From: info AT hoekstra-uitgeverij DOT nl (Richard Bos)
Newsgroups: comp.lang.c,comp.os.msdos.djgpp
Subject: Re: hex calculation routine!?
Date: Mon, 04 Sep 2000 11:15:31 GMT
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Damian Yerrick <Bullcr_pd_yerrick AT hotmail DOT comRemoveBullcr_p> wrote:

> Posted and emailed to ^Hawk^ <Hawk AT DarkSoft DOT Net>:
> 
> >Has anybody a routine to add/subtract etc. hex-values for me? Thnx in
> >advance! =;)
> >
> >ie. 800F0F0F+8F0=??? or 800F0F0F-8F0=???.
> 
> /* assuming unsigned long is a 32-bit or larger data type */

It always is in ISO C; ULONG_MAX is at least 2**32-1.

> unsigned long foo = 0x800f0f0f
> unsigned long bar = 0x8f0
> 
> printf("%lx - %lx = %lx\n", foo, bar, foo - bar);
> 
> IIRC, %x prints an int as hexadecimal, and %lx prints a long as
> hexadecimal.  Crossposted to comp.lang.c for further comment on this
> portable language issue.

Yup. Note also that in C, you have hex constants (in the source), but no
hex _values_. You have integer values; by the time printf() or any other
function gets to deal with them, representation is irrelevant. You could
just as easily do this:

  printf("%ld - %ld = %lx\n", foo, bar, foo - bar);
  printf("%lx - %lx = %ld\n", foo, bar, foo - bar);

(%ld specifies long decimal output)

or any other combination of decimal or hex output. Ditto for values
originally specified as decimal constants.

Richard