Category Archives: Color BASIC

Pop-Up windows in Color BASIC?

In 1987, I wrote a simple routine to display a pop-up window on the 32-column text screen of the CoCo. I am not sure what I wrote this for, but it appears I did use this routine (or a version of it) in my TIDY DISK program a year later:

Tidy Disk – help screen 1

It is sometimes interesting to look back on something I did 35 years ago and see how (or if) I would do it differently today. So let’s do that.

On the Color Computer Archive site is a disk image containing two versions of this routine. The first was done on 8/8/1987, and it looks like a 1.1 was created the next day. I am pleased to see I was retaining versions back then.

Using the wonderful toolshed utility decb, I was able to get a text version of the tokenized BASic program easily:

$decb dir WNDO.DSK
Directory of: WNDO.DSK,

WNDO     BAS  0  B  1
WNDO11   BAS  0  B  1

$decb list -t WNDO.DSK,WNDO.BAS
999 END
1000 REM WINDOW SUBROUTINE
1001 REM S-START W-IDTH
1002 REM A$-TEXT L-ENGTH
1005 PRINT@S,STRING$(W+1,140)CHR$(141);:FORA=1TOL:PRINT@A*32+S,CHR$(133)STRING$(W,32)CHR$(128);:NEXTA:PRINT@32*L+S,CHR$(131)STRING$(W+1,128);:B=1
1010 FORA=W TOW-8STEP-1:IFMID$(A$,A,1)=" "THENPRINT@S+1+32*B,LEFT$(A$,A);:A$=RIGHT$(A$,LEN(A$)-A)ELSENEXTA:PRINT@S+1+32*B,A$;:GOTO1020
1015 B=B+1:GOTO1010
1020 PRINT@S+32*(L-1)+(W/2-5),"press any key";
1025 IFINKEY$=""THEN1025ELSERETURN

$ decb list -t WNDO.DSK,WNDO11.BAS
999 END
1000 REM WINDOW SUBROUTINE 1.1
1001 REM  BY ALLEN C. HUFFMAN
1002 REM      8/8, 8/9/87
1003 C=4
1005 PRINT@S,CHR$(140)STRING$(W,140)CHR$(141);:FORA=1TOL:PRINT@A*32+S,CHR$(133)STRING$(W,32)CHR$(133+16*C);:NEXTA:PRINT@32*L+S,CHR$(131)STRING$(W,131+16*C)CHR$(135+16*C);:B=1
1010 FORA=W TOW-8STEP-1:IFMID$(A$,A,1)=" "THENPRINT@S+1+32*B,LEFT$(A$,A);:A$=RIGHT$(A$,LEN(A$)-A)ELSENEXTA:PRINT@S+1+32*B,A$;:GOTO1020
1015 B=B+1:GOTO1010
1020 PRINT@S+32*(L-1)+W/2-6,"press"CHR$(128)"any"CHR$(128)"key";
1025 IFINKEY$=""THEN1025ELSERETURN

It appears the first version had help in the comments, and the second replaced that with my name and version dates. Looking at a WinMerge diff of the two versions, I see this:

It looks like I basically added color support rather than just a black drop shadow, and must have fixed a bug where width needed 1 added to it. The only other change was printing “PRESS ANY KEY” with black blocks between each word in stead of spaces. I can therefore tell I used at least this 1.1 version in my TIDY DISK program.

The subroutine requires the user to set a few variables before calling it:

  • S – PRINT @ position for the top left of the pop-up.
  • W – width of the pop-up (in characters).
  • L – length (height) of the pop-up (in lines).
  • A$ – message to display inside the pop-up. The message will be word-wrapped by the routine.

It appears I also allowed the user to set the color. In line 1003, the routine sets it to 3 (white), but the user could also set C ahead of time and then GOSUB to 1005 to bypass that line and use their own color. So I’ll add:

  • C – color of the drop shadow (use GOSUB 1005) if setting C.

GOSUB 1000 uses a default color, or set it yourself and GOSUB 1005. Nice.

10 C=6:S=39:W=16:L=10
20 A$="HELLO, 1987. THIS IS A COCO WINDOW POP-UP."
30 GOSUB 1005

The routine itself is not very smart. It doesn’t calculate anything. The user has to specify top left, width and height and do all that work. If you want a nicely centered box, you have to figure that out. But, if you wanted to make windows at different positions, I guess it allowed that flexibility. (I expect that’s why I wrote it this way.)

We can do better.

How many lines is it, anyway?

For most uses, a centered pop-up seems good enough. The issue is knowing how many lines the message is going to take to display after being word-wrapped. The current routine scans the string starting at “width” and moving backwards until it finds a space character. It then uses LEFT$ to print just that portion, and then trims that part off using RIGHT$.

To know how many lines it will take, we have to scan the string first to count them. Then we do it again to print them. This will slow things down, but make things easier on the user.

Anatomy of a Word Wrap

There are some good old articles here about word wrap routines, with some very fast and clever ones contributed by others (way better than mine). Check out those articles for better methods that what I am about to present.

Because I need to scan the line twice — one time to count the lines, and the second for the actual word wrap — I cannot mess with the user’s string. Instead, I’ll just use variables that sell me the Start Position (SP) within the string and End Position (EP) of the width-sized chunk I am going to scan. I’ll use a Position (P) variable and start at the end and scan backwards looking for a space. If I find one, I can count that as a line (increase the line count) and then move to the next character (to skip the space) and reset the Start Position to that. The process will repeat.

Since I do not want to duplicate code, I’ll embed the “print this section of the string” code in this routine, but use another variable to determine if I actually print it or now. I’m calling it Show (SH), and if SH=0 then it just counts, otherwise it prints as it counts each line.

It’s messy, but it works.

Here is a diagram showing the process:

Anatomy of a Word Wrap routine…

When the space is located, it knows that from SP to that space position is something to print (a line).

Easy peasy.

Except it took me way too much brane power to figure this out, and I am sure it could be massively improved.

The routine will use the user’s width (W) variable to know how wide to make the box, and adjust the text space inside accordingly (W-2, leaving room for the borders).

The user can specify C for color, or if it’s not set (0) it will be assigned a color.

The starting location (S) must also be set. This is the top left of the pop-up box, in PRINT@ format (i.e. 0 is the top left of the screen, 32 is the start of the second line, etc.)

Here is what I came up with, along with a few lines at the start to test it out. It will keep displaying random sized pop-ups with the same text over and over and over. I used this to make sure it would work with the various size and color valuies.

NOTE: If you send in a width that is too narrow, it won’t work properly. Ideally I should check for that and have a minimum size. (I’ll try to add that in the final version, since I know it needs to be at least wide enough to fit “PRESS ANY KEY”.)

10 A$="THIS IS A LONG STRING THAT I WANT TO USE TO TEST WORD WRAPPING IN MY WINDOW ROUTINE."
20 W=RND(10)+20:S=48-W/2:C=RND(8)-1
30 CLS:GOSUB 1000
40 GOTO 20

999 END
1000 REM WINDOW SUBROUTINE 2.0
1001 REM  BY ALLEN C. HUFFMAN
1002 REM 8/8, 8/9/87 & 7/27/22
1993 REM    S-START W-IDTH
1004 REM    A$-TEXT C-COLOR
1005 REM
1010 IF C=0 THEN C=4
1020 SH=0:GOSUB 1120
1030 PRINT@S,CHR$(140)STRING$(W-2,140)CHR$(141);
1040 FOR A=1 TO LC+1
1050 PRINT@S+32*A,CHR$(133)STRING$(W-2,32)CHR$(133+16*C);
1060 NEXTA
1070 PRINT@S+32*(LC+1),CHR$(131)STRING$(W-2,131+16*C)CHR$(135+16*C);
1080 SH=1:GOSUB 1120
1090 PRINT@S+32*LC+W/2-7,"press"CHR$(128)"any"CHR$(128)"key";
1100 IF INKEY$="" THEN 1100 ELSE RETURN
1110 ' CALC LINE COUNT/PRINT
1120 LC=1:SP=1:LN=LEN(A$)
1130 EP=SP+W-2:P=EP
1140 IF EP>LN THEN 1190
1150 IF MID$(A$,P,1)<>" " THEN P=P-1 ELSE 1170
1160 IF P>SP THEN 1150 ELSE 1190
1170 IF SH<>0 THEN PRINT@S+32*LC+1,MID$(A$,SP,P-SP);
1180 LC=LC+1:SP=P+1:GOTO 1130
1190 IF SH<>0 THEN PRINT@S+32*LC+1,MID$(A$,SP);
1200 LC=LC+1:RETURN
1210 RETURN

One bad thing about this is all the variables I am using. And it could also be made faster by combining lines and such.

Here are some things I would like to do next:

  • Check for a minimum width (if W<x then W=x).
  • Combine lines to make it smaller and faster.
  • Rename variables to all start with Z.

Z? Since my subroutine uses many variables, the calling program would have to make sure to not use any of them since the subroutine would change them. Today I like the approach of starting all my subroutine variables with Z, and just noting that “Zx variables are reserved for subroutines.” That way I can use any variables I want in my main program — other than the ones that are specifically used to pass values in to the subroutine — without worrying about a conflict.

I guess that means there will be a part 2. Until then, please comment with any ideas you have on improving this.

Until next time…

Pac-Man maze in 64x32 semigraphics.

Drawing a maze level on the CoCo – part 5

See Also: part 1, part 2, part 3, part 4 and part 5.

ASCII HEX data to maze graphics

As our spiraling trip down the rabbit hole continues, we finally return to drawing stuff instead of converting stuff.

We now have DATA statements that contain 2-byte hex numbers. Each hex number represents a byte (0-255) and each of those bytes represents 8 maze blocks on the screen.

For use in a Color BASIC program, we might want to take this HEX data and convert it in to string data (containing the final CHR$ values) for fast printing on the screen. If we were doing this in 6809 assembly, we might have the binary data contained in the executable, and convert it to a buffer that can then be copied to the screen as needed.

We’ll start with a simple (verbose) Color BASIC version. Since I am using the XRoar emulator to CLOAD this text file, I had to shorten the DATA lines a bit since they are now on higher lines, thus instead of loading “1DATA” it is loading “360DATA” taking up more bytes. I just moved some hex values to the next line.

I also adjusted my parser a bit. In my binary example, I wanted to handle binary numbers of any length. i.e, “111” would be 7, and “11111111” would be 255. To do this, I started with 0, and each time I saw a 1 I added one. For each new digit, I shifted the bits left by multiplying by 2. That way if it parsed “111” it would be “1 *2 +1 *2 +1” and so on for longer values.

For the maze, everything has to be a byte, so I took the opposite approach by starting with a bit value of 128 (representing the high bit in an 8-bit byte). Each time I found a new digit, I divided the bit value by 2 and either added it (if the new digit was a “1”, or “X” in the case of the maze) or not. That would make a maze of “XXXXXXXX” be “128 + 16 + 32 + 16 + 8 + 4 + 1” giving 255. If the maze data ran short, so the last few bytes were just “XXXX” it would be “128 + 64 + 32 + 16” which is &HF0 in hex (11110000), exactly what we want. If I had done it the other way, it would end up with 1111 (&H0F) and maze data would look like this:

GOOD:
-------- -------- ----
XXXXXXXX XXXXXXXX XXXX &HFF &HFF &HF0 -> 11111111 11111111 11110000

BAD:
-------- -------- ----
XXXXXXXX XXXXXXXX XXXX &HFF &HFF &H0F -> 11111111 11111111 00001111

I hope that makes sense. Anyway, here is the program I came up with… It will scan the hex values and build an array of strings (LN$) for each line. That way, the converted maze can be quickly printed after it has been generated once.

0 REM HEXMAZE.BAS
10 ' STRING SPACE IS MW*(MH+1)
20 CLEAR 28*32
30 ' SIZE OF MAZE IN DATA
40 MW=28:MH=31
50 ' ROOM FOR CONVERTED MAZE
60 DIM MZ$(MH):LN=0
70 CLS
80 ' READ A LINE OF HEX DATA
90 READ A$:IF A$="" THEN 340
100 ' EVERY 2 CHARS IS A HEX
110 FOR A=1 TO LEN(A$) STEP 2
120 ' COVERT HEX TO A VALUE
130 V=VAL("&H"+MID$(A$,A,2))
140 ' CONVERT BITS TO BLOCKS
150 BT=128
160 IF (V AND BT)=0 THEN B$=CHR$(128) ELSE B$=CHR$(175)
170 ' ADD BLOCK/SPACE TO STRING
180 MZ$(LN)=MZ$(LN)+B$
190 ' IF LEN >= MAZE W, NEXT
200 IF LEN(MZ$(LN))>=MW THEN 280
210 ' SHIFT BIT RIGHT
220 BT=INT(BT/2)
230 ' IF MORE BITS, REPEAT
240 IF BT>0 THEN 160
250 ' GET NEXT CHARACTER
260 GOTO 310
270 ' ELSE NEW MAZE LINE
280 PRINT MZ$(LN)
290 LN=LN+1
300 ' NEXT CHARACTER
310 NEXT
320 ' MORE LINES TO DO?
330 IF LN<MH THEN 90
340 END
350 REM MAZE AS HEX DATA
360 DATAFFFFFFF080060010BDF6FBD0BDF6FBD0BDF6FBD080000010BDBFDBD0BDBFDBD081861810FDF6FBF005F6FA0005801A0005BFDA00FDA05BF000204000FDA05BF005BFDA0005801A0005BFDA00FDBFDBF080060010BDF6FBD0BDF6FBD08C000310EDBFDB70EDBFDB7081861810BFF6FFD0BFF6FFD080000010
370 DATAFFFFFFF0,""

This program works, but it’s very slow, taking about 45 seconds to run. But, it’s verbose enough to (hopefully) explain what is going on.

By removing spaces, swapping out a integer values for faster HEX values, changing a CHR$() to a variable and adding a bit more string space, removing INT (and making one adjustment needed due to that), and combining lines, it can be sped up to just over 32 seconds.

0 REM HEXMAZE2.BAS
10 ' STRING SPACE IS MW*(MH+1)
20 CLEAR 28*32+200
30 ' SIZE OF MAZE IN DATA
40 MW=28:MH=31
50 ' ROOM FOR CONVERTED MAZE
60 DIM MZ$(MH):LN=0:BL$=CHR$(175)
70 CLS
90 READA$:IFA$=""THEN340
110 FORA=1TOLEN(A$)STEP2:V=VAL("&H"+MID$(A$,A,&H2)):BT=&H80
160 IF(V ANDBT)=0THENB$=BK$ELSEB$=BL$
180 MZ$(LN)=MZ$(LN)+B$:IFLEN(MZ$(LN))>=MW THEN280
220 BT=BT/2:IFBT>=1THEN160
260 GOTO310
280 PRINTMZ$(LN):LN=LN+1
310 NEXT:IFLN<MH THEN90
340 END
350 REM MAZE AS HEX DATA
360 DATAFFFFFFF080060010BDF6FBD0BDF6FBD0BDF6FBD080000010BDBFDBD0BDBFDBD081861810FDF6FBF005F6FA0005801A0005BFDA00FDA05BF000204000FDA05BF005BFDA0005801A0005BFDA00FDBFDBF080060010BDF6FBD0BDF6FBD08C000310EDBFDB70EDBFDB7081861810BFF6FFD0BFF6FFD080000010
370 DATAFFFFFFF0,""

And this is before translating that blocky maze into the “rounded corners” and such we want. This may be far too slow to do in BASIC since most folks don’t want to “Please wait 3.5 minutes while loading…” like a program my Commodore 64 friend once showed me. Assembly language helper routines might be needed for the impatient players out there…

Blocky to less blocky

After this code converts HEX values to maze data in a string array, we can process those strings and change the characters to have the “rounded” corners and such. Here is what I came up with, based on the original version that PEEKed and POKEed displayed bytes on the screen. This time, it’s scanning strings using MID$() and altering CHR$() values.

Side Note: Does everyone here know that, while you can’t use LEFT$ or RIGHT$ to change the left or right portion of a string, you can use MID$ to change characters? If you have A$=”1234567890″ and you want to change two characters starting at the third one, you can do MID$(A$,3,2)=”XX” and you get “12XX567890”. I’ll be using that…

Here is what I came up with… Instead of PEEKing memory, I used MID$(). It’s quite the mess, and very slow since I went back to the first version of the hex-to-maze print routine, just for clarity.

0 REM HEXMAZE3.BAS
10 ' STRING SPACE IS MW*(MH+1)
20 CLEAR 28*32
30 ' SIZE OF MAZE IN DATA
40 MW=28:MH=31
50 ' ROOM FOR CONVERTED MAZE
60 DIM MZ$(MH):LN=0
70 CLS
80 ' READ A LINE OF HEX DATA
90 READ A$:IF A$="" THEN 340
100 ' EVERY 2 CHARS IS A HEX
110 FOR A=1 TO LEN(A$) STEP 2
120 ' COVERT HEX TO A VALUE
130 V=VAL("&H"+MID$(A$,A,2))
140 ' CONVERT BITS TO BLOCKS
150 BT=128
160 IF (V AND BT)=0 THEN B$=CHR$(128) ELSE B$=CHR$(175)
170 ' ADD BLOCK/SPACE TO STRING
180 MZ$(LN)=MZ$(LN)+B$
190 ' IF LEN >= MAZE W, NEXT
200 IF LEN(MZ$(LN))>=MW THEN 280
210 ' SHIFT BIT RIGHT
220 BT=INT(BT/2)
230 ' IF MORE BITS, REPEAT
240 IF BT>0 THEN 160
250 ' GET NEXT CHARACTER
260 GOTO 310
270 ' ELSE NEW MAZE LINE
280 PRINT MZ$(LN)
290 LN=LN+1
300 ' NEXT CHARACTER
310 NEXT
320 ' MORE LINES TO DO?
330 IF LN<MH THEN 90

340 ' CONVERT LINES
350 BL$=CHR$(128):PRINT
360 FOR LN=0 TO MH-1
370 FOR C=1 TO LEN(MZ$(LN))
380 IF MID$(MZ$(LN),C,1)=BL$ THEN 510
390 V$=CHR$(175)
400 IF LN>0 THEN U$=MID$(MZ$(LN-1),C,1):UR$=MID$(MZ$(LN-1),C+1,1) ELSE U$=BL$:UR$=BL$
410 IF LN>0 THEN IF C>1 THEN UL$=MID$(MZ$(LN-1),C-1,1) ELSE ELSE UL$=BL$
420 IF C>1 THEN L$=MID$(MZ$(LN),C-1,1) ELSE L$=BL$
430 IF C<LEN(MZ$(LN)) THEN R$=MID$(MZ$(LN),C+1,1) ELSE R$=BL$
440 IF LN<MH-1 THEN D$=MID$(MZ$(LN+1),C,1) ELSE D$=BL$
450 IF LN<MH-1 THEN IF C>1 THEN DL$=MID$(MZ$(LN+1),C-1,1) ELSE ELSE DL$=BL$
460 IF LN<MH-1 THEN IF C<MW THEN DR$=MID$(MZ$(LN+1),C+1,1) ELSE ELSE DR$=BL$
470 IF U$<>BL$ THEN IF L$<>BL$ THEN GOSUB 600
480 IF U$<>BL$ THEN IF R$<>BL$ THEN GOSUB 700
490 IF D$<>BL$ THEN IF L$<>BL$ THEN GOSUB 800
500 IF D$<>BL$ THEN IF R$<>BL$ THEN GOSUB 900
510 NEXT:PRINT MZ$(LN):NEXT
520 END

600 ' UP and LEFT set
610 IF UL$<>BL$ THEN V$=CHR$(ASC(V$) AND NOT 8)
620 IF R$=BL$ THEN IF DR$=BL$ THEN IF D$=BL$ THEN V$=CHR$(ASC(V$) AND NOT 1)
630 MID$(MZ$(LN),C,1)=V$:RETURN

700 ' UP and RIGHT set
710 IF UR$<>BL$ THEN V$=CHR$(ASC(V$) AND NOT 4)
720 IF L$=BL$ THEN IF DL$=BL$ THEN IF D$=BL$ THEN V$=CHR$(ASC(V$) AND NOT 2)
730 MID$(MZ$(LN),C,1)=V$:RETURN

800 ' DOWN and LEFT set
810 IF DL$<>BL$ THEN V$=CHR$(ASC(V$) AND NOT 2)
820 IF U$=BL$ THEN IF UR$=BL$ THEN IF R$=BL$ THEN V$=CHR$(ASC(V$) AND NOT 4)
830 MID$(MZ$(LN),C,1)=V$:RETURN

900 ' DOWN and RIGHT set
910 IF DR$<>BL$ THEN V$=CHR$(ASC(V$) AND NOT 1)
920 IF L$=BL$ THEN IF UL$=BL$ THEN IF U$=BL$ THEN V$=CHR$(ASC(V$) AND NOT 8)
930 MID$(MZ$(LN),C,1)=V$:RETURN

1000 REM MAZE AS HEX DATA
1010 DATAFFFFFFF080060010BDF6FBD0BDF6FBD0BDF6FBD080000010BDBFDBD0BDBFDBD081861810FDF6FBF005F6FA0005801A0005BFDA00FDA05BF000204000FDA05BF005BFDA0005801A0005BFDA00FDBFDBF080060010BDF6FBD0BDF6FBD08C000310EDBFDB70EDBFDB7081861810BFF6FFD0BFF6FFD080000010
1020 DATAFFFFFFF0,""

Testing on the XRoar emulator shows this whole process takes about 161 seconds.

That’s so slow. Perhaps removing some of the CHR$() stuff and changing strings to values (like the originally PEEK version), skipping some of the extra IFs (like, if you are on line 0, there is no up, up left, or up right; if you are on the last line, there is no bottom, bottom left, or bottom right) and some other minor things might help…

340 ' CONVERT LINES
350 BL=128:PRINT
360 FOR LN=0 TO MH-1
365 LL=LEN(MZ$(LN))
370 FOR C=1 TO LL
380 IF ASC(MID$(MZ$(LN),C,1))=BL THEN 510
390 V=175

395 IF LN=0 THEN U=BL:UR=BL:UL=BL:GOTO 420
400 U=ASC(MID$(MZ$(LN-1),C,1)) ELSE U=BL
405 IF C<LL THEN UR=ASC(MID$(MZ$(LN-1),C+1,1)) ELSE UR=BL
410 IF C>1 THEN UL=ASC(MID$(MZ$(LN-1),C-1,1)) ELSE UL=BL

420 IF C>1 THEN L=ASC(MID$(MZ$(LN),C-1,1)) ELSE L=BL
430 IF C<LL THEN R=ASC(MID$(MZ$(LN),C+1,1)) ELSE R=BL

435 IF LN=MH-1 THEN D=BL:DL=BL:DR=BL:GOTO 470
440 D=ASC(MID$(MZ$(LN+1),C,1)) ELSE D=BL
450 IF C>1 THEN DL=ASC(MID$(MZ$(LN+1),C-1,1)) ELSE DL=BL
460 IF C<LL THEN DR=ASC(MID$(MZ$(LN+1),C+1,1)) ELSE DR=BL

470 IF U<>BL THEN IF L<>BL THEN GOSUB 600
480 IF U<>BL THEN IF R<>BL THEN GOSUB 700
490 IF D<>BL THEN IF L<>BL THEN GOSUB 800
500 IF D<>BL THEN IF R<>BL THEN GOSUB 900
510 NEXT:PRINT MZ$(LN):NEXT
515 PRINT TIMER,TIMER/60
520 END

600 ' UP and LEFT set
610 IF UL<>BL THEN V=(V AND NOT 8)
620 IF R=BL THEN IF DR=BL THEN IF D=BL THEN V=(V AND NOT 1)
630 MID$(MZ$(LN),C,1)=CHR$(V):RETURN

700 ' UP and RIGHT set
710 IF UR<>BL THEN V=(V AND NOT 4)
720 IF L=BL THEN IF DL=BL THEN IF D=BL THEN V=(V AND NOT 2)
730 MID$(MZ$(LN),C,1)=CHR$(V):RETURN

800 ' DOWN and LEFT set
810 IF DL<>BL THEN V=(V AND NOT 2)
820 IF U=BL THEN IF UR=BL THEN IF R=BL THEN V=(V AND NOT 4)
830 MID$(MZ$(LN),C,1)=CHR$(V):RETURN

900 ' DOWN and RIGHT set
910 IF DR<>BL THEN V=(V AND NOT 1)
920 IF L=BL THEN IF UL=BL THEN IF U=BL THEN V=(V AND NOT 8)
930 MID$(MZ$(LN),C,1)=CHR$(V):RETURN

Well, that brings it down to 155 seconds and that simply won’t do. There are many other optimizations to this BASIC program that could be done, and I bet we could even double the speed. But even then, that’s far too long to wait for a maze level to be drawn.

It looks like an assembly language routine might be our only option. We’ll take a break for now, but revisit this later after some discussion on assembly language. At some point. Eventually.

Until next time…

Pac-Man maze in 64x32 semigraphics.

Drawing a maze level on the CoCo – part 4

See Also: part 1, part 2, part 3, part 4 and part 5.

ASCII character data to 8-bit ASCII HEX data.

Let’s jump right in and write the program that writes the DATA statements for this maze.

It should generate a string that will start with a line number, the keyword “DATA”, and then add 2-digit hex values to it until it reaches that magic 250 character limit. When that happens, we write it out to cassette or disk, and start a new line. (Of course, this could also be adjusted to make each line 31 characters or less, to fit nicely on the CoCo’s screen. That’s the magic of computer generated code — you can easily make it change the output rather than having to do it by hand.)

I also want to make it get as many bytes as possible on the line, so I’ll start with single digit line numbers, and increment by 1. “1DATA” in ASCII gives me four extra bytes than “1000 DATA” and, when they get tokenized (either by loading from tape/disk in ASCII, or typing them in) they will take up the same amount of room since line numbers are all stored as two byte values in the program code. The result can then be RENUMbered to start at whatever line is needed for the final program.

Here is my very slow, very large version:

10 CLEAR 500
20 ' MAX OUTPUT LINE LENGTH
30 INPUT "MAX LINE LENGTH";MX
40 IF MX=0 OR MX>250 THEN MX=250
50 ' GET/SET DEVICE NUMBER
60 DV=0
70 PRINT "OUTPUT TO D)ISK, P)RINTER"
80 INPUT "S)CREEN OR T)APE";A$
90 IF A$="D" THEN DV=1 ELSE IF A$="P" THEN DV=-2 ELSE IF A$="S" THEN DV=0 ELSE IF A$="T" THEN DV=-1 ELSE 70
100 ' CREATE OUTPUT FILE
110 OPEN "O",#DV,"MAZEDATA.BAS"
120 ' LINE NUMBER/RESET STRING
130 LN=1:LN$=""
140 ' READ LINE OF MAZE DATA
150 READ A$:IF A$="" THEN 470
160 PRINT A$
170 ' START AT FIRST CHARACTER
180 CH=1
190 ' INIT BIT VALUE
200 BT=128
210 ' RESET OUTPUT VALUE
220 V=0
230 ' IF X, ADD 1 (SET BIT)
240 IF MID$(A$,CH,1)="X" THEN V=V+BT
250 PRINT MID$(A$,CH,1);
260 ' SHIFT BIT LEFT
270 BT=INT(BT/2)
280 ' IF MORE BITS, DO NEXT CH
290 IF BT>0 THEN CH=CH+1:GOTO 240
300 ' CREATE 2-DIGIT HEX VALUE
310 HX$=HEX$(V):IF V<16 THEN HX$="0"+HX$
320 PRINT ,HX$;CH
330 ' START LINE IF NEEDED
340 ' TRIM FIRST CHAR
350 IF LN$="" THEN LN$=STR$(LN)+"DATA":LN$=MID$(LN$,2)
360 ' ADD HEX VALUE
370 LN$=LN$+HX$
380 'IF LINE FULL, WRITE IT OUT
390 IF LEN(LN$)>=MX-2 THEN PRINT #DV,LN$:LN$="":LN=LN+1
400 ' GO TO NEXT CH
410 CH=CH+1
420 ' IF PAST END, NEW LINE
430 IF CH>LEN(A$) THEN 150
440 ' OTHERWISE, NEXT BYTE
450 GOTO 200
460 ' WRITE ANY REMAINING DATA
470 PRINT #DV,LN$
480 CLOSE #DV
490 END
1000 DATA "XXXXXXXXXXXXXXXXXXXXXXXXXXXX"
1010 DATA "X            XX            X"
1020 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1030 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1040 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1050 DATA "X                          X"
1060 DATA "X XXXX XX XXXXXXXX XX XXXX X"
1070 DATA "X XXXX XX XXXXXXXX XX XXXX X"
1080 DATA "X      XX    XX    XX      X"
1090 DATA "XXXXXX XXXXX XX XXXXX XXXXXX"
1100 DATA "     X XXXXX XX XXXXX X     "
1110 DATA "     X XX          XX X     "
1120 DATA "     X XX XXXXXXXX XX X     "
1130 DATA "XXXXXX XX X      X XX XXXXXX"
1140 DATA "          X      X          "
1150 DATA "XXXXXX XX X      X XX XXXXXX"
1160 DATA "     X XX XXXXXXXX XX X     "
1170 DATA "     X XX          XX X     "
1180 DATA "     X XX XXXXXXXX XX X     "
1190 DATA "XXXXXX XX XXXXXXXX XX XXXXXX"
1200 DATA "X            XX            X"
1210 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1220 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1230 DATA "X   XX                XX   X"
1240 DATA "XXX XX XX XXXXXXXX XX XX XXX"
1250 DATA "XXX XX XX XXXXXXXX XX XX XXX"
1260 DATA "X      XX    XX    XX      X"
1270 DATA "X XXXXXXXXXX XX XXXXXXXXXX X"
1280 DATA "X XXXXXXXXXX XX XXXXXXXXXX X"
1290 DATA "X                          X"
1300 DATA "XXXXXXXXXXXXXXXXXXXXXXXXXXXX"
1310 DATA ""

Let’s walk through the code…

  • Line 10 – We need more than the default 200 bytes of string space.
  • Lines 20-40 – Just for fun, I decided to let the user choose how long the output lines will be. 32 would give DATA statements that fit on the screen, or maybe 80 for a printer. 250 is the largest allowed.
  • Lines 50-90 – Just to be fancy, the user can choose Disk, Printer, Screen or Tape output. When writing this, I learned you can still use OPEN “O”,#-2,”FILE” even on the printer or screen! The filename is ignored and no error is generated. Very cool.
  • Lines 100-110 – Create the output file.
  • Lines 120-130 – LN is the initial line number for the DATA statements, and LN$ will be the buffer for the line we are creating.
  • Lines 140-160 – Read a line of MAZE DATA and display it. (This program is slow, so I thought some output would be fun to look at.)
  • Lines 170-180 – CH will be the current character of the line we are checking to be an “X” or not.
  • Lines 190-200 – BT is the bit value. For 8-bits, we start with the high bit, which is a value of 128. Unlike my previous example, I think this might be a simpler approach for making the value as we scan 8 characters.
  • Lines 210-220 – V will be the output value (8 characters turned in to a number).
  • Lines 230-250 – If the current character is “X”, add the BT value to V. We then print the character that is being processed.
  • Lines 260-270 – To get to the next bit value, we divide by 2. 128 becomes 64, 64 becomes 32, etc.
  • Lines 280-290 – As long as BT>0, we have more bits to do so we can move to the next CH character of the line. GOTO 240 resumes checking the character.
  • Lines 300-320 – If out of bits, we make a HEX$ out of V. If it is a value less than 16 (0-15 would be 0-F in hex), we add a leading zero to make it always two characters. We then print that to the screen.
  • Lines 330-350 – If LN$ is empty, we need to start it with a line number and “DATA”. When using STR$(x) to turn a number into a string, a leading space will be added. This means 1 becomes ” 1″. If this was negative, that would be used for the minus sign — i.e., -1 would be “-1”. Since we know our number will be positive, we can just get rid of the first character. Using a trick I recently learned about, MID$() will give the right side of a string starting with a position if you leave out the third parameter (length). Thus, instead of RIGHT$(LN$,LEN(LN$)-1), I can do MID$(LN$,2) to give me everything from character 2 on. Wish I knew that in the 80s! (Hat tip to Robin at 8-Bit Show and Tell since I think that is where I learned this.)
  • Lines 360-370 – We add the HX$ to the end of our LN$ (which is either the one we just started, or one that is slowly growing as we keep adding values).
  • Lines 380-390 – If the LN$ reaches our max (minus 2, since we just added a 2 character hex value), we print it out to the selected device, then reset LN$ and increment our line number.
  • Lines 400-410 – Move to the next CH character position.
  • Lines 420-430 – If CH is past the end of the length of our A$ data, go to line 150 to read the next maze DATA.
  • Lines 440-450 – If there’s still more characters in the line, go back to 150 and get the next one.
  • Lines 460-480 – If done, we need to output whatever is left. The earlier output only did that when the line reached max size. This takes care of the final line that is less than that. We then close the file. (Or, if this is printer or screen, this doesn’t do anything.)

Done! Whew…

Running this using a max of 250 produces this in the file:

1DATAFFFFFFF080060010BDF6FBD0BDF6FBD0BDF6FBD080000010BDBFDBD0BDBFDBD081861810FDF6FBF005F6FA0005801A0005BFDA00FDA05BF000204000FDA05BF005BFDA0005801A0005BFDA00FDBFDBF080060010BDF6FBD0BDF6FBD08C000310EDBFDB70EDBFDB7081861810BFF6FFD0BFF6FFD080000010FFFF
2DATAFFF0

Our maze data converts to just two hex values more than what would fit on one line of BASIC. So close! To make it look prettier (but take up more program space) we could have done it with shorter lines:

1DATAFFFFFFF080060010BDF6FBD0BDF6FBD0BD
2DATAF6FBD080000010BDBFDBD0BDBFDBD08186
3DATA1810FDF6FBF005F6FA0005801A0005BFDA
4DATA00FDA05BF000204000FDA05BF005BFDA00
5DATA05801A0005BFDA00FDBFDBF080060010BD
6DATAF6FBD0BDF6FBD08C000310EDBFDB70EDBF
7DATADB7081861810BFF6FFD0BFF6FFD0800000
8DATA10FFFFFFF0

But whatever style you choose, you can then CLOAD/LOAD this ASCII BASIC program and RENUM it to 1000 or wherever you want it to go.

And now that we have our 8-bits maze DATA, we need to write code to read, parse, and display it.

Until next time…

Drawing a maze level on the CoCo – part 3

See Also: part 1, part 2, part 3, part 4 and part 5.

Converting data to other data

So far, we’ve talked about how to represent a simple maze level, and how to make the maze level have more detail than the data used to represent it. We still need to code a way to represent the maze as bits rather than ASCII characters. Let’s try to get that one figured out.

First, we’ll use the DATA statements shown in part 2 and make a program that scans them and outputs the hex bytes we’ll want for DATA statements.

1000 DATA "XXXXXXXXXXXXXXXXXXXXXXXXXXXX"
1010 DATA "X            XX            X"
1020 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1030 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1040 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1050 DATA "X                          X"
1060 DATA "X XXXX XX XXXXXXXX XX XXXX X"
1070 DATA "X XXXX XX XXXXXXXX XX XXXX X"
1080 DATA "X      XX    XX    XX      X"
1090 DATA "XXXXXX XXXXX XX XXXXX XXXXXX"
1100 DATA "     X XXXXX XX XXXXX X     "
1110 DATA "     X XX          XX X     "
1120 DATA "     X XX XXXXXXXX XX X     "
1130 DATA "XXXXXX XX X      X XX XXXXXX"
1140 DATA "          X      X          "
1150 DATA "XXXXXX XX X      X XX XXXXXX"
1160 DATA "     X XX XXXXXXXX XX X     "
1170 DATA "     X XX          XX X     "
1180 DATA "     X XX XXXXXXXX XX X     "
1190 DATA "XXXXXX XX XXXXXXXX XX XXXXXX"
1200 DATA "X            XX            X"
1210 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1220 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1230 DATA "X   XX                XX   X"
1240 DATA "XXX XX XX XXXXXXXX XX XX XXX"
1250 DATA "XXX XX XX XXXXXXXX XX XX XXX"
1260 DATA "X      XX    XX    XX      X"
1270 DATA "X XXXXXXXXXX XX XXXXXXXXXX X"
1280 DATA "X XXXXXXXXXX XX XXXXXXXXXX X"
1290 DATA "X                          X"
1300 DATA "XXXXXXXXXXXXXXXXXXXXXXXXXXXX"
1310 DATA ""

The idea would be to build bytes by scanning 8 characters and setting a 1 bit if there is an “X”, or a 0 bit if there is a ” ” space. In C I’d do this using bit shifting. BASIC doesn’t have that. Except it kinda does.

I have previously done some YouTube videos about bit manipulation in BASIC, but basically you just divide by 2 to shift right, or multiply by 2 to shift left. Here is a sample that will take an ASCII binary string and turn it in to a numeric variable:

0 REM BITPARSE.BAS
10 INPUT "BINARY STRING";A$
20 BT=0
30 ' SCAN LEFT TO RIGHT
40 FOR A=1 TO LEN(A$)
50 ' SHIFT LEFT
60 BT=BT*2
70 ' IF 1, ADD 1 BIT
80 IF MID$(A$,A,1)="1" THEN BT=BT+1
90 NEXT
100 PRINT A$;" =";BT
110 GOTO 10

This lets me type a binary string like “111” and see that it represents 7. “11111111” will show 255. You can even go beyond 8 bits, though I did not test to see what the upper limit of BASIC’s floating point variables are before they start showing bogus values.

BASIC Bit Parsing

If it’s easy to build a value out of ASCII ones and zeros, it should be just as easy to do it looking for ASCII Xs and spaces. The only new part will be stopping every 8 characters to output the number, then starting over with the next 8 characters. Here is the BITPARSE program updated to do just that. Changes are bolded.

0 REM MAZE2NUM.BAS
10 READ A$:IF A$="" THEN END
15 FOR B=1 TO LEN(A$) STEP 8
20 BT=0
30 ' SCAN LEFT TO RIGHT
40 FOR A=B TO B+7
50 ' SHIFT LEFT
60 BT=BT*2
70 ' IF 1, ADD 1 BIT
80 IF MID$(A$,A,1)="X" THEN BT=BT+1
90 NEXT
100 PRINT MID$(A$,B,8);TAB(8);" =";BT
105 NEXT
110 GOTO 10

1000 DATA "XXXXXXXXXXXXXXXXXXXXXXXXXXXX"
1010 DATA "X            XX            X"
1020 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1030 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1040 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1050 DATA "X                          X"
1060 DATA "X XXXX XX XXXXXXXX XX XXXX X"
1070 DATA "X XXXX XX XXXXXXXX XX XXXX X"
1080 DATA "X      XX    XX    XX      X"
1090 DATA "XXXXXX XXXXX XX XXXXX XXXXXX"
1100 DATA "     X XXXXX XX XXXXX X     "
1110 DATA "     X XX          XX X     "
1120 DATA "     X XX XXXXXXXX XX X     "
1130 DATA "XXXXXX XX X      X XX XXXXXX"
1140 DATA "          X      X          "
1150 DATA "XXXXXX XX X      X XX XXXXXX"
1160 DATA "     X XX XXXXXXXX XX X     "
1170 DATA "     X XX          XX X     "
1180 DATA "     X XX XXXXXXXX XX X     "
1190 DATA "XXXXXX XX XXXXXXXX XX XXXXXX"
1200 DATA "X            XX            X"
1210 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1220 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1230 DATA "X   XX                XX   X"
1240 DATA "XXX XX XX XXXXXXXX XX XX XXX"
1250 DATA "XXX XX XX XXXXXXXX XX XX XXX"
1260 DATA "X      XX    XX    XX      X"
1270 DATA "X XXXXXXXXXX XX XXXXXXXXXX X"
1280 DATA "X XXXXXXXXXX XX XXXXXXXXXX X"
1290 DATA "X                          X"
1300 DATA "XXXXXXXXXXXXXXXXXXXXXXXXXXXX"
1310 DATA ""

This was substantially easier to do than I expected.

Text to bits to numbers…

And I got lucky, thanks to something BASIC does I was not aware of. The strings being parsed at not multiples of 8 characters. Each line is 28 characters long. Since I parse each one 8 characters at a time, I should be getting:

  1. MID$(A$,1,8) – first 8 characters.
  2. MID$(A$,9,8) – second 8 characters.
  3. MID$(A$,17,8) – third 8 characters
  4. MID$(A$,25,8) – fourth 8 characters … but there are only 2 left in the line!

MID$() does something that saved me a ton of work. Say you have a ten character string, and you want to print 4 characters starting at position 2:

A$="1234567890"

PRINT MID$(A$,2,4)
2345

But what if you wanted to print four characters starting at position 9? There are only two characters left in the string:

A$="1234567890"

PRINT MID$(A$,9,4)
90

We asked for four… we got two … no error was given. What is BASIC returning? I think this is kind of a bug, because you can do this:

A$="1234567890"

PRINT MID$(A$,25,4)

Tada! Even though you asked for a character beyond the end of the string, it does not error. It just returns … nothing. Literally nothing — the empty string.

A$="1234567890"

IF MID$(A$,200,4)="" THEN PRINT "NOTHING"
NOTHING

And since our parser is looking for an “X” to set a bit, and anything else for 0, nothing counts as anything else and it just works.

Nice. (And I’m too embarrassed to show the overcomplex version I came up with the first time I tried to do this…)

Hexing

It is to make the program print out HEX values by using HEX$, but there is a problem… HEX$ only prints the characters it needs to. If you print HEX$(255) you will get FF. If you print HEX$(15) you will get F. We want our hex values to always be two digits, so we have to add some extra code to make sure any single digit values (0-15) have a 0 added to the start of them.

The simplest way is probably this — assuming we are just dealing with 2-digit hex values (0-255):

V=15
HX$=HEX$(V):IF LEN(HX$)<2 THEN HX$="0"+HX$

Or, since we know only values 0-15 will be one hex digit, this may be both smaller and faster:

V=15
HX$=HEX$(V):IF V<16 THEN HX$="0"+HX$

Or we could go all complicated and build a string like “00” and use MID$ to insert the new string, either one or two characters long.

V=15:V$=HEX$(V)
HX$="00":MID$(HX$,LEN(HX$)-LEN(V$)+1)=V$

That one is cool because you can make HX$=”00″ for two digits, or “0000” for four digits. While the first two are smaller (and probably faster), they are hard-coded for 2 digit hex values.

Which is all we are using here, so I’ll just go with one of those.

Make the computer do the work…

We could now just output all the hex values and put them in DATA statements. Perhaps we copy them down from the screen, or maybe we have the program create a tape or disk text file and write them to it. We could even make them write out as if they were an ASCII program that could be loaded back in to BASIC.

That sounds fun, so we’ll do that.

Here is a simple program that will create a text file on tape (or disk) and write out a few lines of BASIC. You can then load this file in using LOAD or CLOAD and it will load as if it were a program saved in ASCII format.

10 OPEN "O",#-1,"DATA.BAS"
20 PRINT #-1,"1000 DATA 1,2,3"
30 CLOSE #-1

You can change the device from #-1 (cassette) to #1 for disk.

LINE LENGTH WARNING: Lines loaded in this was have a limit of no more than 250 characters. That’s all the room the BASIC input buffer has. If you have a line longer than that, the end of it will be cropped off. Here is the longest line you could load this way (100 1’s, 100 2’s, then 32 dashes, and some numbers so I could count how far it went easily on the screen).

1DATA11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111112222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222--------------------------------1234567890AB

A space after the line number can always be left out of ASCII BASIC program files, and BASIC shows a space there when you LIST it anyway. But if you want a cleaner look with a space between “DATA” and your data, just remember 250 is the limit.

With this in mind, we should be able to generate a string that starts with “1DATA” then packs as many 2 character HEX values as will fit up to 250.

Tangent about line lengths…

Something interesting… If you try typing in BASIC, you will see it stops you after 249 characters (7 full lines plus 25 more characters). I don’t know why they are different.

Also, if you type a long line like this, and press ENTER, the line gets tokenized and takes up less space (assuming it has any BASIC keywords in it like “DATA”). You may then be able to EDIT the line, ‘X’ to the end and type more characters. This is referred to as “packing” the line. However, if you do this EDIT trick, and type until it stops you again, the final character gets dropped. I’m betting that has something to do with the 250 versus 249 difference somewhere in the ROM. I’ll have to look in to that some day.

Also also, you can start the lines out as one digit line numbers, like 1, 2, 3, etc. That gives you more room than if you used line “1000”. Then, you can rename it later, and the tokenized line will still have your full data. Line numbers are represented by 2-bytes once the program is tokenized so it doesn’t matter if it’s line 1 or line 60000 it takes up the same amount of space.

But I digress…

In the next installment, I’ll come up with some kind of program that makes the program so we don’t have to.

Until then…

Drawing a maze level on the CoCo – part 2

See Also: part 1, part 2, part 3, part 4 and part 5.

Making a maze of blocks look less blocky

Okay, where were we?

Right. We wanted to take this…

Pac-Man maze in ASCII

…and turn it in to this…

Pac-Man maze in 64x32 semigraphics.
Pac-Maze in 64×32 semigraphics.

…while storing it like this…

DATA FFFFFFF0

To round or not to round…

If the maze were stored as bits representing blocks that were either set or unset (32×16 blocks will fit on the CoCo 32 column screen), we could make the program handle changing those blocks in to ones with “rounded” corners, or open areas in the middle or large blocks.

First, I needed to decide what the rules are for this, so I did some drawings.

For each corner (top left, top right, bottom left, bottom right) it seemed safe to round if the block above, beside, and diagonally from the corner were empty (empty red square). BUT, only round if the opposite sides were NOT vacant (red X). If that second part isn’t there, a single block with a path all around it would have all of its corners rounded. That’s fine on high resolution graphics (it would be a circle?), but when each block is made up of 2×2 mini squares, you can’t remove each corner or you’ll have nothin’ left!

As I found out.

Next, there are large areas made up of double blocks. On the Pac-Man maze these are larger “open” areas with thin walls around them, like the four along the top of the arcade Pac-Man game:

Arcade Pac-Mac (image from Wikipedia)

Don’t fill me in…

To figure out what the rules were for this, I did more drawings…

This one took me a bit longer to figure out. Above, each mini square of block is represented by the graph. Each white box is 2 blocks by 2 blocks.

After way too much trial and error, I worked it out that any corner that has an occupied square above, beside, and diagonally qualifies to be removed.

At first I tried to brute-force this in BASIC. It was way too long, and way too slow.

Then I decided as I was scanning each block location, I’d use variables to PEEK the contents of what was up, down, left or right from the blocks, as well as up-left, up-right, down-left and down-right from it. Using variables greatly sped things up (though they are still way too slow).

Once I had those variables, I could apply some simply logic.

I noticed that in both CORNERS and CENTERS, there is a need to check for an OCCUPIED block beside and either above or below. I started there, since I could do that check once, then dispatch to a second routine to determine if it was doing a corner, a center, or both.

I ran in to many bugs during all of things, and learned I had to pass each block through all the checks… Originally, I’d find a corner, remove it, and move to the next block. That would leave other corners that should have been removed, or center bits…

Finally, I came up with this (slow) proof-of-concept:

0 REM MAZEVERT.BAS

10 CLS:FOR I=0 TO 15:READ A$:PRINT@2+32*I,A$;:NEXT
20 BL=96
30 FOR M=1024 TO 1535
40 IF PEEK(M)=BL THEN 140
50 POKE M,175:V=175
60 IF M>1055 THEN U=PEEK(M-32):UR=PEEK(M-31) ELSE U=BL:UR=BL
70 IF M>1056 THEN UL=PEEK(M-33) ELSE UL=BL
80 L=PEEK(M-1):R=PEEK(M+1)
90 DL=PEEK(M+31):D=PEEK(M+32):DR=PEEK(M+33)

100 IF U<>BL THEN IF L<>BL THEN GOSUB 200
110 IF U<>BL THEN IF R<>BL THEN GOSUB 300
120 IF D<>BL THEN IF L<>BL THEN GOSUB 400
130 IF D<>BL THEN IF R<>BL THEN GOSUB 500
140 NEXT
150 FOR M=1024 TO 1535:IF PEEK(M)=96 THEN POKE M,128
160 NEXT
170 GOTO 170

200 ' UP and LEFT set
210 IF UL<>BL THEN V=V AND NOT 8
220 IF R=BL THEN IF DR=BL THEN IF D=BL THEN V=V AND NOT 1
230 POKE M,V:RETURN

300 ' UP and RIGHT set
310 IF UR<>BL THEN V=V AND NOT 4
320 IF L=BL THEN IF DL=BL THEN IF D=BL THEN V=V AND NOT 2
330 POKE M,V:RETURN

400 ' DOWN and LEFT set
410 IF DL<>BL THEN V=V AND NOT 2
420 IF U=BL THEN IF UR=BL THEN IF R=BL THEN V=V AND NOT 4
430 POKE M,V:RETURN

500 ' DOWN and RIGHT set
510 IF DR<>BL THEN V=V AND NOT 1
520 IF L=BL THEN IF UL=BL THEN IF U=BL THEN V=V AND NOT 8
530 POKE M,V:RETURN

999 GOTO 999

1000 DATA "XXXXXXXXXXXXXXXXXXXXXXXXXXXX"
1010 DATA "X            XX            X"
1020 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1030 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1040 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1050 DATA "X                          X"
1060 DATA "X XXXX XX XXXXXXXX XX XXXX X"
1070 DATA "X XXXX XX XXXXXXXX XX XXXX X"
1080 DATA "X      XX    XX    XX      X"
1090 DATA "XXXXXX XXXXX XX XXXXX XXXXXX"
1100 DATA "     X XXXXX XX XXXXX X     "
1110 DATA "     X XX          XX X     "
1120 DATA "     X XX XXXXXXXX XX X     "
1130 DATA "XXXXXX XX X      X XX XXXXXX"
1140 DATA "          X      X          "
1150 DATA "XXXXXX XX X      X XX XXXXXX"
1160 DATA "     X XX XXXXXXXX XX X     "
1170 DATA "     X XX          XX X     "
1180 DATA "     X XX XXXXXXXX XX X     "
1190 DATA "XXXXXX XX XXXXXXXX XX XXXXXX"
1200 DATA "X            XX            X"
1210 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1220 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1230 DATA "X   XX                XX   X"
1240 DATA "XXX XX XX XXXXXXXX XX XX XXX"
1250 DATA "XXX XX XX XXXXXXXX XX XX XXX"
1260 DATA "X      XX    XX    XX      X"
1270 DATA "X XXXXXXXXXX XX XXXXXXXXXX X"
1280 DATA "X XXXXXXXXXX XX XXXXXXXXXX X"
1290 DATA "X                          X"
1300 DATA "XXXXXXXXXXXXXXXXXXXXXXXXXXXX"

The program has three phases:

  1. Draw the ASCII maze (or at least what will fit on the screen for this test).
  2. Scan the screen and convert non-blanks to semigraphics blocks, then check to see if a corner needs to be removed (because it is a corner or in the center of a larger group of blocks).
  3. Scan the screen one more time, changing any blank space blocks to black.

The end result is this:

Pac-Man maze in 64x32 semigraphics.

The whole process takes just over 40 seconds. It could be made faster by going in to double speed mode with POKE 65495,0 (or 65497,0 on a CoCo 3), but even then it would still be way too long for the user to wait for a game screen to draw.

Plus, this doesn’t even have the code to convert the less-memory hex string representation of the maze into the actual maze.

Which means we have more to do…

To be continued…

Pac-Man maze in 64x32 semigraphics.

Drawing a maze level on the CoCo – part 1

See Also: part 1, part 2, part 3, part 4 and part 5.

A maze of ideas

Ready to fall down a rabbit hole, exploring ways to represent a maze screen efficiently?

Neither was I.

Somewhere on this site is an article series I wrote about programming a simple maze type game on the CoCo. In it, I presented an ASCII version of the Pac-Man maze. The arcade maze looks like this:

Actual size (224×228) Pac-Man screen.

And you can learn a ton about how the game works on the Pac-Man Dosier website. But basically, the game screen is a grid of 28 x 36 tiles. In the arcade original, each tile is 8×8 pixels. With the CoCo 1/2s text screen being 32×16, this means we could fit 28 tiles horizontally with room to space, but we’d be 20 tiles short vertically. (Pac-Man used a monitor mounted sideways to get this aspect ratio.) My solution was to scroll the maze up and down. I also removed the top and bottom rows that showed score and such, so this will be using only 31 of the games original 36 rows.

A portion displayed in ASCII text would look like this:

Pac-Man maze in ASCII

Though it wouldn’t look like the arcade, this is an accurate representation of the 28×36 tiles of the original game (well, 28×16 seen any any time). There are elements of Pac-Man that couldn’t really be recreated with this low resolution (such as the way Pac can hug to corners to go around them faster), but at least the maze layout could be accurate.

But ASCII is ugly, so we’d probably want to use the semigraphics characters (128-255). Using CHR$(175) for a blue block looks like this:

Pac-Man maze in 32x16 semigraphics.
Pac-Man maze in 32×16.

That’s better, but still a bit ugly. The background could at least be made black:

Pac-Man maze in 32×16 with black background.

That’s more better, but it looks more reminiscent of blocky Atari VCS games like Adventure.

By taking advantage of the 64×32 resolution semigraphics, the corners could be at least “rounded” a bit (if round meant “still square”), and the centers of the areas could be opened up like the arcade original:

Pac-Man maze in 64x32 semigraphics.
Pac-Man maze in 64×32 semigraphics.

At this resolution, I’m not sure if we can really do any better than that ;-)

The maze data could be represented as ASCII strings, either directly PRINTed on the screen, or read from DATA statements to be PRINTed or POKEd:

Pac-Maze DATA statements.

This would make creating new mazes and levels as easy as typing. But storing 28 x 26 characters is 1008 bytes of memory. Imagine trying to do that on a 4K Color Computer or a 5K Commodore VIC-20. (The highest resolution graphics screen on a CoCo 1/2 takes up 6144 bytes, compared to the 512 bytes of the text screen.)

Here’s the ASCII DATA:

1000 DATA "XXXXXXXXXXXXXXXXXXXXXXXXXXXX"
1010 DATA "X            XX            X"
1020 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1030 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1040 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1050 DATA "X                          X"
1060 DATA "X XXXX XX XXXXXXXX XX XXXX X"
1070 DATA "X XXXX XX XXXXXXXX XX XXXX X"
1080 DATA "X      XX    XX    XX      X"
1090 DATA "XXXXXX XXXXX XX XXXXX XXXXXX"
1100 DATA "     X XXXXX XX XXXXX X     "
1110 DATA "     X XX          XX X     "
1120 DATA "     X XX XXXXXXXX XX X     "
1130 DATA "XXXXXX XX X      X XX XXXXXX"
1140 DATA "          X      X          "
1150 DATA "XXXXXX XX X      X XX XXXXXX"
1160 DATA "     X XX XXXXXXXX XX X     "
1170 DATA "     X XX          XX X     "
1180 DATA "     X XX XXXXXXXX XX X     "
1190 DATA "XXXXXX XX XXXXXXXX XX XXXXXX"
1200 DATA "X            XX            X"
1210 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1220 DATA "X XXXX XXXXX XX XXXXX XXXX X"
1230 DATA "X   XX                XX   X"
1240 DATA "XXX XX XX XXXXXXXX XX XX XXX"
1250 DATA "XXX XX XX XXXXXXXX XX XX XXX"
1260 DATA "X      XX    XX    XX      X"
1270 DATA "X XXXXXXXXXX XX XXXXXXXXXX X"
1280 DATA "X XXXXXXXXXX XX XXXXXXXXXX X"
1290 DATA "X                          X"
1300 DATA "XXXXXXXXXXXXXXXXXXXXXXXXXXXX"

If we tried to use numeric DATA statements to represent the screen blocks, it would be 4 times longer since every entry would be a three digit number with a comma:

DATA 175,175,175,175,175,175,175,175,175,175,175,175,175,175,175,175,175,175,175,175,175,175,175

But, if we were just using blocks and spaces, we could have a number represent eight of them. Each number could represent a byte (0-255) and each byte contains 8 bits, therefore each number could represent 8 character blocks.

In a DATA statement with numbers, the top row would look like:

1000 DATA 255,255,255,240

The storage space for “255,255,255,240” is 15 bytes which is less than the 28 string characters (plus quotes, if needed — like if the line started with spaces or had special characters in it like a comma or ‘ REM abbreviation, which this won’t).

I suppose we could use numbers representing 16-bit values (0-65535) and have only two numbers to represent that line:

1000 DATA 65535,65520

That brings us down to 11 bytes to represent the 28 block row.

The 4K CoCo BASIC did not have hexadecimal numbers, but on Extended BASIC machines we could convert to HEX:

1000 DATA &HFF,&HFF,&HFF,&HF0

…but that is 19 bytes. While that would be better than ASCII, it would be worse than two decimal values. Doubling up to 16 bit values results in:

1000 DATA &HFFFF,&HFFF0

That gets us down to 12 bytes so it looks like needing that &H at the start loses out — 65535 versus &HFFFF. Even worse when 1 has to be three times larger as &H1. Though 15 is only one by larger as &HF. But always larger.

Hex values could be written as a string without the &H, and that could be added by the READ routine:

1000 DATA FF,FF,FF,F0

…which gives us 11 bytes, matching the two decimal values. But reading those would require some extra code which might negate the savings:

READ A$:V=VAL("&H"+A$)

…versus just reading a decimal value…

READ V

Using hex without the “&H” requires an extra 15 bytes of code space (actually, maybe a little less, since the “VAL” keyword probably tokenizes in to one or two bytes).

But, we could also pack the hex values together:

1000 DATA FFFFFFF0

That would only take 8 bytes to represent a row. But even more code would be needed to parse that:

READ A$:FOR A=1 TO LEN(A$) STEP 2:V=VAL("&H"+MID$(A$,A,2)):NEXT

The code gets even longer. However, if we had enough data (and 36 rows of data probably is enough), the code space savings should be smaller overall. And clever programming could be done to have one massive hex string and decode it knowing how many bytes go to each line.

But this only gets us to displaying the maze in ASCII or as colored blocks. If we wanted to get that “rounded” corner look of a 64×32 display, we’d need twice as many bytes.

Or we could just make the computer do it for us.

To be continued…

Color BASIC and 16-bits and AND and OR and NOT

During #SepTandy2020, I posted a CoCo-related video each day of the month. A few of them dealt with bit operations in Color BASIC. This was when I fist learned that Color BASIC could not deal with more than 15-bit values. You can do AND/OR/NOT operations

These work:

PRINT 32767 AND 1
PRINT 32767 OR 1
PRINT NOT 32676

…but these will return ?FC ERROR:

PRINT 32768 AND 1
PRINT 32768 OR 1
PRINT NOT 32678

8-bits before 16-bits

8-bits is handled just fine. Here is an example program that will display an 8-bit value (0-255) as binary. It is very slow:

0 ' slow8bit.bas
10 FOR Z=0 TO 255
20 GOSUB 1000:PRINT
30 NEXT
40 END
1000 REM SLOW: 8-BIT Z AS BINARY
1010 FOR BT=7 TO 0 STEP-1
1020 IF Z AND 2^BT THEN PRINT "1"; ELSE PRINT "0";
1030 NEXT:RETURN

If you looked at TIMER, that one would count to 7339 (about 122.2 seconds). And here is a version that is substantially faster because it pre-calculates the AND bit values in an array:

0 ' fast8bit.bas
5 DIM BT(7):FOR BT=0 TO 7:BT(BT)=INT(2^BT):NEXT
10 FOR Z=0 TO 255
20 GOSUB 1000:PRINT
30 NEXT
40 END
1000 REM FAST: 8-BIT Z AS BINARY
1010 FOR BT=7 TO 0 STEP-1
1020 IF Z AND BT(BT) THEN PRINT "1"; ELSE PRINT "0";
1030 NEXT:RETURN

TIMER for that one would count to 1372 (about 22.8 seconds).

When I was recording those videos, I decided to make versions that handled 16-bit values. That is when I discovered you could not to that in Color BASIC.

16-bit workaround

Color BASIC seems to treat bit values as signed, so instead of 0 to 65535, it is really -32768 to 32767. That led me to a workaround where, if the high bit was set, I would subtract 65536 from the value and do the AND/OR/NOT on a negative value. And it worked!

0 ' fast16bit.bas
5 DIM BT(15):FOR BT=0 TO 14:BT(BT)=INT(2^BT):NEXT:BT(15)=-32768
10 FOR Z=0 TO 65535
20 GOSUB 1000:PRINT
30 NEXT
40 END
1000 REM SLOW: 16-BIT Z AS BINARY
1005 IF Z>32767 THEN Z=Z-65536
1010 FOR BT=15 TO 0 STEP-1
1020 IF Z AND BT(BT) THEN PRINT "1"; ELSE PRINT "0";
1030 NEXT:RETURN

The above version uses powers-of-two for the first 15 bits (1, 2, 4, 8, 16, etc.) then for bit 16 it uses -32768, which will work. When the value is passed in, if it is larger than 32767, it is turned in to a negative value which would be a 16-bit number with the high bit set, indicating negative.

I have no idea if this is “supposed” to work, or even be allowed (AND/OR/NOT on a negative), but it does.

And what about AND? (And OR, and NOT?)

For something I am working on, I wanted to use AND to get the most significant and least significant bytes of a 16-bit value. I could not, so I came up with various workarounds:

  1. To get the MSB of a 16-bit value, shift the 16-bit value right 8 bits (divide by 256).
  2. To get the LSB of a 16- bit value, subtract the MSB*256 from the original 16-bit value.
10 W=&HFFEE
20 M=INT(W/256)
30 L=W-M*256
40 PRINT HEX$(W),HEX$(M);" ";HEX$(L)

This worked fine, but required swapping variables around to preserve the original values.

In my project, I then wanted to clear the top 5-bits of the word, so I would clear them in the MSB value:

M=M AND &H07

…and then I could re-build the word variable using “W=M*256+B”.

Then I wondered: Could I somehow use the same trick I used for 16-bit binary numbers to make AND, OR and NOT just work on the value?

It turns out, yes. Yes I could.

I started with a subroutine that would take a 16-bit value in in Z and convert it, as needed, to a negative value that represented the desired bits. I did this this by manually looping through the bits and adding the appropriate bit value from the same array I used in the fast 16-bit routine shown earlier:

5 DIM BT(15):FOR BT=0 TO 14:BT(BT)=INT(2^BT):NEXT:BT(15)=-32768
...
1000 ' CONVERT 16-BIT Z TO ZC
1010 ZC=0:IF Z>32767 THEN Z=Z-65536 ELSE ZC=Z:RETURN
1020 FOR BT=15 TO 0 STEP-1
1030 IF Z AND BT(BT) THEN ZC=ZC+BT(BT)
1040 NEXT:RETURN

I could then set a value to Z and call this subroutine and then use ZC with AND/OR/NOT:

10 V=65535:A=&HFF00
20 PRINT V;"AND";A;"=";
30 Z=V:GOSUB 1000:VC=ZC
40 Z=A:GOSUB 1000:AC=ZC
50 Z=VC AND AC

Yuck. But it worked. Except, after I AND the two converted values, the value I get would be a negative if it was using more than 15-bits. If I wanted to display that as HEX, I’d get another ?FC ERROR because you cannot HEX$ a negative.

Thus, I needed a routine to convert such a 16-bit negative value back to a normal positive number:

1100 ' CONVERT ZC BACK TO Z
1110 IF Z<0 THEN ZC=Z+65536 ELSE ZC=Z
1120 RETURN

The end result was a multistep routine I could use to AND, OR or NOT and 16-bit value:

0 ' AND16.BAS
5 DIM BT(15):FOR BT=0 TO 14:BT(BT)=INT(2^BT):NEXT:BT(15)=-32768

10 V=65535:A=&HFF00
20 PRINT V;"AND";A;"=";
30 Z=V:GOSUB 1000:VC=ZC
40 Z=A:GOSUB 1000:AC=ZC
50 Z=VC AND AC:GOSUB 1100:PRINT ZC

999 END
1000 ' CONVERT 16-BIT Z TO ZC
1010 ZC=0:IF Z>32767 THEN Z=Z-65536 ELSE ZC=Z:RETURN
1020 FOR BT=15 TO 0 STEP-1
1030 IF Z AND BT(BT) THEN ZC=ZC+BT(BT)
1040 NEXT:RETURN

1100 ' CONVERT ZC BACK TO Z
1110 IF Z<0 THEN ZC=Z+65536 ELSE ZC=Z
1120 RETURN

And it works!

But is there a better way?

A better way

To avoid all the swapping with temporary variables, it might be easier to just make subroutines for AND, OR and NOT that handled two passed-in variables and returned the results in another.

I created a generic subroutine that would take V1 and V2, and convert V1 (if negative) and create a new value based on the bits of V2:

4000 ' CONVERT V1 AND V2
4010 IF V1>32767 THEN V1=V1-65536
4020 IF V2>32767 THEN V2=V2-65536
4030 ZZ=0:FOR BT=15 TO 0 STEP-1
4040 IF V2 AND BT(BT) THEN ZZ=ZZ+BT(BT)
4050 NEXT:V2=ZZ:RETURN

The program could set V1 to a value and V2 to a value for the bit operation, then GOSUB 4000. Then the program could do “V=V1 AND V2” or “V=V1 OR V2” or “V=NOT V2” to perform the operation.

Then, a GOSUB to 5000 would convert V back to the positive value (if needed):

5000 ' CONVERT V BACK, IF NEEDED
5010 IF V<0 THEN V=V+65536
5020 RETURN

This allowed a slightly less ugly way to do AND, OR and NOT on 16-bit values:

0 ' ANDORNOT16.BAS
5 DIM BT(15):FOR BT=0 TO 14:BT(BT)=INT(2^BT):NEXT:BT(15)=-32768

10 V1=&HFFFF:V2=&HFF00
20 PRINT HEX$(V1);" AND ";HEX$(V2),"=";
30 GOSUB 4000:V=V1 AND V2:GOSUB 5000:PRINT HEX$(V)

40 V1=&H8000:V2=&H0001
50 PRINT HEX$(V1);" OR  ";HEX$(V2),"=";
60 GOSUB 4000::V=V1 OR V2:GOSUB 5000:PRINT HEX$(V)

70 V2=&HF00F
80 PRINT "     NOT ";HEX$(V2),"=";
90 GOSUB 4000:V=NOT V2:GOSUB 5000:PRINT HEX$(V)

999 END

1000 ' AND 16-BIT V1 WITH V2
1010 GOSUB 4000:V=V1 AND V2:GOSUB 5000:RETURN

2000 ' OR 16-BIT V1 WITH V2
2010 GOSUB 4000:V=V1 OR V2:GOSUB 5000:RETURN

3000 ' NOT 16-BIT V1
2010 GOSUB 4000:V=NOT V2:GOSUB 5000:RETURN

4000 ' CONVERT V1 AND V2
4010 IF V1>32767 THEN V1=V1-65536
4020 IF V2>32767 THEN V2=V2-65536
4030 ZZ=0:FOR BT=15 TO 0 STEP-1
4040 IF V2 AND BT(BT) THEN ZZ=ZZ+BT(BT)
4050 NEXT:V2=ZZ:RETURN

5000 ' CONVERT V BACK, IF NEEDED
5010 IF V<0 THEN V=V+65536
5020 RETURN

It still ain’t pretty, but it still works. (Though has not been extensively tested, so your milage may vary.)

Until next time…

DEF FN for Fun and Profit

In a comment to my recent article about VARPTR, Rogelio Perea chimed in…

Still learning something new about the CoCo even after all the years, excellent article!

Would there ever be one about DEFFN? :-)

–  Rogelio Perea II, 7/19/2022

Challenge accepted.

DEF FN: Definitely Fun?

There are many features of Microsoft Color BASIC that went right over my head when I was learning to program as a junior high school kid. DEF FN was certainly one of them.

My first CoCo came with Extended Color BASIC, and that was required to use DEF FN. But what is is? According to Getting Started with Extended Color BASIC:

Extended Color BASIC has one numeric function, DEF FN, that is unlike any others we’ve talked about so far. DEF FN lets you create your own mathematical function. You can use your new function the same as any of the available functions (SIN, COS, TAN, and so on). Once you’ve used DEF FN to define a function, you may put it to work in your program by attaching the prefix FN to the name you assign to the new function. Here is the syntax for DEF FN:

DEF FN name (variable list) – formula

name is the name you assign to the function you create.
variable list contains one “dummy variable” for each variable to be used by the function.
formula defines the operation in terms of the variables given in the variable list

Note: Variable names that appear in formula serve only to define the formula; they do not affect program variables that have the same name. You may have only one argument in a formula call; therefore, DEF FN must contain only one variable.

You may use DEF FN only in a program, not in the immediate mode.

– Getting Started with Color BASIC, page 177.

I think my work here is done.

Until next time…

Okay, maybe not.

DEF FN: Define Function

There are commands in BASIC, such as PRINT and GOTO, and then there are functions that return values, such as SIN, ABS and INT.

PRINT SIN(42)
-.916521549

X=-42
PRINT ABS(X)
42

X=4.2
PRINT INT(X)
4

DEF FN allows you to create your own function that returns a value.

A simple example

We start with a simple example. What if you wanted a function that added 42 to any value you passed in? I will call this function “UA” (for “Ultimate Answer”) and define it as follows:

10 DEF FNUA(N)=N+42
20 PRINT FNUA(10)

If I run that, it will print 52. The new “function” FNUA(x) returns whatever you pass in with 42 added to it. Tada!

Function names seem to follow the same rules as variable names so you can have one or two letters (A to Z, AA to ZZ) or a letter and a number (A0 to Z9).

In the parens is a parameter that also follows the same rules of a variable. The parameter is just a placeholder so you have something to use in your new formula on the right of the equal sign. If you did DEF FNA(X) then after the equal you’d have whatever you want to happen to X. When the user calls it with FNA(123) then 123 goes to where the X was in the formula after the equals… Neat.

It’s kind of like a C function with only one parameter, now that I think about it/

// fnua(n) - return n plus 42
int fnua(float n)
{
    return n+42;
}

int main()
{
    float x = fnua(10);

    printf ("x = %f\n", x);

    return EXIT_SUCCESS;
}

You can have DEF FNA(X) or DEF FNZZ(AB) or any combination.

What can you use it for?

The manual gives an example of doing math with a long floating point value so you don’t have to have that value in your code everywhere you want to use it.

7 DEF FNR(X)=X/57.28577951
6 AA=FNR(AA): AB=FNR(AB): AC=FNR(AC)

At least I’m not the only one with code snippets that don’t really do anything.

Of course, for something like this you could have placed that number in a variable which would be faster than parsing a long number (in most cases). Still, if the formula itself is big, it could save typing, I guess.

Benchmarks, anyone?

I was curious if it would be any faster for other types of maths…

Suppose I want to use PRINT@ to put a character on the text screen, but want to use PEEK to see if it “hits” another character on the text screen. I would use the PRINT@ positions of 0-511, but I’d want to PEEK the screen memory locations of 1024-1535.

10 CLS:P=0
20 FOR I=1 TO 20:PRINT@RND(511)-1,"X";:NEXT
30 PRINT @P,"*";
40 A$=INKEY$:IF A$="" THEN 40
50 IF A$="W" THEN NP=P-32:GOTO 90
60 IF A$="A" THEN NP=P-1:GOTO 90
70 IF A$="S" THEN NP=P+32:GOTO 90
80 IF A$="D" THEN NP=P+1
90 IF NP<0 THEN 40
100 IF NP>510 THEN 40
110 IF PEEK(NP+1024)<>96 THEN 40
120 PRINT @P," ";
130 P=NP
140 GOTO 30

Above is a program that will clear the screen and put X characters in random positions. The player is an asterisk in the top left of the screen. Using the keys W (up), A (left), S (down) and D (right) will move the asterisk around the screen. Checks ensure you cannot go off the top or bottom of the screen, or run in to an X.

Line 110 does the conversion from PRINT@ position to PEEK position. Maybe we can do this:

0 ' defusr2.bas
10 CLS:P=0
15 DEF FNP(X)=X+1024
20 FOR I=1 TO 20:PRINT@RND(511)-1,"X";:NEXT
30 PRINT @P,"*";
40 A$=INKEY$:IF A$="" THEN 40
50 IF A$="W" THEN NP=P-32:GOTO 90
60 IF A$="A" THEN NP=P-1:GOTO 90
70 IF A$="S" THEN NP=P+32:GOTO 90
80 IF A$="D" THEN NP=P+1
90 IF NP<0 THEN 40
100 IF NP>510 THEN 40
110 IF PEEK(FNP(NP))<>96 THEN 40
120 PRINT @P," ";
130 P=NP
140 GOTO 30

The behavior is the same, and now I get the benefit of it not having to parse “NP+1024” each time. Parsing decimal values is slow. On Extended BASIC I could have replaced 1024 with hex &H400 and that would have been faster. For more speed, I could have used a variable and had “NP+S” or whatever.

But is it faster?

0 ' defbench1.bas
5 P=0
10 TIMER=0
20 FOR A=1 TO 1000
30 Z=P+1024
40 NEXT
50 PRINT TIMER

Doing “P+1024” 1000 times prints 481.

0 ' defbench2.bas
5 P=0
6 DEF FNP(X)=X+1024
10 TIMER=0
20 FOR A=1 TO 1000
30 Z=FNA(P)
40 NEXT
50 PRINT TIMER

That version prints 603. Not faster. The overhead of FN seems to be more than just doing it directly. In a way, this makes sense because in-lining code in assembly or C is faster than calling it as a subroutine/function.

But let’s try more… We could also include other things in the function, like the PEEK itself.

0 ' defbench3.bas
5 P=0
6 DEF FNP(X)=PEEK(X+1024)
10 TIMER=0
20 FOR A=1 TO 1000
30 Z=FNP(P)
40 NEXT
50 PRINT TIMER

That one prints 690, and if you had just done “30 Z=PEEK(P+1024)” it would have been 572.

Cool, but definitely slower. I believe this is because it’s still storing that formula in a buffer somewhere and executing it as if you’d just typed it in. Indeed, if you change the formula to use hex:

0 ' defbench3.bas
5 P=0
6 DEF FNP(X)=PEEK(X+&H400)
10 TIMER=0
20 FOR A=1 TO 1000
30 Z=FNP(P)
40 NEXT
50 PRINT TIMER

…it prints 480, which means it’s parsing that formula each time instead of converting it to a number when you define it.

Pity. It would have been really cool if this had been a way to bypass the detokenizer stuff in BASIC and speed up routines like that.

FN Conclusion

DEF FN is slower. I don’t see any example where it speeds anything up. With that said, I expect I’d use DEF FN if…

  • I had a custom function I used in multiple places, and wanted to maintain it in just one spot. This would be more efficient, perhaps, than making it a GOSUB subroutine.
  • I needed to save some bytes of BASIC memory.
  • I wanted to write an example that fix things on small lines on the screen, just for appearances.

For anything else, I think I’d just put it in and make the program a bit faster.

Where else might one use this?

More examples

Sebastian Tepper commented to another post with a more useful example for DEF FN. There are many times when we use two PEEKs to get a 16-bit value from two bytes in memory:

V=PEEK(25)*256+PEEK(27)

Sebastian mentioned using DEF FN for this:

DEF FNP(AD)=PEEK(AD)*256+PEEK(AD+1)

V=FNP(25)

That’s very cool, and something I could see myself using. I wish I had thought of that when I was doing all my “Color BASIC memory location” articles, since I typed the double PEEK lines over and over and over…

But that’s not all…

The defined function can also use other variables. You can only pass one in, but any variable that is declared at the time the FN is used may be used. You could do this:

10 DEF FNA(X)=X*Y
20 Y=10
30 PRINT FNA(Y)
RUN
50

Above, you pass in a value for “X” and it returns the value multiplied by whatever Y is at that moment. You could create more complex formulas that way, but it kind of defeats the purpose.

Speaking of defeating the purpose, here are more silly examples. Using it to replicate PEEK:

10 DEF FNP(X)=PEEK(X)
20 PRINT FNP(1024)

How about returning if something is true or not? Like, is a memory location outside screen memory?

…but that sure looks like it would be really slow. Still neat, though. You could do:

IF FNL(L) THEN PRINT"NOT IN SCREEN MEMORY"

It does make the code look neater — especially if something is being checked multiple times (such as checking four ghost in Pac-Man or whatever).

What other examples can you think of? Leave a comment.

Until next time…

Color BASIC string abuse – part 2

See also: part 1 and part 2.

In the first part, the following Color BASIC program was shared:

5 CLEAR 100*255
10 MX=99
20 DIM A$(MX):AL=0:DL=-1
30 SZ=RND(255):SZ=255
40 A$=STRING$(SZ,"X")
50 GOSUB 1000:GOTO 30
999 END

1000 REM
1001 REM ADD NEW STRING
1002 REM
1010 IF AL=DL THEN GOSUB 2000
1020 GOSUB 3000:IF Z<1024 THEN GOSUB 2000
1030 PRINT "ADDING";LEN(A$)"BYTES AT";AL;Z
1040 A$(AL)=A$
1050 AL=AL+1:IF AL>MX THEN AL=0
1060 IF DL=-1 THEN DL=0
1070 RETURN

2000 REM
2001 REM DELETE OLD STRING
2002 REM
2010 PRINT ,"DELETING";DL
2020 A$(DL)=""
2030 DL=DL+1:IF DL>MX THEN DL=0
2040 GOSUB 5000
2050 RETURN

3000 REM
3001 REM GET FREE STRING SPACE
3002 REM
3010 Z=(PEEK(&H23)*256+PEEK(&H24))-(PEEK(&H21)*256+PEEK(&H22))
3020 RETURN

5000 REM
5001 REM PAUSE
5002 REM
5010 PRINT"[PAUSE]";
5020 IF INKEY$="" THEN 5020
5030 PRINT STRING$(7,8);:RETURN

The program allocates enough string space to hold 100 strings of 255 bytes each. It then starts adding line after line until it detects string memory is getting low. When that happens, the oldest line is deleted (set to “”). The process continues…

The “gut feeling” was that this program should have been able to hold 100 full sized strings, but since it did use a temporary A$ (created to be 255 strings line, and the string we would be adding to the array), it seemed logical that it would have to start purging lines maybe a few entries before the end.

But instead, it starts deleting lines at around entry 47. And, the memory usage being printed out shows it drops by 510 byes each time instead of 255.

510 is an interesting number. That is 255 * 2. That makes it seem like each time we add a 255 byte string, we are using twice that memory.

And we are!

Strings may live again to see another day

The key to what is going on is in the main loop starting at line 30. We create a new A$ and set it to a string of 255 X’s. That string has to be stored somewhere, so it goes in to string memory. Then we do the GOSUB and add a string to the array, which copies our A$ in to the array A$(AL) entry.

When we go back to 30 for the next round, we create A$ again. The old copy of A$, in string memory, is deallocated, and a new A$ is created at the current string memory position. BASIC does not see if the old string space is large enough to be re-used. It just moves on to a new allocation of string space.

It looks like this… And note that strings fill from the end of the memory and move lower. Let’s say we have 16 bytes of reserved string memory (just to keep things fitting on this screen).

FRETOP points to where reserved string memory begins. MEMSIZ is the end of string memory. If we had done CLEAR 16, then FRETOP would be MEMSIZ-16.

STRTAB is where the next string will be added. It looks like this:

FRETOP                                     MEMSIZ
 |                                            |
[.][.][.][.][.][.][.][.][.][.][.][.][.][.][.][.]
                                              |
                                           STRTAB 

Let’s say we create a 4-byte A$=STRING(4, “X”) (“XXXX”) that we plan to add to an array. It will be stored in string memory:

FRETOP                                     MEMSIZ
 |                                            |
[.][.][.][.][.][.][.][.][.][.][.][.][X][X][X][X]
                                  |
                               STRTAB 

Later in the code, we assign that to the array, such as A$(0)=A$. Now A$(0) gets a copy of A$, and it looks like this (using lowercase ‘x’ to represent the copy of A$ that was put in A$(0)):

FRETOP                                     MEMSIZ
 |                                            |
[.][.][.][.][.][.][.][.][x][x][x][x][X][X][X][X]
                      |
                   STRTAB 

When we go back to do this again, a new A$ is created, and it gets stored next. The old string data is still there, but A$ has been updated to point to the new entry.

FRETOP                                     MEMSIZ
 |                                            |
[.][.][.][.][X][X][X][X][x][x][x][x][X][X][X][X]
          |
       STRTAB

…and so on. As you can see, the way this loop was written, it is creating a new A$ every time through, copying it to the array (a new entry for that) and so on. That is why we see 510 each time through instead of just 255.

Now, if the string was short, we could have done A$=”XXXXXXXXXXXXX”. If we did that, the string would exist in program space and not in string memory. But we wanted a 255 byte string, and you can’t type a line that long in BASIC so STRING$() was used instead, which requires putting the string in string memory.

However, since in THIS version we are just using the same 255 character A$ over and over again, let’s make one change so we don’t create it every time through the loop. Just change the GOTO 30 in line 50 to be GOTO 50:

30 SZ=RND(255):SZ=255
40 A$=STRING$(SZ,"X")
50 GOSUB 1000:GOTO 50

Now the program will create one A$, which will store at the start of string memory, and then loop over and over just making new entries in the array.

That small change will instantly change the results to be more like we might have expected. Now we get all the way to entry 94 before any string deleting happens:

And, from looking at that screen, each number is dropping by 255 bytes as we expected.

By the time it reached line 94, it saw that there were less than 1024 bytes of free string space left. 1024 would have held another four 255 byte strings, meaning actually had enough memory to have gotten to line 98 — just one line short of our max 99 before it rolls over. And that memory is where the initial 255-byte A$ is stored.

Tada! Mystery solved.

But wait! There’s more…

The reason I chose 1024 as a threshold was to allow for other temporary string use in the program. Things like LEFT$, MID$, STRING$ all make temporary strings. When you add two strings together it creates a third string that combines the first two. Be sure to check out my string theory article for more details on this — I learned quite a bit when researching it. I also learned that some things required strings that I did not expect to. Fun reads. Helps put you to sleep.

If I modify line 1020 to check for 255 bytes remaining instead of 1024, then re-run, I get this:

…and that is as perfect as it gets. Array is filled with strings 0-98, plus the temporary string, which is a total of 100 strings of 255-bytes each — and that is how much memory we set aside with CLEAR!

Now how much would you pay?

And because this program is self-aware when it comes to knowing how much string space is there, it can actually operate with much less string space. It will just delete old strings sooner.

You can change the CLEAR in line 5 to something smaller, like 2000, and it will still work. But, 2000/255 is 7, so it has room for the A$ plus six array entries. I expect it would DELETE every 6 lines. Let’s try…

Bingo! After lines 0-5 (six lines) it deleted and old one, then since everything was now full, it had to delete every time it added something new.

And the point is…?

Well, let’s just say I wish I knew about this back in 1983 when I wrote my cassette-based Bulletin Board System, *ALLRAM*.

I always wanted to write version 2.0.

Until next time…

Color BASIC string abuse – part 1

See also: part 1 and part 2.

This program demonstrates Color BASIC’s string garbage collection. As strings are manipulated, string memory used will continue to grow until BASIC decides to clean up the string space and move things around a bit.

The program creates a string array and fills each entry up with string data. Using code that checks how much string space is remaining, it will delete old strings if string memory is low.

Watching it run reveals some curious things…

The Program

On an Extended or Disk BASIC system, you will need to do a PCLEAR 1 to get enough memory to run it.

  • Initialization
    • Line 5 – We CLEAR enough string space to hold 100 lines at their largest size of 255 characters. Note that when you use DIM to set an array size, it is base-0. Doing a DIM A$(99) gives you A$(0) to A$(99) — 100 elements. Thus, the CLEAR uses 100, but the DIM in the next line uses 99.
    • Line 10 – MX represents the maximum number of lines in the array.
    • Line 20 – DIM A$(MX) creates an array of 0-99 entries (100). AL is set to 0, and will be the line (array entry) we will ADD next. DL will be used to track the next line (array entry) we need to delete. DL is set to -1 for “nothing to delete yet.”
  • Main Loop
    • Line 30 – SZ is set to the length of the string we want to add. It has an unused RND at the start, but then SZ is hard-coded to 255. The program was designed to test random lengths of strings, but for this demo, we will be overriding that and making every string the maximum size.
    • Line 40 – GOSUB 1000 goes to a routine that will ADD a line. Then we GOTO 30 to create a new line and do it again.
  • Add New String subroutine
    • Line 1010 checks to see if the ADD line has caught up to the DELETE line. If it has, we GOSUB 2000 to handle deleting the delete line.
    • Line 1020 – GOSUB 3000 will return the amount of free string space in Z. If Z is less than 104, GOSUB 2000 is called to delete a string.
    • Line 1030 – It prints how long of a string is being added to which line, followed by the current free string space before the add.
    • Line 1040 – The string is added to the A$ array at position AL.
    • Line 1050 – AL (add line) is incremented by 1, moving it to the next line of the array. If AL is larger than the MX (max array entries), it will wrap around to entry 0.
    • Line 1070 – Returns.
  • Delete Old String subroutine
    • Line 2010 – Print which line is about to be deleted (set to “”).
    • Line 2020 – The A$ entry at position DL is set to “”, releasing the string memory it was using.
    • Line 2030 – DL (delete line) is incremented by one. If DL is larger than MX (max array entries), it will wrap around to entry 0.
    • Line 2040 – GOSUB 5000 is a pause routine so we can see what just happened.
  • Get Free String Space routine
    • Line 3010 – Z is calculated as the difference between the FRETOP (start of string storage) memory location and the STRTAB (start of string variables) in the reserved string area. This gives us how many bytes of unused string memory is available.
    • Line 2030 – Returns.
  • Pause subroutine
    • Line 510 – Print the message “[PAUSE]” with no carriage return at the end.
    • Line 5020 – If INKEY$ returns nothing (no key pressed), keep checking.
    • Line 5030 – Prints a string of 7 CHR$(8)s (backspace character) which will erase over the “[PAUSE]” prompt.

Based on the intent, one might think that running this would fill strings up to around entry 100, and then it would start deleting the old string

But is that what it will do?

Let’s run it and find out.

5 CLEAR 100*255
10 MX=99
20 DIM A$(MX):AL=0:DL=-1
30 SZ=RND(255):SZ=255
40 A$=STRING$(SZ,"X")
50 GOSUB 1000:GOTO 30
999 END

1000 REM
1001 REM ADD NEW STRING
1002 REM
1005 IF AL=DL THEN GOSUB 2000
1006 GOSUB 3000:IF Z<1024 THEN GOSUB 2000
1010 PRINT "ADDING";LEN(A$)"BYTES AT";AL;Z
1020 A$(AL)=A$
1030 AL=AL+1:IF AL>MX THEN AL=0
1040 IF DL=-1 THEN DL=0
1050 RETURN

2000 REM
2001 REM DELETE OLD STRING
2002 REM
2010 PRINT ,"DELETING";DL
2020 A$(DL)=""
2030 DL=DL+1:IF DL>MX THEN DL=0
2035 GOSUB5000
2040 RETURN

3000 REM
3001 REM GET FREE STRING SPACE
3002 REM
3010 Z=(PEEK(&H23)*256+PEEK(&H24))-(PEEK(&H21)*256+PEEK(&H22)):RETURN

5000 REM
5001 REM PAUSE
5002 REM
5010 PRINT"[PAUSE]";
5020 IF INKEY$="" THEN 5020
5030 PRINT STRING$(7,8);:RETURN

Running the Program

After a PCLEAR 1 and RUN, the program starts filling lines and we can see string memory going down. When it gets to line 47, there is only 1275 bytes of string space left.

And if we check those values, the difference between each one isn’t the 255 we might have expected. We clearly added a new 255 byte string, but memory went from 7905 to 7395 (510 bytes less) and continued down to 1785 to 1275 (510 bytes less). It appears each time we add a 255 byte string, it takes 510 bytes of string memory.

As we get to line 47, we must have had less than 1024 bytes of string memory left and the DELETE LINE subroutine is called. It deletes the oldest line, which is currently line 0. That should free up 255 bytes of string memory.

After pressing a key, we see the next few entries:

After deleting 0, memory has gone from 1275 to 756 (5120 bytes), so DELETE is called again. This time it deletes line 1.

We press a key and let it scroll a bit more until the next DELETE:

Here we see that, at some point, some string cleanup was done and our free string space grew larger. The trend of reducing by 510 bytes for each ned 255 byte string added continues…

And the process repeats, until eventually we roll over at line 99.

From here on, things get a bit more predictable, with a DELETE happening almost every line — though sometimes every two lines.

Eventually things settle in to a pattern of basically DELETING for every line ADDED.

Is it broken? Is it just poorly implemented? Or is it behaving exactly like it is supposed to?

Tune in next time for the exciting conclusion…