array. True pointers to arrays, when subscripted or incremented, step over entire arrays, and are generally only useful when operating on arrays of arrays, if at all. (See question 2.8 above.) When people speak casually of a pointer to an array, they usually mean a pointer to its first element. If you really need to declare a pointer to an entire array, use something like "int (*ap)[N];" where N is the size of the array. (See also question 10.3.) If the size of the array is unknown, N can be omitted, but the resulting type, "pointer to array of unknown size," is useless. 2.10: How can I dynamically allocate a multidimensional array? A: It is usually best to allocate an array of pointers, and then initialize each pointer to a dynamically-allocated "row." The resulting "ragged" array can save space, although it is not necessarily contiguous in memory as a real array would be. Here is a two-dimensional example: int **array = (int **)malloc(nrows * sizeof(int *)); for(i = 0; i < nrows; i++) array[i] = (int *)malloc(ncolumns * sizeof(int)); (In "real" code, of course, malloc would be declared correctly, and each return value checked.) You can keep the array's contents contiguous, while making later reallocation of individual rows difficult, with a bit of explicit pointer arithmetic: int **array = (int **)malloc(nrows * sizeof(int *)); array[0] = (int *)malloc(nrows * ncolumns * sizeof(int)); for(i = 1; i < nrows; i++) array[i] = array[0] + i * ncolumns; In either case, the elements of the dynamic array can be accessed with normal-looking array subscripts: array[i][j]. If the double indirection implied by the above schemes is for some reason unacceptable, you can simulate a two-dimensional array with a single, dynamically-allocated one-dimensional array: int *array = (int *)malloc(nrows * ncolumns * sizeof(int)); However, you must now perform subscript calculations manually, accessing the i,jth element with array[i * ncolumns + j]. (A macro can hide the explicit calculation, but invoking it then requires parentheses and commas which don't look exactly like multidimensional array subscripts.) Finally, you can use pointers-to-arrays: int (*array)[NCOLUMNS] = (int (*)[NCOLUMNS])malloc(nrows * sizeof(*array)); , but the syntax gets horrific and all but one dimension must be known at compile time. With all of these techniques, you may of course need to remember to free the arrays (which may take several steps) when they are no longer needed. You must also be extremely cautious when passing dynamically-allocated arrays down to other functions, if those functions are also to accept conventional, statically- allocated arrays (see question 2.8). 2.11: Here's a neat trick: if I write int realarray[10]; int *array = &realarray[-1]; I can treat "array" as if it were a 1-based array. A: This technique, though attractive, is not strictly conforming to the C standards (in spite of its appearance in Numerical Recipes in C). Pointer arithmetic is defined only as long as the pointer points within the same allocated block of memory, or to the imaginary "terminating" element one past it; otherwise, the behavior is undefined, _even if the pointer is not dereferenced_. The code above could fail if, while subtracting the offset, an illegal address were generated (perhaps because the address tried to "wrap around" past the beginning of some memory segment). References: ANSI Sec. 3.3.6 p. 48, Rationale Sec. 3.2.2.3 p. 38; K&R II Sec. 5.3 p. 100, Sec. 5.4 pp. 102-3, Sec. A7.7 pp. 205-6. 2.12: I passed a pointer to a function which initialized it: main() { int *ip; f(ip); return 0; } void f(ip) int *ip; { static int dummy; ip = &dummy; *ip = 5; } , but the pointer in the caller was unchanged. A: Did the function try to initialize the pointer itself, or just what it pointed to? Remember that arguments in C are passed by value. The called function altered only the passed copy of the pointer. You'll want to pass the address of the pointer (the function will end up accepting a pointer-to-a-pointer). 2.13: I have a char * pointer that happens to point to some ints, and I want to step it over them. Why doesn't ((int *)p)++; work? A: In C, a cast operator does not mean "pretend these bits have a different type, and treat them accordingly;" it is a conversion operator, and by definition it yields an rvalue, which cannot be assigned to, or incremented with ++. (It is an anomaly in pcc- derived compilers, and an extension in gcc, that expressions such as the above are ever accepted.) Say what you mean: use