From:	"Andrew Davidson" <andrew AT lemure DOT freeserve DOT co DOT uk>
Newsgroups:	comp.os.msdos.djgpp
Subject:	How does DJGPP store its structs?
Date:	Wed, 31 Mar 1999 04:00:17 +0100
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To:	djgpp AT delorie DOT com
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Reply-To:	djgpp AT delorie DOT com

I need to create a rather complex structure but have, so far, been unable to
achieve this.

The structure looks like this:

typedef struct {
    unsigned char callbyte;
    signed long offset;
    unsigned char retbyte;

    // various other longs/shorts/chars etc
} scratch;

This probably needs a little explaining. Callbyte (should be located at byte
0 of a scratch structure) contains the value 0xe8 (near 32bit call with
offset - in machine code) and retbyte contains 0xc3 (return from near call
in machine code). 'offset' is then written to within the C code forming a
small section of self modifying code. This code is called by passing a
pointer to the struct to a simple 1 line asm() 'call'. My problem is that,
although I would expect callbyte, offset, and retbyte to occupy the first 6
bytes of memory, they appear to be each occupying 4 bytes for a total of 12
bytes of memory.

The structure is used in the following way within my code...

typedef struct {
    scratch *scratchmem;
    char *workmem;

    // various other longs/shorts.chars etc
} mainstruct;

mainstruct *mystruct;

I malloc memory for mystruct, scratchmem, and workmem, and I set all
appropriate variables.
I've tested the structs and all stuff done from within C code works fine.
Is this 12 byte thing the result of the way I'm trying to use the scratch
struct (from within another struct) or just a limitation of djgpp? I have a
feeling that djgpp might be trying to optimise the struct  so is there any
way I can dissable this optimisation while keeping all the other forms of
optimisation available?

Andrew

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