Re: A83: 83P format verification
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Re: A83: 83P format verification
In a message dated 04-03-99 7:49:32 PM Eastern Standard Time, AtmaZ@aol.com
writes:
> OK, I think I got an idea of how the 83P file is stored, but I need somebody
> else to look on it.
>
> Right now I'm going to assume than every character in the ANSI/ASCII set is
> represented by one byte. (I think...somebody want to proof this?)
>
> Bytes 1-8: Always "**TI83**".
> Bytes 9-10: In DIAMONDS.83P (my example) they are ASCII values 26 and 10,
> respectively. I'm not too sure what they represent; I think they can be
> ignored.
> Bytes 11-50: Comment field. I think.
> Bytes 51-60: ??? In my editor, they were ASCII values 46 (".") , 95 ("_"),
> 85
> ("U"), 169, 8, 11, 0, 154, 8, and 6. I have no idea if this represents
> anything significant.
> Bytes 61-68: Name of the program.
> Bytes 69-end: Program data.
>
> Actually, all I'm trying to get to is the program data, but I need to make
> sure of this so I don't access the wrong stuff.
>
> Thanks in advance,
>
You should probably work with an easier program, like:
xor A
Or a small BASIC program even.
Bytes $0B to $3A are the comment
Bytes $3C to second to last byte is the program
The remaining bytes ($2B and the second to last byte) are one more than normal
if protected (there seems to be no other pattern)
I figured this out in a few seconds by making a very small basic program and
hex-editing it.
-David