From: j DOT aldrich6 AT genie DOT com
Message-Id: <199605091349.AA117349769@relay1.geis.com>
Date: Thu,  9 May 96 13:41:00 UTC 0000
To: djgpp AT delorie DOT com
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Subject: Re:  more sizeof questions

Reply to message 5683427    from KAGEL AT DG1 DOT BLO on 05/09/96  8:31AM


>Not quite.  Actually, each field does NOT need word-alignment.  Only objects
>whose size is >= 4 bytes (long, int, float, long long & double [did I miss
>any?])

long double  :)

>require word (4byte) address alignment.  The member TotalMemory is not
>padded as both it and the char array Info require 2byte alignment.  The extra
>two bytes are added at the end of the structure to insure that each element of
>an array of _VGAInfoBlock's will each have their first member, VESASignature,
>aligned.

That doesn't make sense.  The last time I checked, 256 was divisible by 4, so
the struct should be word-aligned without any padding, right?  Adding 2 bytes
would make the size 258, which is NOT aligned.

BTW, there's something I meant to mention about the original post...

   >typedef unsigned char 		BYTE;
   >typedef unsigned short 	WORD;
   >typedef unsigned long		DWORD;

This is not a good way to define WORD and DWORD - the whole reason
that int is different sizes on different machines/compilers is that they use a
different word size.  On a 16-bit compiler, a word is 16 bits, and on a 32-bit
compiler, a word is 32 bits.  According to the ANSI C spec, there is no
guarantee that any given type will be any given number of bytes, only that
they guarantee at least a certain minimum range of values.  Heck, ANSI
C even allows for the possibility of using one's complement negatives
instead of two's complement, which would blow away any possibility
of cross-compatibility between two such programs.  Not that the above
code won't work on 99.9% of C compilers, but it's a little misleading.

John