From: "Damian Yerrick" <die DOT spammers AT pineight DOT 8m DOT com>
Newsgroups: comp.os.msdos.djgpp
Subject: Re: From Bytes to Int and Char
Date: Sun, 15 Aug 1999 20:36:53 -0500
Organization: Rose-Hulman Institute of Technology
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Campbell, Rolf [SKY:1U32:EXCH] <cp1v45 AT americasm01 DOT nt DOT com> wrote in message
news:37B466D7 DOT 958F09E5 AT americasm01 DOT nt DOT com...
> Nick wrote:
>
> > I am trying to load some data from a files, the 6 and 7th bytes are an
> > Integer and so are the 11th and 12th.  How do I make them into an INT?
This
> > is what I used earlier
> >
> > char ver, type;
> > int sz, tz;
> >
> > sz = file_buffer[6] + file_buffer[7];
> > is that right?
>
>     No.  Here's the right way to do it.  This code is not edian safe.
>
> sz = *(short*)(file_buffer+6);
>
> --
>      -Rolf Campbell (39)3-6318
>

/* That still isn't byte-order safe. The byte-order-safe way to do it is by
shifting each individual byte. The following code works on all platforms,
but make sure that file_buffer points to _unsigned_ char data or you'll get
nasty sign extension that screws this up. */

/* if file was written in big endian (Motorola-order) format */
sz = ((short)file_buffer[6] << 8) + file_buffer[7];

/* if file was written in little endian (Intel-order) format */
sz = ((short)file_buffer[7] << 8) + file_buffer[6];

/* Oh, by the way, the same kind of trick can be used to read 4-byte ints
(on 16-bit platforms such as real mode DOS, change int to long) */

/* if file was written in big endian (Motorola-order) format */
score = ((int)file_buffer[16] << 24) + ((int)file_buffer[17] << 16) +
((int)file_buffer[18] << 8) + file_buffer[19];

/* if file was written in little endian (Intel-order) format */
score = ((int)file_buffer[19] << 24) + ((int)file_buffer[18] << 16) +
((int)file_buffer[17] << 8) + file_buffer[16];

This should clear things up.

Damian Yerrick (comp DOT os DOT msdos DOT djgpp AT pineight DOT 8m DOT com)
Download DOSArena from http://come.to/yerrick