From: ryot AT bigfoot DOT com (George Ryot)
Newsgroups: comp.os.msdos.djgpp
Subject: Re: From Bytes to Int and Char
Message-ID: <37ba82f5.7514501@news.clara.net>
References: <rfXs3.4$bZ1 DOT 1603 AT typhoon01 DOT swbell DOT net> <37B466D7 DOT 958F09E5 AT americasm01 DOT nt DOT com> <37B4DFD1 DOT 4D66 AT surfsouth DOT com> <37B57EB0 DOT 492A93AE AT unb DOT ca>
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Date: Sat, 14 Aug 1999 14:54:48 GMT
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To: djgpp AT delorie DOT com
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Reply-To: djgpp AT delorie DOT com

Endlisnis <s257m AT unb DOT ca> wrote:

> Chris Holmes wrote:
> 
> > > > 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);
> >
> >   (sorry... have to say it) They aren't endians, they're native
> > arrangements.  Please, be PC with your PC.  <grin>
> 
>     I don't know what you are talking about.   The word endian means the native
> order of bytes in a multibyte contiguous variable type.
> 
> >   One solution (kinda tricky because you need to make sure they are
> > stored on a 2 byte boundary or hack around my trick, which is a hack
> > itself).
> >   Given: void *buffer;
> >   make: char *c_buffer=(char *)buffer;
> >   make: short *i_buffer=(short *)buffer;
> >   then: short i = i_buffer[BYTE_OFFSET / 2];
> >   (this should be a little more endian safe)
> 
>     I don't see how, because it's doing the exact same thing in a more obsure way.
> I was trying to say that if the file was stored on a big-endian machine, then using
> a little-endian like DJGPP to read the file in the way I mentioned wouldn't work.

Me thinks some programmers just get stuck in hack mode determined to
produce unsafe code for the sake of a few microseconds. ;-))

How would Nick extract the byte pairs he needs using that last hack? 

The following (boring?) solution should do the trick:

    unsigned short sz, tz;
    sz = (file_buffer[5]  << 8) & file_buffer[6];
    tz = (file_buffer[10] << 8) & file_buffer[11];

Note that I have assumed you really do mean 6th, 7th, 11th & 12th
bytes (file_buffer[0] would be the first byte) and that the 16-bit
integers are stored high byte low byte.  I have also assumed that the
values are unsigned (always positive), things might get complicated
otherwise.  You can use int instead of short if you prefer.
-- 
george