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-
- ╟AME ─ESIGN FOR THE ├-64
- ┬Y ╥OBERT ┴LONSO
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- <CONTINUED FROM PREVIOUS ARTICLE>
-
-
- ├ONSECUTIVE ┬YTES
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- ┼ACH CHARACTER DESIGN IS MADE OF
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- EIGHT CONSECUTIVE MEMORY LOCATIONS.
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- ┼ACH OF THESE LOCATIONS IS CALLED A
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- BYTE. ┼ACH BYTE HAS EIGHT BITS THAT
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- CAN BE INDIVIDUALLY TURNED ON AND OFF
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- BY USING ╨╧╦┼S. ┘OU CAN FORM A LETTER
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- OR CHARACTER BY TURNING ON SOME BITS
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- AND TURNING SOME OTHERS OFF TO FORM
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- PATTERNS. ╘RY TO IMAGINE AN EIGHT-BY-
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- EIGHT SET OF LIGHT BULBS THAT YOU CAN
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- TURN ON OR OFF AT THE FLICK OF A
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- SWITCH. ╠ET'S SAY THAT YOU WANTED TO
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- MAKE A BOX OUT OF THE EIGHT-BY-EIGHT
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- GRID. ╘O DO THIS, YOU WOULD HAVE TO
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- TURN ON THE LIGHTS AROUND THE BORDER
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- OF THE GRID. ╒SING THE SAME
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- TECHNIQUE, YOU CAN DESIGN JUST ABOUT
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- ANYTHING YOU CAN IMAGINE.
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- ╘HE FOLLOWING TABLE SHOWS YOU HOW
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- THE LETTER ┴ IS FORMED. ╘HE NUMBERS
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- THAT YOU SEE TO THE FAR RIGHT OF THE
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- DESIGN ARE DERIVED FROM THE DIAGRAM BY
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- ADDING TOGETHER THE VALUES THAT THE
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- BLOCKS REPRESENT. ┼ACH BLOCK
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- REPRESENTS A NUMBER, AS SHOWN, BASED
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- ON THE BINARY NUMBER SYSTEM.
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- ├OMPUTERS USE THE BINARY SYSTEM
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- BECAUSE INTERNALLY THEY CAN UNDERSTAND
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- ONLY TWO STATES OF BEING--ON AND OFF.
-
-
-
- ┬ITS OF A ┬┘╘┼
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- 128 64 32 16 8 4 2 1 ╓┴╠╒┼
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- ...**... 0 0 0 1 1 0 0 0 24
- ..****.. 0 0 1 1 1 1 0 0 60
- .** **. 0 1 1 0 0 1 1 0 102
- .******. 0 1 1 1 1 1 1 0 126
- .** **. 0 1 1 0 0 1 1 0 102
- .** **. 0 1 1 0 0 1 1 0 102
- .** **. 0 1 1 0 0 1 1 0 102
- ........ 0 0 0 0 0 0 0 0 0
-
-
- ╔N THE ABOVE DIAGRAM, AN ASTERISK
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- REPRESENTS A BINARY ONE AND A PERIOD
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- REPRESENTS A BINARY ZERO. ┬Y ADDING
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- UP THE VALUES OF EACH PLACE THAT IS
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- "ON" STARTING FROM THE RIGHT AND GOING
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- TOWARDS THE LEFT, YOU CAN DERIVE THE
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- NUMBER YOU WILL NEED IN YOUR PROGRAM'S
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- DATA STATEMENTS.
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- ╘O CREATE ANY NEW CHARACTER, YOU
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- WILL HAVE TO DESIGN AN EIGHT-BY-EIGHT
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- GRID AND FILL IN THE BLOCKS THAT YOU
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- WANT TURNED ON IN THE SCREEN IMAGE OF
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- THE CHARACTER.
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- ╞OR THE COMPUTER TO KNOW WHAT YOU
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- WANT IT TO DESIGN ON THE SCREEN, YOU
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- WILL HAVE TO PUT THE NUMBERS THAT
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- CORRESPOND TO EACH ROW OF THE DESIGN
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- IN DATA STATEMENTS. ╠ET'S SAY YOU
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- WANTED TO DO THE BOX CHARACTER ╔
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- MENTIONED BEFORE. ╘HE FIRST ROW WOULD
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- HAVE TO HAVE A VALUE OF 255 BECAUSE
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- EVERY BIT IS LIT AND THE SECOND ONE
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- WOULD HAVE TO HAVE A VALUE OF 129
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- BECAUSE THE FIRST AND LAST BITS ARE
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- LIT. ╘HE VALUE IS 129 BECAUSE THE
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- RIGHT-MOST BIT HAS A VALUE OF ONE AND
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- THE LEFT-MOST BIT HAS A VALUE OF 128.
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- ┴LL THE OTHER ROWS EXCEPT THE EIGHTH
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- ONE WILL ALSO HAVE A VALUE OF 129.
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- ╘HE EIGHTH ROW WILL BE 255 BECAUSE ALL
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- THE BITS ARE ON. ╘HE BOX CHARACTER
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- WOULD THUS BE REPRESENTED WITH THE
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- FOLLOWING LINE:
-
-
- 90 ─┴╘┴ 255,129,129,129,129,129,
- 129,255
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- ╘HE BOX CHARACTER'S GRID
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- REPRESENTATION WOULD LOOK LIKE THIS:
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-
- ┬╔╬┴╥┘ ─┼├╔═┴╠
- ╓┴╠╒┼ ╓┴╠╒┼
-
- ******** 11111111 255
- *......* 10000001 129
- *......* 10000001 129
- *......* 10000001 129
- *......* 10000001 129
- *......* 10000001 129
- *......* 10000001 129
- ******** 11111111 255
-
-
-
- ┴FTER YOU HAVE CONVERTED ALL YOUR
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- CHARACTERS INTO DATA STATEMENTS, YOU
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- CAN THE ╨╧╦┼ THEM INTO MEMORY
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- LOCATIONS WHERE YOUR NEW CHARACTER SET
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- IS LOCATED. ┘OU'LL RECALL THAT THE
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- CHARACTER SET WE ARE WORKING ON IS
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- LOCATED BETWEEN 12288 AND 12799. ╔F
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- YOU WRITE A SHORT ╞╧╥-╬┼╪╘ LOOP THAT
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- ╨╧╦┼S THE DATA INTO THE PROPER MEMORY
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- LOCATIONS, YOU WILL GET YOUR NEW
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- CHARACTER SET. ╘HE ╞╧╥-╬┼╪╘ LOOP THAT
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- WILL ╨╧╦┼ OUR BLOCK CHARACTER INTO THE
-
- '@' SYMBOL'S LOCATION IS:
-
-
- 70 ╞╧╥ ╪=0 ╘╧ 7: ╥┼┴─ ┴:
- ╨╧╦┼ 12288+╪,┴ : ╬┼╪╘ ╪
-
-
- ╔F YOU WOULD LIKE TO EXPERIMENT WITH
-
- THE WHOLE PROGRAM UP TO NOW, PRESS THE
- \OAD"GAME PROG-2",8
- '\'. ┬ECAUSE OF THE NATURE OF THIS
-
- PROGRAM, IT DOES ╬╧╘ RETURN TO
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- ╠╧┴─╙╘┴╥. ┴FTER RUNNING THE PROGRAM,
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- PLEASE TURN YOUR COMPUTER OFF BEFORE
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- RE-ENTERING ╠╧┴─╙╘┴╥.
-
- --< CONTINUED IN THE NEXT ARTICLE >---
-