After migrating from Perforce to Git (BitBucket, in our case) at work, I learned that Git does not support any embedded source code keywords for replacement on check in. In our case, we use things like:
On Submit, Perforce will replace those keywords with useful information, such as the username ($Author$) that did the submit, the data and time of the submit ($DateTime$) and the filename ($File$). I find this very useful when looking at source code outside of Perforce, since it tells me how new or old the code is. (Anyone who’s ever had to print out a ton of code for a group code review knows how easy it is to end up looking at the wrong version of the file… Usually not discovered until someone finds a bug you know you already fixed ;-)
Since Git does not support this, I wanted to at least search/replace “$File$” to be the actual filename of the source file. I am sure there are many ways to do this, but I ended up using a PowerShell script, based on code I found some web searches (probably on Stack Exchange or similar):
In this case, this code specifically targets “.h” files at or below the directory you run the script in. I expect you can make a multi-filter that does .c and .h at one time, but I only needed to do this once so I ran it like this, then edited the “*.h” to be “*.c” and ran it again. (You’d change it to whatever your source file extensions are, like .cs or whatever.)
The [system.io.file] tip came from someone who noticed the other output would always add a blank line at the end of the file. This method re-writes the file as-is.
WARNING: I did notice that some files get messed up if they contain special characters. It would put some garbage (when viewing in a text editor) in place of things like weird apostrophes and such, so if you use this, make sure to diff your files before checking them back in and revert any goofs that “-replace” causes. I had to revert about one dozen blocks in my code.
I also had to run a command to grant my Windows 11 machine permission to even execute a PowerShell script.
Hope this helps someone else, and saves them a few hours of research…
The new InstaX3 was announced on 9/8/2022, and made instantly available on Amazon. It shipped with beta firmware, but had a 1.0.00 update available to install during activation.
If you have found any bugs, please leave a comment with the version and details and I will add them to this list. As workarounds are discovered, I will update this list.
As a new version of firmware is released, these bugs will be re-tested. When they work for some, and not for others, a note will be added to that effect.
Initially, the camera shipped with pre-1.0.00 beta software. It would prompt to upgrade to 1.0.00 on activation from the app.
2022-9-9 – v1.0.00
TBA – various crashes, settings being changed, etc.
2022-09-19 – v1.0.04
TBA
Open WiFi – a poorly implemented WiFi system has the camera broadcast itself as a WiFi hotspot to anyone within range, and allows users that know the default WiFi password all X3 cameras have to access and download any files on the memory card from a web browser… or worse. (Suggested by commenter, yt)
Raw notes… will be cleaned up and made purty with more details, soon.
X3: up to 30 seconds between taking photos via app, and more timing notes.
Some notes on timing, for those who want to compare against your existing camera. This is with the current firmware that the camera will install when you activate it (v1.0.00). Recommended Sandisk Extreme 32GB card.
App (on iPhone 13 Pro):
360 Photo, 72MP, 2:1 – there is nearly 4-5 second delay between the time you press the on screen button and the time the X3 clicks. It takes a total of about 15 seconds before the UI updates and you can take the next photo.
360 Photo, 18MP, 2:1 – 4-5 delay, and a total of about 9 seconds.
360 HDR Photo – 3-4, then about 13 seconds total.
150 Photo, 36MP, 16:9 – 3-4, about 11 seconds total.
150 Photo, 9MP, 16:9 – 3-4, about 8 seconds total.
I did have one instance where it took almost 30 seconds to be ready for the next shot.
Camera Button:
Using the button on the camera is almost instant (within a second) and ready for the next photo in about 6-7 seconds total.
In 360 Photo mode, 72MP, pressing the button on the camera makes the click sound between 2-3 seconds later, and it takes a total of 14 seconds before the screen comes back on for the next photo.
In 360 Photo mode, 18MBP, it takes about a second to take the picture, and a total of about 5-6 before you can take the next shot.
In 360 HDR Photo mode, 18MP, it takes about a second to take the picture, and a total of 9-10 seconds before you can take the next shot.
Some of this feels like the timer is on, which isn’t being shown in the app or on the camera. earlier, I used the Quick button and had selected that mode. It seems it may be remembering settings that have since been turned off.
This is just a (hopefully) short article following up on some tangents I went on while learning about Color BASIC string memory and VARTPR and other related (and nonrelated) items.
The story so far…
Color BASIC has reserved string memory that is located at the very top of the 32K that is accessible by BASIC. There are three 16-bit pointers maintained by BASIC that contain the starting address for this memory, the address where the next string will be inserted (sorta), and the end of this memory.
The Color BASIC Unraveled book called them FRETOP, STRTAB and MEMSIZ.
You can get the address in BASIC by using PEEK(x)*256+PEEK(y) with these memory locations:
FRETOP – 33 and 34
STRTAB – 35 and 36
MEMSIZ – 39 and 40
When digging in to this, I noticed the the MEMSIZ location returned on a 32K CoCo was 32766, and on a 16K it was 16382, and on a 4K CoCo it was 4094. The actual last byte of RAM was one byte higher (32767, 16383, and 4095), making me wonder if this was a mistake.
Indeed, when I checked on a CoCo 3, which had newer patches to BASIC done by Microware, the 32K value “correctly” reported 32767 as would have expected.
… That missing byte is needed so VAL can work on a string stored at the very top of string space. It needs to temporarily put a NUL byte after the string so the number parser knows when to stop. After executing, VAL restores the byte to its original value. Things would fail if the byte after the string data was ROM though. On the Coco3, that byte is in RAM so it didn’t need an extra reserved byte. I suspect, however, that Microware didn’t know that, or thought it was a bug in the original, or just didn’t notice when they replaced the now useless memory scan with a simple LDX. So, it’s not a bug in the original ROM and, accidental or not, … the behavior on the Coco3 [is] also correct.
– William Astle, 7/14/2022
This led me back to the ROM disassembly, where I did indeed locate where the VAL command will load a byte, store 0 there, then restore the original value later.
Color BASIC Unraveled, page B38.
Prove it or lose it.
Since Man cannot code on faith alone, Art “ADOS” Flexser provided a simple way to prove this to be true:
You can demonstrate the need for the extra byte on a 32/64K CoCo 1/2 by the following:
Change the top of string space to $7FFF rather than the usual $7FFE by POKE 40, 255:CLEAR, upping the LSB of MEMSIZ by one.
Then try PRINT VAL(“&H5”) and you get the rather surprising answer of 94! Why 94? Because that’s decimal for hex 5E, and the “E” has arisen from the first byte of Extended Basic at $8000.
On a CoCo 3, the top of string space is already at $7FFF, so you just need to put it into ROM/RAM mode with POKE&HFFDE,0 before entering PRINT VAL(“&H5”) to get the answer of 94.
Art Flexser, 7/17/2022
And there you have it — the reason why MEMSIZE is not actually pointing to the end of usable BASIC memory on a CoCo 1 or 2, and why it still works on a CoCo 3 after being changed to actually point to the end of usable BASIC memory.
The more you know …
Thanks, William and Art. Your knowledge is greatly apprecaited.
From the “wish I knew then what I (almost) know now” department..
On the Color Computer, the first 1K (1024 bytes) of RAM is reserved for the BASIC ROM code to use. In this article, we will look at some memory locations that pertain to our BASIC code, variables, arrays and strings.
Here are the locations as described in the Color BASIC Unraveled book:
Color BASIC Unraveled, page A1.
Each location contains a 16-bit number which is the address in memory of that information. Here are the values we are interested in, translated to decimal:
TXTTAB – 25 and 26
VARTAB – 27 and 28
ARYTAB – 29 and 30
ARYEND – 31 and 32
FRETOP – 33 and 34
STRTAB – 35 and 36
MEMSIZ – 39 and 40
To turn two bytes in to the 16-bit address, we take the PEEK of the first byte, multiply it by 256, then add the PEEK of the second byte. For TXTTAB (Beginning of BASIC program) we would do:
PRINT PEEK(25)*256+PEEK(26)
That value returned will be different depending on the configuration of the Color Computer.
Let’s dig in to what each of these locations means.
TXTTAB: Beginning of BASIC program – 25/26 (&H19/&H1A)
This location represents where in RAM the BASIC program starts.
PRINT PEEK(25)*256+PEEK(26)
On a CoCo with only Color BASIC (no Extended or Disk BASIC), storage for a BASIC program will be right after the 512 bytes of screen memory. The CoCo’s screen is stored at 1024 to 1535, so BASIC programs load at 1536.
START END SIZE DESC
----- ----- ----- --------------
0 1023 1024 ROM USE
1024 1535 512 TEXT SCREEN
1536 32767 31232 BASIC PROGRAM
With Extended BASIC added, some RAM after the text screen is used for high resolution graphics screens. By default, 6K is reserved on startup, but more or less can be specified using the PCEAR command. Since the smallest graphics screen using 1.5K (1536 bytes), allocation is done in multiples of that size. Thus, on startup, Extended BASIC has four 1.5K pages reserved for graphics. This means BASIC would start at 7668 on a CoCo with Extended BASIC.
START END SIZE DESC
----- ----- ----- --------------
0 1023 1024 ROM USE
1024 1535 512 TEXT SCREEN
1536 7679 6144 HI-RES GFX
7680 32767 25088 BASIC PROGRAM
With Disk BASIC, there is an additional 2K of RAM used just after screen memory at 1536, with the high resolution graphics screens after that.
START END SIZE DESC
----- ----- ----- --------------
0 1023 1024 ROM USE
1024 1535 512 TEXT SCREEN
1536 3583 2048 DISK ROM USE
3584 9727 6144 HI-RES GFX
9728 32767 23040 BASIC PROGRAM
NOTE about “START”: As I calculated this, I see that the starting location for BASIC is actually one byte higher than I expected. On a 4K CoCo, I expected BASIC to start at 1536, but PEEK(25)*256+PEEK(27) shows 1537. On an Extended BASIC CoCo, those peeks show 7681 instead of 7680. And on a Disk BASIC system, they show 9729 instead of 9728. I am not sure why it is off by one, but I’m sure someone smarter than I will have the answer in a comment.
NOTE about “END”: You will notice in my table the end location for BASIC is listed as 32766. That would only be true if the machine has 32K or 64K. If the machine had only 4K or 16K, it would be less. (Basically, MEMSIZE – START OF BASIC PROGRAM). And then it’s also off by by one byte for some reason (4094 on a 4K machine, 16382 on a 16K machine, and 32766 on a 32K machine). I’ll have to look in to this. Maybe I’m the one who is off…
NOTE about BASIC Memory: You may note I just said a 32K and 64K CoCo will show the same location for the end of BASIC memory. This is because reasons. See this article for a discussion about getting the most memory for BASIC, or just read this excerpt:
64K NOTE: The reason BASIC memory is the same for 32K and 64K is due to legacy designs. The 6809 processor can only address 16-bits of memory space (64K). The BASIC ROMs started in memory at $8000 (32768, the 32K halfway mark). This allowed the first 32K to be RAM for programs, and the upper 32K was for BASIC ROM, Extended BASIC ROM, Disk BASIC ROM and Program Pak ROMs. Early CoCo hackers figured out how to piggy-pack 32K RAM chips to get 64K RAM in a CoCo, but by default that RAM was “hidden” under the ROM address space. In assembly language, you could map out the ROMs and access the full 64K of RAM. But, since a BASIC program needed the BASIC ROMs, only the first 32K was available.
Note the use of parenthesis around each calculation. I was originally not using them, and was getting bad results because I made, as William Astle noted, “an elementary arithmetic error.”
Math is hard. Just ask Barbie.
Variable Table
The variable table is a series of 7 byte entries. The first two bytes are the variable name.
Variable Name (bytes 1 and 2) – If a variable is “LN”, the first two bytes would be “LN”. But, if the variable is a string such as LN$, the first byte would be “L” and the second byte would be the value of “L” + 128 (high bit set).
Variable Value (bytes 3-7) or String Descriptor (bytes 3, and 5-6)
If the variable is a number, the next five bytes are a floating point representation of the value.
If the variable is a string (high bit of the second name byte is set), the five bytes will be a string descriptor entry that contains the size and location of the string data bytes. A string descriptor only uses three of those five bytes. The first byte will be the length of the string (0-255), the next byte is unused, then the third and fourth bytes are the address of the string data in memory (located in reserved string space, to be discussed later in this article).
Here is a short program that will print out all the variables. The variables MUST be declared before this code runs. If any new variables are created inside the FOR/NEXT loop, it will not work as expected since the table will have been changed.
0 ' SHOWVARS.BAS
5 DIM VS,VE,L,VN$
10 VS=PEEK(27)*256+PEEK(28)
20 VE=PEEK(29)*256+PEEK(30)
30 FOR L=VS TO VE-5 STEP 7
40 VN$=CHR$(PEEK(L))+CHR$(PEEK(L+1) AND 127)
50 IF PEEK(L+1) AND 128 THEN VN$=VN$+"$"
60 PRINT VN$;
70 FOR I=0 TO 4
80 PRINT TAB(5*I+6);PEEK(L+2+I);
90 NEXT:PRINT
100 NEXT
Running this will show a list of all the variables in use, and the five bytes after the name.
Above, the VN$ entry shows that it is a 3 byte string located in memory at location 127*256+236 (32748). More on this later in the article.
You can test this code further by declaring more variables before line 10, such as with the DIM statement, or just adding declarations like A=42 or ZZ$=”TOP”.
6 DIM A,B,C,D,A$,B$,AB$,ZZ$
7 AB$="ABBA":ZZ$="TOP"
ARYTAB: Start of Arrays – 29/30 (&H1D/&H1E)
This location represents where arrays will be stored.
PRINT PEEK(29)*256+PEEK(30)
And if we know the start…
ARYEND: End of Arrays (+1) – 31/32 (&H&1F/&H20)
And this location represents where arrays end (though it returns one byte past the last entry of the arrays).
PRINT PEEK(31)*256+PEEK(32)
Arrays
This part gets a bit complicated and messy, so unless you plan on doing stuff with arrays, feel free to skip this section completely… :)
Arrays get stored in their own block of memory. Each array entry starts with a 5-byte header that contains the name of the array (2-bytes), the length of the array (2-bytes) and how many dimensions the array is (1-byte). It looks like this:
ARRAY TABLE ENTRY HEADER (5 bytes)
--------------------
[ ][ ][ ][ ][ ] - numeric array "NM"
N M AB CD #D - ABCD = memory used (header to end)
#D = number of dimension
DIM(X)=1 DIM(X,Y)=2
This is followed by one or more 2-byte entries that contain the dimension size for each dimension. A one dimensional array such as DIM X(10) would have two bytes representing 11. Remember that in BASIC, DIMensions are base 0. DIM X(10) gives you X(0) through X(10) for a total of 11 entries.
Since an array has to have at least one entry, the header will really always have at least 7 bytes. Here is an example for a two dimension array like DIM X(10,20)
ARRAY TABLE ENTRY HEADER (5 bytes + at 2 bytes per each dimension)
------------------------------------
[ ][ ][ ][ ][ ][ ][ ][ ][ ]- numeric array "LN"
N M AB CD #D D1 D1 D2 D2 - ABCD = memory used
- #D = number of dimensions
- D1D1 = number of dimension 1 entries
- D2D2 = number of dimension 1 entries
And, dimensions are backwards from how they are entered in BASIC. If you had this:
DIM AB(1,2,3,4)
You would have an array table entry that would be 15 bytes and look like this:
[65][66] [xx][xx] [4] [0][5] [0][4] [0][3] [0][2] [0][1]
[65][66] - name "AB"
[xx][xx] - total bytes used from header to end
[4] - four dimensions
[0][5] - DIM (x,x,x,4)
[0][4] - DIM (x,x,3,x)
[0][3] - DIM (x,2,x,x)
[0][2] - DIM (1,x,x,x)
After each header are a series of 5-byte entries (one for each element in the array) in the same format as variables – either 5-bytes representing a floating point number, or 5-bytes that make up the string descriptor (size and location of string data).
The blocks of entries are in the order listed in the array table. i.e, if you have DIM(1,2) you will first have 3 5-byte entries for the second dimension DIM(x,2) followed by 2 5-byte entries for the first dimension DIM(1,x).
I think.
Here is a program that shows the Array Table:
0 ' SHOWARRAYS.BAS
5 DIM SA,EA,L,AN$,I,AL
6 DIM X(1),Y(10),XY(2,3),ZZ(10,15,20),A$(10),UL$(10,3)
10 SA=PEEK(29)*256+PEEK(30)
20 EA=PEEK(31)*256+PEEK(32)
30 IF SA=EA THEN PRINT "NO ARRAYS":END
40 PRINT "ARRAYS FROM";SA;"TO";EA
45 PRINT "NAME TSIZE DIM .. .. .. .."
50 L=SA
60 AN$=CHR$(PEEK(L))+CHR$(PEEK(L+1) AND 127)
70 IF PEEK(L+1) AND 128 THEN AN$=AN$+"$"
80 PRINT AN$;
90 AL=PEEK(L+2)*256+PEEK(L+3)
100 PRINT TAB(4);AL;
110 PRINT TAB(11);PEEK(L+4);
120 FOR I=0 TO PEEK(L+4)-1
130 PRINT TAB(14+I*4);PEEK(L+5+I*2)*256+PEEK(L+5+I*2+1);
140 NEXT:PRINT
150 L=L+AL:IF L<EA THEN 60
And, with a bit of work, a program could be written to dump the 5-byte entrees for each array. I’ll add that to my “TODO” list…
FRETOP: Start of String Storage – 33/34 (&H21/&H22)
This location represents where reserved string memory begins. String space is reserved using the CLEAR command, with 200 bytes allocated by default. Strings are stored at the top of memory. Thus, on startup, this should print 200 bytes less than the end of BASIC memory:
PRINT PEEK(33)*256+PEEK(34)
If you change the amount of reserved string space, this number will change accordingly. Here we start out with 200 bytes reserved for strings, then change that to 0 bytes, then to 500 bytes:
If you wanted to know how much room is available for a BASIC program plus variables, you could take this value and subtract the start of BASIC location:
However, there is still some extra overhead so this value won’t match with what PRINT MEM shows. I guess it’s better to just PRINT MEM:
Maybe we’ll figure out what those “missing” 17 bytes are being used for.
STRTAB: Start of String Variables – 35/36 (&H22/&H24)
This location represents where the next string will be stored. Sorta.
PRINT PEEK(35)*256+PEEK(36)
Didn’t we just cover this? Not exactly. FRETOP shows where the string memory starts, but strings are actually stored from the top of memory, and grow downward. Meaning, if you have 200 bytes reserved, and place a ten byte string there (A$=”1234567890″), it will be located at the END of that 200 bytes, not at the start. This value points to where strings are stored within the reserved area.
If string memory were set to 16 bytes by using CLEAR 16, string memory would look like this:
At this point, we could add another one byte string, but anything longer would result in an ?OS ERROR (out of string space).
Here is a test program (and notice we are adding +”” to the end of each string. If that isn’t done, the string will be stored inside the BASIC program itself, and not in string memory.)
MEMSIZE should never move, as it is the highest memory location available to BASIC. FRETOP will move, depending on how many bytes are reserved for strings using the CLEAR command.
After the two strings in the previous example were added, it would look like this:
With these memory locations, we can determine how large a BASIC program is. We can detect how much string space is reserved, and calculate how much is being used or is free. We can figure out how much variable storage is being used, as well as array storage.
Most of these locations are never meant to be set by a program, but there is one exception. The CLEAR command can also be used to prevent BASIC from growing past a certain memory location. If you use a second parameter such as CLEAR 200,16000, it changes the MEMSIZ to that value.
The CLEAR command erases all variables, so there would be nothing for BASIC to relocate. Instead, this just changes MEMSIZE and adjusts the string values FRETOP and STRTOP and accordingly.
While it would be possible to manually POKE those six values and accomplish the same thing, there could always be unintended consequences. (But for fun, a BASIC program could be written that moves the string memory somewhere else, then POKEs the values to update it so BASIC keeps running.)
Conclusion
There might be some interesting things one could do by monitoring string space closely… Perhaps all the research I did for this article may be leading somewhere…
Until then…
BONUS CONTENT
Here is the program I used to calculate the tables in this article.
The C programming language has a few standard library functions that deal with strings, namely strcpy() (string copy) and strcat() (string concatenate).
Microsoft BASIC has similar string manipulation features built in to the language. For example, to copy an 8 character “string” in to a buffer (array of chars):
In BASIC, you do not need to allocate space for individual strings. Color BASIC allows doing whatever you want with a string provided it is 255 characters or less, and provided the total string space is large enough. By default, Color BASIC reserves 200 bytes for string storage. If you wanted to strcpy() “12345678” to a string variable, you would just do:
BUFFER1$="12345678"
Yes, that’s legal, but Color BASIC only recognizes the first two characters of a string name, so in reality, that is just like doing:
BU$="12345678"
If you need more than the default 200 bytes, the CLEAR command will reserve more string storage. For example, “CLEAR 500” or “CLEAR 10000”.
“CLEAR 500” would let you have five 100 character strings, or 500 one character strings.
And, keep in mind, strings stored in PROGRAM MEMORY do not use this space. For example, if you reduced string space to only 9 bytes, then tried to make a 10 byte string direct from BASIC:
CLEAR 9
A$="1234567890"
?OS ERROR
The wonderful “?OS ERROR” (Out of String Space).
BUT, if strings are declared inside PROGRAM data, BASIC references them from within your program instead of string memory:
…you would get a 48 character string (8 characters, added 6 times).
In C, using strcat(buffer, buffer) with the same buffer has a doubling effect each time, just like A$=A$+A$ does in BASIC each time.
And, adding a bunch of strings together like…
A$=A$+A$+A$+A$+A$+A$+A$+A$ '64 characters
…could also be done in three doubling steps like this:
A$=A$+A$:A$=A$+A$:A$=A$+A$ ' 64 characters
Two different ways to concatenate strings together to make a longer string. Which one should we use?
Must … Benchmark …
In my code, I just added my 8 character A$ up 8 times to make a 64 character string. Then I started thinking about it. And here we go…
Using my standard benchmark program, we will declare an 8 character string then add it together 8 times to make a 64 character string. Over and over and over, and time the results.
0 REM strcat1.BAS
5 DIM TE,TM,B,A,TT
10 FORA=0TO3:TIMER=0:TM=TIMER
20 FORB=0TO1000
30 A$="12345678"
40 A$=A$+A$+A$+A$+A$+A$+A$+A$
70 NEXT
80 TE=TIMER-TM:PRINTA,TE
90 TT=TT+TE:NEXT:PRINTTT/A:END
This produces an average of 1235.
Now we switch to the doubling three times approach:
0 REM strcat2.BAS
5 DIM TE,TM,B,A,TT
10 FORA=0TO3:TIMER=0:TM=TIMER
20 FORB=0TO1000
30 A$="12345678"
40 A$=A$+A$:A$=A$+A$:A$=A$+A$
70 NEXT
80 TE=TIMER-TM:PRINTA,TE
90 TT=TT+TE:NEXT:PRINTTT/A:END
This drops the time down to 888!
Adding separately three times versus add together eight times is a significant speed improvement.
Based on what I learned when exploring string theory (and being shocked when I realized how MID$, LEFT$ and RIGHT$ worked), I believe every time you do a string add, there is a new string created:
“A$=A$+A$+A$+A$+A$+A$+A$+A$” creates eight strings along the way.
Earlier this week, Insta360 introduced a new camera – the X3. The X3 is the latest 360 action camera, and builds upon the feature set of the ONE X2 which came before it (and the ONE X before it, and so on).
Insta360 X3
One of the features that was brought over from the ONE X2 is the unprotected WiFi hotspot that is active any time the camera is powered up. This allows anyone within WiFi range the ability to connect to the camera and browse (or download) any photos or videos that are stored on it:
In addition to this, telnet is still enabled, with the root account having no password. Just telnet to that IP address and log in as root to have full access to the camera’s file system:
telnet 192.168.42.1
It is disappointing to see this unsecure access method continued in the next generation of their camera, but the company has posted that they are aware of this and are working on it.
Earlier in December, @BASIC10Liners on Twitter retweeted a tweet from Tony Lyon. The original tweet contained a photograph of a ZX81 BASIC program listing found in ‘Computer & Video Games’ magazine. The game ran on a 1K machine, and was a Space Invaders-style game, featuring a single invader.
This tweet was picked up and shared in the Tandy CoCo group on Facebook, and led to a number of folks porting it over to the CoCo. Jim Gerrie also ported it to the MC-10:
Jim also did the research to find the issue that the full article appeared in:
For the past few weeks, folks have been sharing optimizations and game improvements on Facebook, and I wanted to highlight one in particular.
Erico Monteiro is known for doing some impressive graphics work using the 8-color 64×32 resolution “mode” of the CoCo. He took the basic framework of the game, then sped it up and added new colors, animations and features.
David Healey, if you are out there, thank you for inspiring so many with your 1981 ZX81 program in that magazine…
While writing up my article on Color BASIC memory, I ran across something I was unaware of. Special thanks to William “Lost Wizard” Astle for his responses on the CoCo mailing list that got me pointed in the right direction…
BASIC memory looks like this:
+-------------+ 0
| SYSTEM USE |
+-------------+ 1024
| TEXT SCREEN |
+-------------+ 1536
| DISK USE |
+-------------+
| HI-RES GFX |
+-------------+
| BASIC PROG |
+-------------+
| VARIABLES |
+-------------+
| ARRAYS |
+-------------+
| FREE MEMORY |
+-------------+
| STRINGS |
+-------------+ 32767
Without Disk BASIC, “DISK USE” is not there, and without Extended BASIC, neither is “HI-RES GFX”.
To determine free memory available to BASIC (for program, variables or arrays), you would subtract the end of “ARRAYS” from the start of “STRINGS.” The locations that store this are at &H1F-&H20 (31-32) and &H21-&H22 (33-34).
Color BASIC Unraveled, page A1.
Each of those two byte locations holds a 16-bit address to some area of RAM. You can get the values by taking the first byte, multiplying it by 256, and adding the second byte.
PRINT "MEMSIZE:";PEEK(&H27)*256+PEEK(&H28)
Here is a program that prints out some pertinent information:
0 ' BASINFO2.BAS
10 ' START OF BASIC PROGRAM
20 BS=PEEK(25)*256+PEEK(26)
30 ' START OF VARIABLES
40 VS=PEEK(27)*256+PEEK(28)
50 ' START OF ARRAYS
60 SA=PEEK(29)*256+PEEK(30)
70 ' END OF ARRAYS (+1)
80 EA=PEEK(31)*256+PEEK(32)
90 ' START OF STRING STORAGE
100 SS=PEEK(33)*256+PEEK(34)
110 ' START OF STRING VARIABLES
120 SV=PEEK(35)*256+PEEK(36)
130 ' TOP OF STRING SPACE/MEMSIZ
140 ST=PEEK(39)*256+PEEK(40)
150 PRINT "PROG SIZE";(VS-BS),"STR SPACE";(ST-SS)
170 PRINT "ARRAY SIZE";(EA-SA)," STR USED";(ST-SV)
180 PRINT " VARS SIZE";(SA-VS)," FREE MEM";(SS-EA)
999 END
And here is that PRINT code, without using any variables. It could be made a subroutine that you could GOSUB to and see status of your memory usage:
There are some bytes missing somewhere. This made me wonder which was correct, so I consulted the PRINT MEM code in Color BASIC Unravelled. It had comments that shed some light on what is going on:
Color BASIC Unraveled, page B34.
“This is not a true indicator of free memory because BASIC requires a STKBUF size buffer for the stack for which MEM does not allow.”
– Color BASIC Unraveled, page B34.
On page A1, I see the definition for STKBUF:
STKBUF EQU 58 STACK BUFFER ROOM
I see code in the ROM that takes this in to consideration, adding the value of ARYEND plus the value of STKBUF.
Color BASIC Unraveled, page B20.
That routine is used to test if enough memory is available for something, and if there isn’t, it returns the ?OM ERROR.
But the difference I get is 12 bytes, not 58. What it BOTSTK? It looks like it is just before the pointer variables I have been working with. I just did not know what it was.
0017 BOTSTK RMB 2 BOTTOM OF STACK AT LAST CHECK
&H17 would be memory location 23. Let’s see where it is by doing…
PRINT PEEK(23)*256+PEEK(24)
That gives me a location about 30 bytes before FRETOP (start of string storage). Here is the program I am using:
0 ' BASPTRS.BAS
10 ' BOTTOM OF STACK
20 SP=PEEK(23)*256+PEEK(24)
30 ' START OF BASIC PROGRAM
40 BS=PEEK(25)*256+PEEK(26)
50 ' START OF VARIABLES
60 VS=PEEK(27)*256+PEEK(28)
70 ' START OF ARRAYS
80 SA=PEEK(29)*256+PEEK(30)
90 ' END OF ARRAYS (+1)
100 EA=PEEK(31)*256+PEEK(32)
110 ' START OF STRING STORAGE
120 SS=PEEK(33)*256+PEEK(34)
130 ' START OF STRING VARIABLES
140 SV=PEEK(35)*256+PEEK(36)
150 ' TOP OF STRING SPACE/MEMSIZ
160 ST=PEEK(39)*256+PEEK(40)
180 PRINT "START OF PROG",BS;(VS-BS)
190 PRINT "START OF VARS",VS;(SA-VS)
200 PRINT "START OF ARRAYS",SA;(EA-SA)
210 PRINT "END OF ARRAYS+1",EA
215 PRINT "BOTTOM OF STACK",SP
229 PRINT "START/STR STORE",SS;(ST-SS)
230 PRINT "START/STR VARS",SV;(ST-SV)
240 PRINT "TOP OF STRINGS",ST
250 PRINT "FREE MEMORY",(SP-EA)
999 END
William clarified a bit of this in a follow-up post.
Actually, currently available memory would be approximately the difference between BOTSTK and ARYEND less the STKBUF amount. MEM neglects to add the STKBUF adjustment in when it calculates the difference. Total usable memory under current PMODE/FILES/CLEAR settings would be the difference between BOTSTK and TXTTAB.
Note that the stack (and consequently the BOTSTK value) will grow downward from FOR loops, GOSUB, and expression evaluation. Expression evaluation can use a surprising amount of stack space depending on how many times it has to save state to evaluate a higher precedence operation, how many function calls are present, etc.
ARYEND is the top of all the memory used by the program itself, PMODE graphics pages, disk buffers and file descriptors, scalar variables, and arrays.
When calculating OM errors, it takes ARYEND, adds the amount of memory requested, adds the STKBUF amount, and then compares that with the current stack pointer. It does the comparison by storing S since you can’t directly compare two registers.
STKBUF is 58 for whatever reason. That’s sufficient to allow for a full interrupt frame (12 bytes) plus a buffer so routines can use the stack for saving registers, routine calls, etc., within reason, without having to continually do OM checks. It does this to prevent corrupting variables when memory is tight. Even so, there may be a few routines in Disk Basic that may still cause the stack to overflow into variable space when memory is very tight.
– William Astle, 7/8/2022
So how can we test? Using the XRoar emulator, I started with a standard 64K CoCo with Disk BASIC. PRINT MEM returns 22832 bytes free.
On startup, a disk-based CoCo has 22823 bytes available for BASIC.
For testing, I’ll create one line of BASIC that is just a REM statement:
10 REM
Now PRINT MEM shows 22817 bytes free. That is 6 bytes less, which is two bytes for where the next line will be (38 7), two bytes for the line number (00 10), the token for REM (130), and a NULL byte (0). Each new line number takes up 5 bytes plus whatever the content of the line is.
Now I want to reserve most of that memory for strings. I’ll do CLEAR 22800. After this, PRINT MEM shows 207… That can’t be right. Shouldn’t it really be 17?
In this case, the original PRINT MEM was already counting 200 bytes reserved for strings. When I did CLEAR 22800, it was release that 200 bytes, then redoing it as 22800 (so, 228600+200).
Let’s start over…
Okay, so let’s redo this. I restarted the CoCo back to where PRINT MEM shows 22823.
The next thing I did was type CLEAR 0 to reserve NO string space, then check PRINT MEM. It shows 23023 — the largest BASIC program I can have on this disk-based machine, unless I deallocate some graphics memory (6K is reserved for that, by default).
Now I enter in the “10 REM” line, and PRINT MEM again. It shows 23017 — 6 bytes less, as expected.
I try to type CLEAR 23017, which gives ?OM ERROR.
I back that off a bit to CLEAR 22960. PRINT MEM shows ?OM ERROR. And now I have hosed BASIC. Even CLEAR 0 now returns ?OM ERROR.
Is this a bug? I can still EDIT my line 10, and LIST it, but while CLEAR by itself works, it changes nothing. I’ve managed to consume so much memory, I can’t run the command I need to give it back.
Let’s start over again…
CoCo reset. CLEAR 0 entered. PRINT MEM shows 22823. 10 REM typed. PRINT MEM shows 23017.
This time I’ll try CLEAR 22950 to give BASIC a bit more room. PRINT MEM shows 67.
If this is accurate, I should be able to add 67 characters to the end of that REM statement.
I type “EDIT 10”, then “X” to extend to the end. I enter ten characters, then ENTER.
10 REM1234567890
PRINT MEM gives me an ?OM ERROR.
Now, the program still LISTS and would run (if it did anything), but clearly other things are broken.
I EDIT 10 and try to add another ten characters.
10 REM12345678901234567890
?OM ERROR.
And now, LINE 10 is completely gone! EDIT must try to “add” a line when it’s done, and if there’s no memory, the line is … erased?
PRINT MEM still shows 67, even though the “10 REM” seems to be gone. Where did that memory go? I try adding a total of 15 characters after the REM:
10 REM123456789012345
That works. PRINT MEM shows ?OM ERROR.
Let’s start over again again.
I reboot the CoCo, then CLEAR 22950. PRINT MEM shows 73 (there is no line number yet).
This “should” mean I could enter a line (5 bytes) that has 73-5=68 characters. But, if I tried to type anything longer than 16, I get ?OM ERROR (REM token is one byte, then 15 characters).
This tells me that when PRINT MEM shows me 67, I really only had 5+16=21 bytes available. 67-21=46 bytes difference.
Looking back at my BASIC POINTERS program output:
…I see the calculated free memory was 22042, versus PRINT MEM showing 22038. That is only four bytes different.
Questions keep stacking up
It is clear you cannot use PRINT MEM to know how much room you will have for a BASIC program. It is more of a guideline, within 50 or so bytes.
The wildcard is probably this mysterious (to me) “stack” that BASIC needs for operation. Depending on what is going on, the stack in-use may be larger or smaller, but the max limit is 58 bytes. And, STKBOT notes that is is “bottom of stack at last check” so it may not be reflecting the actual stack at the particular moment when the code looked at it.
I really don’t have any conclusions from this. When I started writing, I expected I’d find a set of PEEKs I could subtract to get a more accurate PRINT MEM value.
