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Loadstar 17
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017.d81
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game design.2
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2022-08-26
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╟AME ─ESIGN FOR THE ├-64
┬Y ╥OBERT ┴LONSO
<CONTINUED FROM PREVIOUS ARTICLE>
├ONSECUTIVE ┬YTES
┼ACH CHARACTER DESIGN IS MADE OF
EIGHT CONSECUTIVE MEMORY LOCATIONS.
┼ACH OF THESE LOCATIONS IS CALLED A
BYTE. ┼ACH BYTE HAS EIGHT BITS THAT
CAN BE INDIVIDUALLY TURNED ON AND OFF
BY USING ╨╧╦┼S. ┘OU CAN FORM A LETTER
OR CHARACTER BY TURNING ON SOME BITS
AND TURNING SOME OTHERS OFF TO FORM
PATTERNS. ╘RY TO IMAGINE AN EIGHT-BY-
EIGHT SET OF LIGHT BULBS THAT YOU CAN
TURN ON OR OFF AT THE FLICK OF A
SWITCH. ╠ET'S SAY THAT YOU WANTED TO
MAKE A BOX OUT OF THE EIGHT-BY-EIGHT
GRID. ╘O DO THIS, YOU WOULD HAVE TO
TURN ON THE LIGHTS AROUND THE BORDER
OF THE GRID. ╒SING THE SAME
TECHNIQUE, YOU CAN DESIGN JUST ABOUT
ANYTHING YOU CAN IMAGINE.
╘HE FOLLOWING TABLE SHOWS YOU HOW
THE LETTER ┴ IS FORMED. ╘HE NUMBERS
THAT YOU SEE TO THE FAR RIGHT OF THE
DESIGN ARE DERIVED FROM THE DIAGRAM BY
ADDING TOGETHER THE VALUES THAT THE
BLOCKS REPRESENT. ┼ACH BLOCK
REPRESENTS A NUMBER, AS SHOWN, BASED
ON THE BINARY NUMBER SYSTEM.
├OMPUTERS USE THE BINARY SYSTEM
BECAUSE INTERNALLY THEY CAN UNDERSTAND
ONLY TWO STATES OF BEING--ON AND OFF.
┬ITS OF A ┬┘╘┼
128 64 32 16 8 4 2 1 ╓┴╠╒┼
...**... 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
REPRESENTS A BINARY ONE AND A PERIOD
REPRESENTS A BINARY ZERO. ┬Y ADDING
UP THE VALUES OF EACH PLACE THAT IS
"ON" STARTING FROM THE RIGHT AND GOING
TOWARDS THE LEFT, YOU CAN DERIVE THE
NUMBER YOU WILL NEED IN YOUR PROGRAM'S
DATA STATEMENTS.
╘O CREATE ANY NEW CHARACTER, YOU
WILL HAVE TO DESIGN AN EIGHT-BY-EIGHT
GRID AND FILL IN THE BLOCKS THAT YOU
WANT TURNED ON IN THE SCREEN IMAGE OF
THE CHARACTER.
╞OR THE COMPUTER TO KNOW WHAT YOU
WANT IT TO DESIGN ON THE SCREEN, YOU
WILL HAVE TO PUT THE NUMBERS THAT
CORRESPOND TO EACH ROW OF THE DESIGN
IN DATA STATEMENTS. ╠ET'S SAY YOU
WANTED TO DO THE BOX CHARACTER ╔
MENTIONED BEFORE. ╘HE FIRST ROW WOULD
HAVE TO HAVE A VALUE OF 255 BECAUSE
EVERY BIT IS LIT AND THE SECOND ONE
WOULD HAVE TO HAVE A VALUE OF 129
BECAUSE THE FIRST AND LAST BITS ARE
LIT. ╘HE VALUE IS 129 BECAUSE THE
RIGHT-MOST BIT HAS A VALUE OF ONE AND
THE LEFT-MOST BIT HAS A VALUE OF 128.
┴LL THE OTHER ROWS EXCEPT THE EIGHTH
ONE WILL ALSO HAVE A VALUE OF 129.
╘HE EIGHTH ROW WILL BE 255 BECAUSE ALL
THE BITS ARE ON. ╘HE BOX CHARACTER
WOULD THUS BE REPRESENTED WITH THE
FOLLOWING LINE:
90 ─┴╘┴ 255,129,129,129,129,129,
129,255
╘HE BOX CHARACTER'S GRID
REPRESENTATION WOULD LOOK LIKE THIS:
┬╔╬┴╥┘ ─┼├╔═┴╠
╓┴╠╒┼ ╓┴╠╒┼
******** 11111111 255
*......* 10000001 129
*......* 10000001 129
*......* 10000001 129
*......* 10000001 129
*......* 10000001 129
*......* 10000001 129
******** 11111111 255
┴FTER YOU HAVE CONVERTED ALL YOUR
CHARACTERS INTO DATA STATEMENTS, YOU
CAN THE ╨╧╦┼ THEM INTO MEMORY
LOCATIONS WHERE YOUR NEW CHARACTER SET
IS LOCATED. ┘OU'LL RECALL THAT THE
CHARACTER SET WE ARE WORKING ON IS
LOCATED BETWEEN 12288 AND 12799. ╔F
YOU WRITE A SHORT ╞╧╥-╬┼╪╘ LOOP THAT
╨╧╦┼S THE DATA INTO THE PROPER MEMORY
LOCATIONS, YOU WILL GET YOUR NEW
CHARACTER SET. ╘HE ╞╧╥-╬┼╪╘ LOOP THAT
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
╠╧┴─╙╘┴╥. ┴FTER RUNNING THE PROGRAM,
PLEASE TURN YOUR COMPUTER OFF BEFORE
RE-ENTERING ╠╧┴─╙╘┴╥.
--< CONTINUED IN THE NEXT ARTICLE >---