On the old school 8-bit home computers, not all of them used ASCII. Commodore used a variation called PETSCII, and the Atari 8-bits used ATASCII. While the trick discussed in this article might work on other systems that have a VARPTR or similar command, this discussion will be specifically about the character set in the Radio Shack Color Computer.
ASCII 65 is the uppercase letter ‘A’
PRINT CHR$(65) A
If you POKE the value of 65 to the first position on the 32×16 text screen (location 1024), you will also see an uppercase 65.
POKE 1024,65
However, the embedded font data in the MC6847 VDG video generator chip does not follow ASCII for all of its characters. For example, CHR$(0) to CHR(31) are non printable characters. On the CoCo, two of them do something special — CHR$(8) will print a backspace and CHR$(13) will print an ENTER:
It would have been nice if the CoCo could have done a beep for CHR$(7) like Apple 2s did, or clear the screen with CHR$(12) like many other systems did, but those are the only two that do anything other than “print nothing” on the CoCo.
If you POKE around a bit…
While you will not see anything if you PRINT those characters, if you POKE those values to the screen memory you will see something. For example, you could POKE characters 0 to 31 to the first row of the 32 column text screen like this:
FOR A=0 TO 31:POKE 1024+A,A:NEXT
The character set in the video chip has 0-31 representing reverse video characters “@” (AT sign) to “<-” (left arrow). We can expand that loop to POKE the first 128 characters onto the video screen:
FOR A=0 TO 127:POKE 1024+A,A:NEXT
But for PRINTing the ASCII characters, we have already established nothing shows up for characters 0-31, but things do PRINT when for 32-128:
FOR A=32 TO 127:PRINT CHR$(A);:NEXT
I put together this sloppy program that will show the differences, 32 characters at a time, of what you get when you PRINT the character values versus POKE the character values:
0 'POKEPRNT.BAS 10 CLS
20 PRINT@0,"PRINT 0-31:" 30 FOR A=0 TO 31:PRINT CHR$(A);:NEXT 40 PRINT@64,"POKE 0-31:" 50 FOR A=0 TO 31:POKE 1120+A,A:NEXT
60 PRINT@128,"PRINT 32-63:" 70 FOR A=32 TO 63:PRINT CHR$(A);:NEXT 80 PRINT@192,"POKE 32-63:" 90 FOR A=0 TO 31:POKE 1248+A,32+A:NEXT
100 PRINT@256,"PRINT 64-95:" 110 FOR A=64 TO 95:PRINT CHR$(A);:NEXT 120 PRINT@320,"POKE 64-95:" 130 FOR A=0 TO 31:POKE 1376+A,64+A:NEXT
140 PRINT@384,"PRINT 96-127:" 150 FOR A=96 TO 127:PRINT CHR$(A);:NEXT 160 PRINT@448,"POKE 96-127:" 170 FOR A=0 TO 31:POKE 1504+A,96+A:NEXT
999 GOTO 999
Looking at this, you can see only the characters 64-95 match between PRINT and POKE.
This means that the “copy screen to a string” concept from my earlier post doesn’t really do what we might expect. It does copy the data, but if we PRINT it back, we do not get back exactly what we started with.
This is the same thing that would happen if you tried to build a string by using PEEK from screen memory. This example prints stuff on the first line of the screen, then builds a string made of up characters using the PEEK value of that first line:
0 'PEEK2STR.BAS 10 CLS 20 PRINT "HELLO, WORLD! THIS IS A TEST." 30 FOR A=1024 TO 1024+31 40 A$=A$+CHR$(PEEK(A)) 50 NEXT 60 PRINT "PEEKED STRING:" 70 PRINT A$
And running that shows this awfulness…
Yuck!
But that’s okay since there is not much use to copying TEXT data and then putting it back with PRINT. PRINT is fast, and we can easily PRINT that text data. Sure, there could be benefits if stuff being PRINTed is doing calculations and such to generate the output, but this trick won’t help there.
However, the semi graphics characters (128-256) are the same between PRINT and POKE.
0 'POKEPRT2.BAS 10 CLS 20 FOR A=0 TO 255 30 PRINT@A,CHR$(A); 40 POKE 1280+A,A 50 NEXT 60 GOTO 60
The top half is the PRINT CHR$ and the bottom half is the POKE:
Since there is no way on the CoCo to type those semi graphics characters into a string (pity, the later MC-10 could do this), we are forced to PRINT them like this:
PRINT CHR$(128);CHR$(128);CHR$(128)
That would print three black blocks. To speed things up, we could pre-generate a string of those three black blocks then we can PRINT that string very fast later:
A$=CHR$(128);CHR$(128);CHR$(128) PRINT A$
And now you know why I chose to do a “splash screen” example for my demo in part 1. I initially tried it using the TEXT characters and quickly remembered why that can’t work (as explained here).
But it’s still a neat trick.
Bonus: This is stupid
For dumb fun, here is a program that makes A$ be whatever is on the first 32 character line of the screen.
When you RUN that, doing a PRINT A$ will show a 32 character line that is whatever was on the first line of the screen. If you do a “CLS” to clear the screen and show “OK” on the top line, then PRINT A$, you will see “OK” followed by 30 reverse @ symbols, which is CHR$(96) — but in video memory, a 96 is an empty block (space).
In a recent post about BASIC INSTR, a few comments were left trying to clarify or justify or explain the behavior of why “” is reported as matching the first character in the search string.
PRINT INSTR("ABC","A") 1
PRINT INSTR("ABC","C") 3
PRINT INSTR("ABC","X") 0
PRINT INSTR("ABC","") 1
Copilot AI at least pretends to know the actual reason, which involves saving a few precious bytes in the early 4K Microsoft ROM. The behavior was then retained for backwards compatibility as Microsoft BASIC expanded. Copilot even reports the behavior continued into VB.NET, though I did not fact check this answer.
LuisCOCO commented:
This is logical, if you search for three letters it searches for that piece but when searching for nothing it starts in the first position and searches for zero characters in the destination since it is useless to search for more letters if you search for fewer letters, that brings a null string as the first data found and compares it with the null to search and gives true
– LuisCOCO
Indeed, without a check for an empty string, the search would stop there. “Compare an X byte string to a Y byte string”. As the scan is performed, I suspect the moment a character does not match, it sets the return value to 0. Without looking at the code (yet), my assumption is that it starts with 1 as the return value and just exits when no more bytes are available to compare (immediately with a zero byte target string) leaving the 1 there. Since code exists to flag it with a zero if it a mismatch is encountered, perhaps they did just want to save a few instructions rather than doing a check then branch to that zero return.
Unless there truly is an intentional reason it works this way.
William “Lost Wizard” Astle, who wrote the 6809 assembly (LWasm) I use quite often, explained some math stuff:
Put another way, the empty string is self-evidently a valid substring of every string at every position within that string so it will naturally match the position where the search starts. This is also mathematically correct when you look at set theory. The null set is a subset of all sets.
– William Astle
It seems math folks are fine with how INSTR works. Though math set theory is not where my mind would have been as a child learning BASIC back in junior high school.
William Astle then sends me on a side quest with this comment…
Incidentally, strstr() in C does exactly the same thing with an empty target string.
– William Astle
Let me C what I can find…
strstr() works similarly to BASIC’s INSTR, except it returns a pointer to where the match was found in memory. If no match was found, it returns a NULL pointer (which is 0, matching INSTR).
BASIC gives back a base-1 index, so 1 for a match at the first character of the search string, or 5 for a match starting at the 5th character of the search string. Strings in MS BASIC used “normal” base-1 counting numbers for the index. If you use MID$(), LEFT$(), RIGHT$() and similar string functions, they all use 1 for the first character in a string.
C uses base-0 for indexes, so if you have a string in C:
char *st = "ABC";
…the first character would be index 0, st[0], while in BASIC:
ST$="ABC"
…the first character would be index 1, MID$(ST$,1,1).
int main (int argc, char **argv) { char *search = "ABC"; char *target = NULL; char *ptr = NULL;
// const char * strstr ( const char * str1, const char * str2 ); // A pointer to the first occurrence in str1 of the entire sequence of // characters specified in str2, or a null pointer if the sequence is // not present in str1.
printf ("search = '%s' (%p)\n", search, search);
target = "A"; ptr = strstr (search, target); if (NULL == ptr) { printf ("target '%s' not found in '%s'\n", target, search); } else { printf ("target '%s' found in '%s' at index %d\n", target, search, (ptr - search)); }
target = "C"; ptr = strstr (search, target); if (NULL == ptr) { printf ("target '%s' not found in '%s'\n", target, search); } else { printf ("target '%s' found in '%s' at index %d\n", target, search, (ptr - search)); }
target = "X"; ptr = strstr (search, target); if (NULL == ptr) { printf ("target '%s' not found in '%s'\n", target, search); } else { printf ("target '%s' found in '%s' at index %d\n", target, search, (ptr - search)); }
target = ""; ptr = strstr (search, target); if (NULL == ptr) { printf ("target '%s' not found in '%s'\n", target, search); } else { printf ("target '%s' found in '%s' at index %d\n", target, search, (ptr - search)); }
return EXIT_SUCCESS; }
Yuck. I should have made a function… This produces:
search = 'ABC' (0x58a8ed9ba008) target 'A' found in 'ABC' at index 0 target 'C' found in 'ABC' at index 2 target 'X' not found in 'ABC' target '' found in 'ABC' at index 0
I realize my example is … not optimal. strstr() is returning a pointer to memory, and my example is subtracting the pointer where the target string was found from the pointer where the search string is. If they are the same, the math is 0 for the first character. But if the function actually returns a 0 (NULL), that means it was not found. I should have considered that before writing this example ;-)
But I digress…
Which came first?
A quick (and non fact-checked) web search about when strstr was added to C reveals:
The strstr function was added to the C standard library as part of the ANSI C standard (C89/C90), which was approved in 1989 and published in 1990.
Well, at least it worked the same way BASIC programmers would have been used to. But I do wonder if there was an earlier implementation of BASIC that introduced INSTR, or similar.
Another quick (and non fact-checked) web search says DEC BASIC-PLUS for the PDP-11 had a similar function and was introduced in 1970.
If I can figure out how to use the PDP-11 online emulator, maybe I can figure this out:
“If B$ is a null string (B$ = “”), the INSTR functions returns the value 1. The null string is a proper substring of any string and is treated conventionally as the first element of A$ in null string search operations. In addition, if both A$ and B$ and null strings, the INSTR function returns the value of 1″
Unfortunately, the 1971 manual does not specifically address what happens with a null/empty string, but we can for sure say it was definitively documented to do that in the 1975 manual, and without any mention to it being a change from the earlier releases, likely was there in 1971.
I think we have a winner. As described, it is clear the intent matches William Astle‘s comment about null being a proper substring of any string. Microsoft implemented it the way it
Conclusion
I guess all I have to say is that I’d have preferred is for INSTR to return 0 if an empty string was passed in.
Unless someone can tell me the benefit of having it return 1, that is. I am a big fan of changing my own personal views when I learn new information. Just ask me about politics sometime. . . #independent
UPDATE: I believe I have found the answer, and will share it in an upcoming post. Until then, keep those comments coming. I learn so much from all of you!
This topic has been discussed here years ago, but every time something reminds me about it, I get annoyed. While my annoyance is triggered by how it works in the CoCo’s Extended Color BASIC, past research showed the behavior was the same even in much later Microsoft Visual BASIC. But why?
INSTR is a command to return the index where a target string is found in a search string. From one of the Getting Started with Extended Color BASIC manuals, it is shown as this:
What the manual did not mention is that it can also return 1 when there is no match. See this example:
Looking for “B” in “ABC”? That’s at position 2. Good.
Looking for “X” in “ABC”? It is not there, so it returns 0. Good.
Looking for “A” in “ABC”? That’s at position 1. Good.
Looking for “” in “ABC”? Apparently “” is found at position 1. Don’t tell that to the “A” there.
Callbacks
I ran into this years ago when I was experimenting with various ways to handle key presses. You could have code block until a key was pressed, and then pass the key to INST and then use ON GOTO/GOSUB to get to the routine. Like this:
0 'INSTR.BAS 10 PRINT "A)BORT, R)ETRY, C)ONTINUE:"; 20 A$=INKEY$:IF A$="" THEN 20 30 LN=INSTR("ARC",A$) 40 IF LN>0 THEN ON LN GOSUB 1000,2000,3000 50 GOTO 10
1000 ' ABORT 1010 PRINT"ABORT":STOP
2000 ' RETRY 2010 PRINT "RETRY":RETURN
3000 ' CONTINUE 3010 PRINT "CONT":RETURN
This was a great technique when dealing with a long list of menu options.
I had tried to optimize this by eliminating the A$ and embedding it inside the INSTR (someone in the comments may have suggested this to me; not sure if I am clever enough to have thought that up):
ON INSTR("ARC",INKEY$) GOSUB 1000,2000,3000
…but if I put that in my code replacing lines 20-40, running it immediately shows me “ABORT” as if INSTR returned 1.
Because INSTR returned 1.
The workaround suggested to me (again, from smart folks in the comments) was maybe to add a bogus value as the first search string character, and have that routine do nothing.
ON INSTR("*ARC",INKEY$) GOSUB 999,1000,2000,3000
However, for my example where I show the prompt again after it returns, it sticks in a loop printing the prompt over and over again. The code thinks the first option is being selected, then calls that routine (the empty routine that is just a RETURN in line 60) and then prints the prompt again.
One quick solution is to not use RETURN and let each function decide where to go back to. When you GOSUB, BASIC has to scan forward (possibly starting at the top of the program if the line number is before the current line being parsed) to find the target. RETURN lets it “pop” back to right after the GOSUB, so that part is faster.
Also, GOSUB routines can be called from different places in the main code and they will return back to where they were called.
If these routines are never called from anywhere but the menu code, and the extra speed to GOTO back is not a problem, this this change makes it work. And, as a bonus, the fake first GOTO line can just be back to the ON INSTR again since it doesn’t need to do anything:
2026-05-03 – Corrected hex value of 64 (thanks MiaM).
Today I was tagged in a Facebook post by MC-10 (well, and CoCo) programmer, Jim Gerrie. He shared a snipped of code he was trying to get working. The concept was to have stuff on the 32 column text screen get copied into a normal string and then be able to PRINT it back.
Jim’s post (with the code that wasn’t working) with code for the MC-10 was this:
10 CLEAR1200:DIMJ,K,A$,B$:GOSUB100 20 A$=””:K=VARPTR(A$):POKEK,255:POKEK+1,0:POKEK+2,64:B$=A$ 50 CLS:PRINTB$; 60 GOTO60 100 CLS1:PRINT”THIS IS LINE ONE” 110 PRINT”THIS IS LINE TWO” 112 PRINT”THIS IS LINE THREE” 113 PRINT”THIS IS LINE FOUR” 114 PRINT”THIS IS LINE FIVE” 115 PRINT”THIS IS LINE SIX” 116 PRINT”THIS IS LINE SEVEN” 117 PRINT”THIS IS LINE EIGHT”:RETURN
Can someone explain why this program doesn’t work on my TRS-80 MC-10?! It should reassign the memory pointer of string variable A$ to the beginning of screen memory (highbyte 64 lowbyte 0) so that I can then just assign A$ to B$, which will allow me to “capture” the first 255 bytes of screen mem.
It works on the TRS-80 MODEL I/III (using its screen start at highbyte 60 lowbyte 0)!
Any help greatly appreciated.
– Jim Gerrie in the Facebook TRS-80 MC-10 Group.
I could immediately see what the program was attempting to do, and it was something that never occurred to me to try. The concept is “simple” now that I see it:
Stuff is placed on the screen (CLS, PRINT, etc.)
A string (A$) is declared (line 20) and then VARPTR is used to get the memory location of the 5-byte string descriptor for that string. At this point, A$ is zero bytes long, but it will point to somewhere inside the program memory just after the first quote in A=”” because that is where the string begins (even if it is zero bytes long).
The string descriptor is modified using POKE to change the length of the string to 255 bytes, then the start location of the string from that location inside program memory to be the start of the text screen. That was the first bug. The location being POKEd was off by one and was not modifying the string start address properly.
After this, A$ is copied into a normal string, thus saving the contents of the first string (screen memory) into the new string (in normal reserved string memory). This is where the second bug was.
Before continuing, If you need a refreshed on VARPTR, start with this article. It will show how the five byte string descriptor is used. Here is a refresh:
STRING DESCRIPTOR (5 BYTES) 0 - LENGTH OF STRING 1 - NOT USED FOR STRINGS 2 - MSB OF ADDR OF STRING IN MEMORY 3 - LSB OF ADDR OF STRING IN MEMORY 5 - ALWAYS 0 FOR A STRING
The first POKE at the address returned by VARTPR (K) was fine, setting the length to 255. But the next two pokes were at K+1 and K+2. For Color BASIC, they should have been at K+2 and K+3. Also, the values being poked were 0 and 64, which is backwards. From a quick search, the MC-10s text screen starts at the 16K mark, $4000 (16384). To verify this, I went to the online MC-10 emulator here:
…and then did POKE 16384,42. That indeed placed an inverted “*” in the top left of the text screen.
The MC-10 is a big endian processor, so the memory location should be MSB ($40) then LSB ($00). $40 in decimal is 64 in decimal (no HEX support on the MC-10, I don’t think). So the actual pokes to make a string start at the top left corner of the text screen should have been 64 and 0 rather than 0 and 64. (That is 64*256+0 to make 16384.)
Adjusting those POKEs to be at the proper spot in VARPTR and swapping the values to MSB/LSB was the first fix.
At that point, I still wasn’t getting it working. This was due to the initial string being a “hard coded” string in BASIC. When A$=”” was declared in BASIC, it made a string that pointed into the program space. I have not looked into why, but forcing the string to be in RAM by doing A$=””+”” was all I needed to change to make this work. (NOTE TO SELF: Explore this and understand what was different about the program-space string.)
Jim posted a corrected version for the MC-10:
0 CLEAR1200:DIMC1,M$,I$,AA$,BB$:GOSUB100:GOTO20 8 M$="":C1=VARPTR(M$):POKEC1,255:POKEC1+2,64:POKEC1+3,0:AA$=M$+"" 9 M$="":C1=VARPTR(M$):POKEC1,255:POKEC1+2,65:POKEC1+3,0:BB$=M$+"":RETURN 20 GOSUB8:CLS:PRINTAA$" "BB$; 60 GOTO60 100 CLS8:PRINT"THIS IS LINE ONE" 110 PRINT"THIS IS LINE TWO" 112 PRINT"THIS IS LINE THREE" 113 PRINT"THIS IS LINE FOUR" 114 PRINT"THIS IS LINE FIVE" 115 PRINT"THIS IS LINE SIX" 116 PRINT"THIS IS LINE SEVEN" 117 PRINT"THIS IS LINE EIGHT" 118 PRINT"THIS IS LINE NINE" 119 PRINT"THIS IS LINE TEN" 120 PRINT"THIS IS LINE ELEVEN" 121 PRINT"THIS IS LINE TWELVE" 122 PRINT"THIS IS LINE THIRTEEN" 123 PRINT"THIS IS LINE FOURTEEN" 124 PRINT"THIS IS LINE FIFTEEN" 125 PRINT"THIS IS LINE SIXTEEN";:RETURN
Meanwhile, I had come up with a silly program that would create some kind of image on the CoCo screen, then capture it in two strings so it could be quickly restored later. Well, almost the entire screen — a string is limited to 255 bytes so two strings captures 510 bytes of the 32×16 screen. If one were to use this trick, it could be adjusted to capture just the number of lines on the screen needed (like, the first 5 lines, or lines 10-20, etc.).
My example looked like this:
0 'SAVESCR1.BAS 10 CLEAR511:DIMS1$,S2$ 11 ' 0 = LEN OF STRING 12 ' 1 = NOT USED FOR STRING 13 ' 2 = MSB OF ADDRESS 14 ' 3 = LSB OF ADDRESS 15 ' 4 = ALWAYS 0 16 ' 1024 = 4*256+0 17 ' 1279 = 4*256+255 20 REM DRAW SCREEN 25 CLS0:C=0:FOR I=0 TO 29 STEP 2 30 SET(I*2,0,C):SET(63-I*2,30,C) 35 SET(0,30-I,C):SET(63,I,C) 40 C=C+1:IF C>7 THEN C=0 50 NEXT 55 PRINT@266,"THIS";CHR$(128)"IS";CHR$(128);"COOL"; 60 'SAVE SCREEN 65 GOSUB 1000 70 'WAIT FOR KEY 75 GOSUB 5000 80 'CLEAR SCREEN 85 CLS 5 90 'WAIT FOR KEY 95 GOSUB 5000 100 'RESTORE SCREEN 105 GOSUB 2000 110 GOTO 70
999 GOTO 999
1000 ' SAVE SCREEN 1005 Z$="":K=VARPTR(Z$):POKEK,255:POKEK+2,4:POKEK+3,0:S1$=Z$+"" 1010 Z$="":K=VARPTR(Z$):POKEK,255:POKEK+2,4:POKEK+3,255:S2$=Z$+"" 1015 RETURN
5000 ' WAIT FOR KEY 5005 IF INKEY$="" THEN 5005 5010 RETURN
I made a “save screen” subroutine at line 1000. My program starts and makes a simple splash screen (colored pixels set around the screen and a text message in the center), then calls the save screen routine which makes a temporary Z$ then modifies it to point to the start of the text screen (1024 – so values 4*256+0) and be 255 bytes long. It then copies that string to S1$. It repeats the process for the next part of the screen, which begins at 1024+255 (1279, so 4*256+255). That is saved in S2$.
Then the main program waits for a keypress, then does a CLS 5 to erase the screen to white, then after another keypress, it calls the “restore screen” subroutine which just prints the two saved strings starting at position 0 on the top left corner of the screen.
AND IT WORKS!
Now blasting bytes to the screen can be as fast as printing a string. I can think of some interesting uses for this, such as “drawing” various levels slowly during initialization, and capturing them to strings so they can be displayed later very fast.
And that gives me an idea for an old project I was playing with some years ago.
But that will have to wait for the next installment…
Just posting this so I can share the link… This was a planning file from my “ASCII Kong” program I am toying with:
---------------------------- 12345678901234567890123456789012 H H BONUS H H 3900 H HXXXXXXX OO H H H OO.....H H H XXXXXXXXXXXXXXXXXXXXXXXXXX H XXXXXXXXXXXXXXXXXXXXXXXXXX H H XXXXXXXXXXXXXXXXXXXXXXXXXX H H H XXXXXXXXXXXXXXXXXXXXXXXXXX H H XXXXXXXXXXXXXXXXXXXXXXXXXX U H XXXXXXXXXXXXXXXXXXXXXXXXXXXX 12345678901234567890123456789012 ---------------------------- ----- @ <=> / \ ----- @ /= |\ ----- @ =\ /| -----
NOTE: I originally started writing this in November 2025, but kept thinking I’d do more work on it. I haven’t gotten around to it, so here you go…
Here is a Color BASIC 6809 assembly quickie… (That ended up not being very quick by the time I finished working through all of this…)
Recently I began working on an assembly language Color BASIC extension that makes certain characters move the cursor around the screen rather than just printing those characters (similar to what my VIC-20 could do). Initially, I created the 6809 assembly routine you could load into memory and EXEC. Next I decided to let it be called from DEF USR so I could pass in parameters and return a status code like A=USR0(-1). Next next I decided I wanted it to still work with EXEC so the user could use it either way–just use defaults with EXEC, or customize things using USR.
Then I ran into a snag…
USRx(n) or EXEC?
If the USR routine ONLY expected a number parameter, the code to handle both USR and EXEC seems easy. When calling a routine with EXEC, the D register will be zero (it seems). If it wasn’t zero, I could then do the JSR INTCNV call which would process the parameter in BASIC and put it in the D register.
; Show if routine is being called with USRx(n) or EXEC
ORGADDR equ $3e00 ; Where program loads in memory.
; Absolute addresses of ROM calls. CHROUT equ $A002 INTCNV equ $B3ED GIVABF equ $B4F4
org ORGADDR
; This code expects to have been called by USRx(x) or EXEC xxxx. start cmpd #0 ; called from EXEC? beq fromexec ; if yes, goto fromexec fromusr jsr INTCNV ; else, get USR number parameter in D pshs d ; save D leax usrmsg,pcr ; display "called from USR" message bsr print puls d ; restore D addd #1 ; add one to D jmp GIVABF ; return back to USR call.
; PRINT subroutine. Prints the 0-terminated string pointed to by X plus CR. print lda ,x+ beq printdone jsr [CHROUT] bra print printdone lda #13 jsr [CHROUT] rts
usrmsg fcc "CALLED FROM USR" fcb 0
execmsg fcc "CALLED FROM EXEC" fcb 0
end
When the routine starts, it checks to see what D is set to. If 0, it assumes it was called from EXEC and jumps to code that just prints “FROM EXEC” then ends.
If not 0, it assumes it was called from USR and the code calls the ROM INTCVT routine to parse the parameter and place it in D, then it prints “FROM USR”, increments D (just so we can verify it passed something back), and returns it back to BASIC.
Here it is in operation:
And all was right in the world… Until I tried just using EXEC by itself. After using it first with the address (“EXEC &H3E00”) BASIC will remembers that address so when you just type “EXEC” next it uses the previous address:
EXEC &H3E00 FROM EXEC
EXEC ?TM ERROR
Making the user always have to provide the EXEC address each time is not optimal. My solution was clearly not a solution.
But wait! There’s more…
I also learned about Sean Conner documenting how USR can also take a string parameter instead of just a number. If you are interested in USR, be sure to check out that link. He also has a cool 6809 compiler (“a09”) I just started playing with. It has some unique features not available in other compilers I have tried.
USRx(n) or USRx(“STRING”)
With this new knowledge, I had an idea to make my USR routine also be able to take a string for a special configuration function. I could let the user specify the four characters that will move the cursor by doing something like A=USR0(“udlr”). But, if you pass in a string and it calls INTCNV, that routine will check the parameter type and, if not a number, return with a ?TM ERROR (type mismatch).
This required me to learn how to tell whether USR was being called with a number or a string.
Under Extended Color BASIC (the original Color BASIC did things differently, see Sean’s page for details), the ROM code sets up some registers when calling the USR function. Sean documented these in his excellent blog post on USR. Basically, register A would be 0 if the USR parameter was a number, or 255 if it was a string. If it was a string, register X would have the address of the string descriptor (the location in memory that VARPTR returns) and register B would be the length of the string.
That is really convenient. Now you can have code that detects if it is being called from USR with a number or a string. My test code looked like this:
; Show if USR is being called with a number or a string.
ORGADDR equ $3e00 ; Where program loads in memory.
; Absolute addresses of ROM calls. CHROUT equ $A002 INTCNV equ $B3ED GIVABF equ $B4F4
org ORGADDR
; This code expects to have been called by USRx(x) or USRx("STRING") start tsta ; A=0 is USR(0), A=255 is USR("...") bne usrstring ; if not 0, goto usrstring usrnumber pshs d,x ; save D and X leax numbermsg,pcr ; display "number" message bsr print puls d,x ; restore D and X jsr INTCNV ; else, get USR number parameter in D addd #1 ; add one to D jmp GIVABF ; return back to USR call.
usrstring leax stringmsg,pcr ; display "string" message bsr print ldd #123 ; load D with return value jmp GIVABF ; return back to USR call.
; PRINT subroutine. Prints the 0-terminated string pointed to by X plus CR. print lda ,x+ beq printdone jsr [CHROUT] bra print printdone lda #13 jsr [CHROUT] rts
stringmsg fcc "STRING" fcb 0
numbermsg fcc "NUMBER" fcb 0
end
And here it is in operation:
Now I know how to detect a USRx(number) versus EXEC, and how to detect a USRx(number) versus a USRx(string). But, this has the same problem if called by EXEC with no address:
EXEC &3E00 NUMBER
EXEC NUMBER ?TM ERROR
It appears that using EXEC with the address after it sets registers up differently than using EXEC with no address (where it uses the last address EXEC used). While both end up at the code path for USRx(number), is seems that plain EXEC thinks it is returning an invalid type and the ?TM ERROR is displayed.
EXEC or EXEC xxxx or USRx(n) or USRx(“STRING”)
Can both routines be combined? On the CoCo mailing list, this all started when I asked: Is there a way to tell if a routine was called from USR versus EXEC? It was Sean’s reply that got me going down this rabbit hole:
Maybe.
Address $9D contains the address EXEC uses to jump to your code, so that should be called address. Also, X will also be this address (implementation detail).
For Color BASIC, you need to know you are running under Color BASIC. Address $112 is the address for USR, so this should point to your code. Also, upon calling, X should be equal to $AA2F and B should be 6 (both are implementation details).
For Extended Color BASIC, you need to know you are running under Extended Color BASIC (16 bits at $8000 are $4558). Addresses $013E through $0150 contain the USRn addresses, so one of these 10 addresses should point to your code. Also, A will equal the contents of address $06. If A=0, then X=$4F; if A=255, then X is pointing elsewhere (the string descriptor).
For Disk Extended Color BASIC, you need to know you are running under Disk Extended BASIC (16 bits at $C000 are $444B). The USRn addresses are now $095F through $0971, but other than that, it’s the same as Extended Color BASIC.
Based on all that, I think the best method might be (completely untested):
mycode cmpx #mycode beq called_by_exec ; otherwise, assume called by USR/USRn
Good luck.
-spc
– Sean Conner via the CoCo Mailing List
This gave me a lot of think about. I did some tests to see what register X looked like when being called by EXEC with or without an address, as well as looking at what was stored in the $9D memory location which is the address EXEC (with no address after it) will use. I created a simple program that would print the value of the X register and the value of $9D so I could test it and see what the pattern was. This code uses an undocumented ROM call that will print the value of the D register. (I learned about this call from Sean’s pages.)
ORGADDR equ $3e00 ; Where program loads in memory.
; Absolute addresses of items in RAM variables. EXECJP equ $9d location of jump address for EXEC
; Absolute addresses of ROM calls. REGDOUT EQU $BDCC ; Convert the value in ACCD into a decimal ; number and send it to CONSOLE OUT.
org ORGADDR
start tfr x,d ; X=D jsr REGDOUT lda #32 ; space jsr [CHROUT] ldd EXECJP ; load D with EXEC address jsr REGDOUT rts
end
Now I could load this into memory, set up a DEFUSR0=&H3E00 and do some tests:
15872 ($3E00) is the start of my user program. EXEC with that address will have both X and the $9D memory location containing that value.
EXEC without an address will have 43947 ($ABAB) in X, and 15872 ($3E00) as the address of the last EXEC address specified. But what is $ABAB? Looking at the Color BASIC Unravelled book, that address is where the EXEC token is:
ABAB FDB EXEC
I did not dive into this, but I expect X was is used for the token scanning and since that was the last thing it found (no address after it to parse) that is what was in the register when it jumps to the user code.
When I tested A=USR0(0), I got a 79 in register X, and $9D still had the last EXEC address used. It then errored out with a ?TM ERROR due to this code not setting up a clean return back to a USR call.
And lastly, A=USR0(“STRING”) put 425 in register X, and $9D was still the last EXEC address used.
Now, had I done the USR calls first, that $9D would not be set up yet and it would look like this:
46154 ($B44A) appears to be the default value EXEC will use. By default, EXEC points to the routine that prints ?FC ERROR:
B44A FDB LB44A ARGUMENT OF EXEC COMMAND - SET TO ‘FC’ ERROR
So on a power cycle, typing EXEC is the same as typing EXEC &HB44A:
EXEC &HB44A ?FC ERROR
Having this value there is not useful for any of my checks since all that means is that the user has not done an EXEC with an address yet.
BUT, now that I see what happens with register X, I should be able to check it, and the $9D exec location and determine if I am being called by EXEC, EXEC xxxx, or a USRx command. Here is my test program:
ORGADDR equ $3e00 ; Where program loads in memory.
; Absolute addresses of items in RAM variables. EXECJP equ $9d location of jump address for EXEC
; Absolute addresses of ROM calls. CHROUT equ $A002
org ORGADDR
; This code expects to have been called by USRx(x). start cmpx #start ; called by "EXEC xxxx"? beq fromexec ; if yes, goto fromexec cmpx #$abab ; called by "EXEC"? bne fromusr ; if no, must be USR. goto fromusr ldx EXECJP ; get EXEC address cmpx #start ; called by "EXEC xxxx"? beq fromexec ; if yes, goto from exec fromusr leax usrmsg,pcr lbsr print rts fromexec leax execmsg,pcr lbsr print rts
; PRINT subroutine. Prints the 0-terminated string pointed to by X plus CR. print lda ,x+ beq printdone jsr [CHROUT] bra print printdone lda #13 jsr [CHROUT] rts
usrmsg fcc "FROM USR" fcb 0
execmsg fcc "FROM EXEC" fcb 0
end
And here is what it does:
I now have code that can properly (?) detect if it was called from EXEC xxxx, EXEC, or USR. This demo does not handle detecting a string parameter to USR, but … I think it proves it is possible to do it.
With a few more lines of assembly, I came up with this test program:
ORGADDR equ $3e00 ; Where program loads in memory.
; Absolute addresses of items in RAM variables. EXECJP equ $9d location of jump address for EXEC
; Absolute addresses of ROM calls. CHROUT equ $A002 INTCNV equ $B3ED GIVABF equ $B4F4
org ORGADDR
; This code can be called by USRx(n), USRx("STRING"), EXEC addr or EXEC. start cmpx #start ; called by "EXEC xxxx"? beq fromexec ; if yes, goto fromexec cmpx #$abab ; called by "EXEC"? bne fromusr ; if no, must be USR. goto fromusr ldx EXECJP ; get EXEC address cmpx #start ; called by "EXEC"? beq fromexec ; if yes, goto from exec fromusr tsta ; A=0? beq donumber ; if yes, number passed in. goto donumber. inca ; inc A so if 255 (string) it will be 0 now. beq dostring ; if A=0 (was 255), string. goto dostring. bra unknown ; else, goto unknown (this should never happen).
donumber leax numbermsg,pcr ; show "number" message bsr print jsr INTCNV ; get number that was passed in addd #1 ; add 1 to D jmp GIVABF ; return new number back to BASIC
dostring leax stringmsg,pcr ; show "string" message bsr print ldd #12345 ; load D with a return value jmp GIVABF ; return that number back to BASIC
unknown leax unknownmsg,pcr ; this should never happen lbsr print ; show "unknown" message rts
; PRINT subroutine. Prints the 0-terminated string pointed to by X plus CR. print lda ,x+ beq printdone jsr [CHROUT] bra print printdone lda #13 jsr [CHROUT] rts
execmsg fcc "FROM EXEC" fcb 0
numbermsg fcc "FROM USR(NUMBER)" fcb 0
stringmsg fcc "FROM USR(STRING)" fcb 0
unknownmsg fcc "UNKNOWN" fcb 0
end
And here is what I get after loading this into memory:
DEF USR0=&H3E00 OK
A=USR0(42) FROM USR(NUMBER) PRINT A 43
A=USR0("STRING") FROM USR(STRING) PRINT A 12345
EXEC &H3E00 FROM EXEC
EXEC FROM EXEC
I think we may have a winner! The important parts are:
start cmpx #start ; called by "EXEC xxxx"? beq fromexec ; if yes, goto fromexec cmpx #$abab ; called by "EXEC"? bne fromusr ; if no, must be USR. goto fromusr ldx EXECJP ; get EXEC address cmpx #start ; called by "EXEC"? beq fromexec ; if yes, goto from exec
If X is the address of the user program, it was called by “EXEC xxx”
If not, then if X is NOT $ABAB, it was called by USR
Else, it was $ABAB, so the EXECJP ($9D) is checked to see if it is the address of the user program. If it is, it is from EXEC.
I hope that makes sense. If not, think of it like this:
X=program start – it was called from EXEC xxxx
X=$ABAB and EXECJP=program start – it was called by EXEC.
Anything else is USR.
Now what I need from you is to double check my work and tell me if I got this right, and if this method can be relied on.
Over the years I have shared many tidbits about Color BASIC.
This is another one.
A recent post by Juan Castro to the Groups.IO Color Computer mailing list caught my attention, mostly because he called me out by name in the subject line:
As a reminder, Color BASIC allows 1 or 2 character variable names. They must start with a letter (A-Z) and the second character can be either letter (A-Z) or number (A-0). BUT, the BASIC interpreter does let you type longer names for variables, but it only honors the first two characters. Here is a screenshot from a past blog post here (which I’d link to if I was not so lazy):
Color BASIC variables may be very long, but only the first two characters are used.
This is a reminder that, if you try to use variables longer than two characters, you have to make sure you always keep the first two characters unique since “LONGVARIABLE” and “LOST” and “LO” are all the same variable to BASIC.
…but not all variable name limits are the same.
To break the rule I just said, in Color BASIC, some variable names are forbidden. A forbidden variable is one you cannot use because it is already reserved for a keyword or token. For example, FOR is a keyword:roar
FOR I=1 TO 10 PRINT I NEXT I
Because of this, even though BASIC only honors the first two characters of a variable name, you still cannot use “FOR” as a variable.
FOR=42 ?SN ERROR
But you can use “FO”, since that is not long enough to be recognized as a BASIC token or keyword.
FO=42 PRINT FO 42
There are a number of two-character tokens, such as “TO” in the FOR/NEXT statement (“FOR I=1 TO 10”), and “AS” in the Disk BASIC FIELD statement (“FIELD #1,5 AS A$”), as well as “FN” which is used in DEF FN.
AS=42 ?SN ERROR
FN=42 ?SN ERROR
TO=42 ?SN ERROR
This means if you wrote something for Color BASIC or Extended Color BASIC that uses “AS” as a variable, that would not work under Disk Extended Color BASIC.
BASIC ignores spaces
In recent years, someone pointed me to the fact that when scanning a BASIC line (either type in directly or when parsing a line of code in a program), spaces get ignored by the scanner. This means:
N M = 42 PRINT N M 42
That one surprised me when I learned it. This is probably why, when printing two variables, a semicolon is required between them:
N = 10 M = 20 PRINT N;M 10 20
And if you had done this (remember to CLEAR between these tests so variables are erased each time):
N = 10 M = 20 NM = 30 PRINT N M 30 PRINT N;M;N M 10 20 30
By the way, if you have ever wondered about that space printed in front of numeric variables when you do things like “PRINT X”, I covered why this happens in an earlier blog and included a simple patch to BASIC that removes that feature.
How to turn a forbidden variable into a non-forbidden one for fun and profit
Well, Juan Casto showed that using this “BASIC ignores spaces” quirk as a way to use forbidden variables. From his post:
Now it seems obvious. BASIC’s interpreter looks for keywords like “FOR” and will not recognize “F O R” or “FO R” as that token. The detokenizer honors the spaces.
But when it comes to variables, the spaces are ignored by the parser, so “T O” will not match as a token for “TO”, but will be processed as a variable “TO”.
Go figure.
Admittedly, space in two-character variable names look silly, but now I can finally update my old *ALLRAM* BBS to use the variable “TO$” for who a message is to:
FR$="ALLEN HUFFMAN" T O$="JUAN CASTRO" SB$="THAT'S REALLY COOL"
I suspect this was discovered by the early pioneers of BASIC, likely soon after the original Color Computer was released in 1980. If you know of a reference to this behavior from some early newsletter or magazine article, please let me know.
And as to Juan … thanks for sending me down a BASIC rabbit hole again…
Just when I thought I was out, they pull me back in.
In part 3 I showed a simple assembly language routine that would uppercase a string.
In part 5, this routine was made more better by contributions from commenters.
Today, I revisit this code and update it to use “what I now know” (thank you, Sean Conner) about being able to pass strings into a USR function without using VARPTR.
First, here is the code from part 5:
* UCASE.ASM v1.01 * by Allen C. Huffman of Sub-Etha Software * www.subethasoftware.com / alsplace@pobox.com * * 1.01 a bit smaller per Simon Jonassen * * DEFUSRx() uppercase output function * * INPUT: VARPTR of a string * RETURNS: # chars processed * * EXAMPLE: * CLEAR 200,&H3F00 * DEFUSR0=&H3F00 * A$="Print this in uppercase." * PRINT A$ * A=USR0(VARPTR(A$)) * ORGADDR EQU $3f00
opt 6809 * 6809 instructions only opt cd * cycle counting
org ORGADDR
start jsr INTCNV * get passed in value in D tfr d,x * move value (varptr) to X ldy 2,x * load string addr to Y beq null * exit if strlen is 0 ldb ,x * load string len to B ldx #0 * clear X (count of chars conv)
loop lda ,y+ * get next char, inc Y ; lda ,y * load char in A cmpa #'a * compare to lowercase A blt nextch * if less, no conv needed cmpa #'z * compare to lowercase Z bgt nextch * if greater, no conv needed lcase suba #32 * subtract 32 to make uppercase leax 1,x * inc count of chars converted nextch jsr [CHROUT] * call ROM output character routine ; leay 1,y * increment Y pointer cont decb * decrement counter bne loop * not done yet ; beq exit * if 0, go to exit ; bra loop * go to loop
exit tfr x,d * move chars conv count to D jmp GIVABF * return to caller
null ldd #-1 * load -2 as error return jmp GIVABF * return to caller
In the header comment you can see an example of the usage, and that it involved using VARPTR on a string to get the string’s descriptor location in memory, then pass that address into the USR function.
Now that I know we can just pass a string in directly, I thought it would be fun (?) to update this old code to use that method. Here is what I came up with. Note that I changed the “*” comments to “;” since the a09 assembly does not support those. If you wanted to run this in EDTASM, you would have to change those back.
; UCASE3.ASM v1.02 ; by Allen C. Huffman of Sub-Etha Software ; www.subethasoftware.com / alsplace@pobox.com ; ; 1.01 a bit smaller per Simon Jonassen ; 1.02 converted to allow passing a string in to USR ; ; DEFUSRx() uppercase output function ; ; INPUT: string ; RETURNS: # chars converted or -1 if error ; ; EXAMPLE: ; CLEAR 200,&H3F00 ; DEFUSR0=&H3F00 ; A$="Print this in uppercase." ; PRINT A$ ; A=USR0(A$) ; PRINT "CHARS CONVERTED:";A ; A=USR0("This is another test"); ; PRINT "CHARS CONVERTED:";A ; ORGADDR EQU $3f00
start jsr CHKSTR ; ?TM ERROR if not a string. ; X will be VARPTR, B will be string length tstb beq reterror ; exit if strlen is 0 ldy 2,x ; load string addr to Y ldx #0 ; clear X (count of chars conv)
loop lda ,y+ ; get next char, inc Y cmpa #'a ; compare to lowercase A blo nextch ; if less, no conv needed cmpa #'z ; compare to lowercase Z bhi nextch ; if greater, no conv needed suba #32 ; subtract 32 to make uppercase leax 1,x ; inc count of chars converted nextch jsr [CHROUT] ; call ROM output character routine decb ; decrement counter bne loop ; not done yet
tfr x,d ; move chars conv count to D bra return
reterror ldd #-1 ; load -1 as error return jmp GIVABF ; return to caller
Here are the changes… In the original version, I have this:
start jsr INTCNV * get passed in value in D tfr d,x * move value (varptr) to X ldy 2,x * load string addr to Y beq null * exit if strlen is 0 ldb ,x * load string len to B ldx #0 * clear X (count of chars conv)
That first jsr INTCNV expects a number parameter and, if not a number, it exits with ?TM ERROR. If it gets past that, the number is in the D register and it gets transferred over to X. In this case, the number is the value returned by VARPTR:
A=USR0(VARPTR(A$))
That value is the address of the 5-byte string descriptor that contains the address of the actual string data and the length of that data. Y is loaded with 2 bytes in from wherever X points which makes Y contain the address of the string data.
After this is a bug, I think. Looking at the comments, I think that “beq null” should be one line lower, like this:
start jsr INTCNV * get passed in value in D tfr d,x * move value (varptr) to X ldy 2,x * load string addr to Y ldb ,x * load string len to B beq null * exit if strlen is 0 ldx #0 * clear X (count of chars conv)
That way, Y is loaded with the address of the string data, then b is loaded with the length of that data, and the branch-if-equal check is now checking B. If the length is 0, it is an empty string so no processing can be done on it. (That’s a bug, right?)
The new code is this:
start jsr CHKSTR ; ?TM ERROR if not a string. ; X will be VARPTR, B will be string length tstb beq reterror ; exit if strlen is 0 ldy 2,x ; load string addr to Y ldx #0 ; clear X (count of chars conv)
The first line is something I learned from Sean Conner‘s excellent writeup on USR. That is an undocumented ROM call which checks is a variable is a string. If it isn’t, it will return back to BASIC with a ?TM ERROR. By having that check there, if the user tries to pass in a number, that error will be seen. As a bonus, if you try to EXEC that code, that, too, will show ?TM ERROR.
After that, B should be the length of the string so tstb checks that to be 0 (empty string) then the rest of the code is similar.
As I write this, I could have altered the order of my new code to do the tstb/beq after the ldy and then it would be closer to how the original worked. But since the original appears buggy, I won’t worry about that.
Now if I load this and set it up, I should see this:
DEF USR0=&H3F00
A=USR0(42) ?TM ERROR
A=USR0("This is a test") THIS IS A TEST
Also, I notice that the value I return can be -1 if you pass in an empty string…
A=USR0("") OK PRINT A -1
…and if it is non-empty, it is only the count of the characters that had to be converted. So “Hello World” converts the “ello” and “orld” for a return value of 8. It does not touch the uppercase “H” and “W” or the space.
I am not sure that is really useful. The code could be modified to return the length of the string it processed, but at least this way you know that a positive non-zero return value means it did do some work.
Since this is a symmetrical pattern, if we can figure out how to draw one quadrant, we can draw the others.
The pattern is 19 characters wide, which contains a center column of asterisks, and a left and right column that are spaces except for the center row of asterisks.
“As if they had planned it,” this means the pattern in each quadrants is 8 characters, matching the number of bits in a byte.
I typed it up to figure out what the bit pattern would be. (Actually, I typed up a bit of it, then pasted that into Copilot and had it tell me the bit pattern.)
That’s a mess, but in the left the “.” would represent the blank space down the left side up to the row of 19 asterisks. After that is the 8-bit pattern with “-” representing a space in the pattern (0 bit) and the “*” representing the asterisk (1 bit).
This let me quickly cobble together a proof-of-concept:
1 READ V 2 A$=STRING$(19,32):MID$(A$,10,1)="*" 3 FOR B=0 TO 7 4 IF V AND 2^B THEN MID$(A$,9-B,1)="*":MID$(A$,B+11,1)="*" 5 NEXT 6 PRINT A$:A$(L)=A$ 7 L=L+1:IF L<8 THEN 1 8 PRINT STRING$(18,42) 9 FOR B=7 TO 0 STEP -1:PRINT A$(B):NEXT 10 DATA 2,81,48,114,9,4,162,81
Line 10 are the 8 rows of byte data for a quadrant of the snowflake.
Line 1 reads the first value from the DATA statement.
Line 2 builds a string of 19 spaces, then sets the character at position 10 (in the center) to an asterisk. Every row has this character set.
Line 3 begins a loop representing each bit in the byte (0-7).
Line 4 checks the read DATA value and ANDs it with the bit value (2 to the power of the the FOR/NEXT loop value). If it is set, the appropriate position in the left side of the string is set to an asterisk, and then the same is done for the right side. To mirror, the left side is center-minus-bit, and the right side is center-plus-bit.
Line 5 is the NEXT to continue doing the rest of the bits.
Line 6 prints the completed string, then stores that string in an A$() array. L has not been used yet so it starts at 0.
Line 7 increments L, and as long as it is still ess than 8 (0-7 for the first eight lines of the pattern) it goes back to line 1 to continue with the next DATA statement.
Line 8 once 8 lines have been done, the center row of 19 asterisks is printed.
Line 9 is a loop to print out the A$() lines we saved, backwards. As they were built in line 6, they went from 0 to 7. Now we print them backwards 7 to 0.
…and there we have a simple way to make this pattern, slowly:
Logiker 2025 pattern on a CoCo.
On a CoCo 3, adding a WIDTH 40 or WIDTH 80 before it would show the full pattern:
Logiker 2025 pattern on a CoCo 3.
My example program can be made much smaller by packing lines together and removing unnecessary spaces. One minor optimization I already did was doing the bits from 0 to 7 which removed the need to use “STEP -1” if counting backwards. Beyond that, this is the raw proof-of-concept idea of using bytes.
Other options folks have used in past challenges included rune-length type encoding (DATA showing how many spaces, then how many asterisks, to make the pattern) so that probably is worth investigating to see if it helps here.
Then, of course, someone will probably figure out a math pattern to make this snowflake.
Hat tip to Erico Monteiro for sending me down another quick benchmarking rabbit hole…
NOTE: This technique will work poorly for ASCII TEXT characters, since the PEEK value is not the same as the PRINT CHR$ value for some characters. It works fine with the graphics blocks (128-255). See the example at the end.
In general, I expect PEEK to be faster than looking up a variable. PEEK only has to process whatever is in the parentheses:
V=PEEK(1024)
Parsing the decimal 1024 can be slow. Using hex is faster (&H400). Using a variable can be even faster (unless there are a ton of variables BASIC has to scan to before finding the target one):
V=PEEK(L)
Erico just showed me technique using an array to store all the characters on the CoCo’s 32 column screen. PRINT@ can be used to put characters on the screen quickly, and when you want to PRINT@ the character somewhere else, you can PRINT@ whatever character used to be there by taking it from the array.
I expected PEEK would be faster than accessing elements of an array so I did a test where I looped through 512 characters using PEEK versus an array:
0' peek-vs-array.bas
10 TIMER=0:FORA=1024 TO 1536 20 Z=PEEK(A) 30 NEXT:PRINT TIMER
40 DIMB(511):TIMER=0:FOR A=0 TO 511 50 Z=B(A) 60 NEXT:PRINT TIMER
At line 10, I loop through all the locations of the 32×16 screen. One by one, Z is set to the value of that location. The value of the loop (1024-1536) matches the POKE/PEEK memory location of the screen.
At line 40, I have an array B() that would be loaded with all the bytes in the screen. The elements of the array (0-511) match the PRINT@ location of the screen.
My results:
123 127
Very close, though the array access is slightly slower. I confirmed PEEK is indeed faster … in this example.
Let’s pretend the loop is a “background” screen and we would PEEK and POKE to restore it versus PRINT@ from the array. (In this example, I am just getting what is there and printing that same thing there again, just for timing.)
0' peek-vs-array2.bas
10 TIMER=0:FOR A=1024 TO 1536 20 Z=PEEK(A):POKEA,Z 30 NEXT:PRINT TIMER
40 DIMB(511):TIMER=0:FOR A=0 TO 511 50 Z=B(A):PRINT@A,CHR$(Z); 60 NEXT:PRINT TIMER
And my results:
210 258
PEEK is faster here, too.
But I have now seen “real code” using this to put a CHR$() player graphic on the screen, and erase it as it moved across the screen (restoring the background as it goes) and the array was faster.
This is another example of why benchmarking a specific item is not always useful. For example, if using PRINT to put things on the screen, you are using the 0-511 location which matches the 0-511 array. If using PEEK, you have one location that is the display screen and another that would be the “saved” background screen. Each time you want to update something, you have to take the location (offset) and add it to the background (to get that one) and the foreground (to get that one). That doubling of math could make it slower versus PRINT@ using 0-511 and CHR$(B(x)) using the same 0-511.
So while PEEK is faster by itself, if you do that and need more math to use it, ARRAYs could win.
50 DIM B(511):TIMER=0:FOR A=0 TO 511 60 Z=B(A):PRINT@A,CHR$(Z); 70 NEXT:PRINT TIMER
The first test loops 0-511 then uses “math” to calculate the background memory location, and more “math” to calculate the foreground memory location. Twice the math, twice the slowness.
The second does not match because the array and PRINT@ location.
333 259
Array is faster than two maths.
But, we can cut that math in half but have the FOR loop go through screen memory, then only add to that to get the background. &H3E00 (15872) background minus &H400 (1024) foreground is &H3A00 (14848). I am using hex because it is quicker for BASIC to parse an &H value than a decimal value.
