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