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From: mpl@sfsup.UUCP (M.P.Lindner)
Newsgroups: comp.lang.c
Subject: Re: typeof isn't enough to define swap correctly
Message-ID: <2019@sfsup.UUCP>
Date: Wed, 9-Sep-87 20:07:52 EDT
Article-I.D.: sfsup.2019
Posted: Wed Sep  9 20:07:52 1987
Date-Received: Sat, 12-Sep-87 00:36:51 EDT
References: <557@rocky.STANFORD.EDU> <1880@sol.ARPA> <109@umigw.MIAMI.EDU>
Organization: AT&T-IS, Summit N.J. USA
Lines: 26
Summary: typeof isn't necessary

In article <109@umigw.MIAMI.EDU>, steve@umigw.MIAMI.EDU (steve emmerson) writes:
> In article <1880@sol.ARPA> crowl@cs.rochester.edu (Lawrence Crowl) writes:
 ...
> >#define swap(a,b) { \
 ...
> >}
> >One problem still remains.  What if typeof(a) != typeof(b)?
> 
> It also won't work with pure array objects (i.e. swapping one array with
> another).  A generic routine should be able to work with *anything*.

Here's one that works WITHOUT TYPEOF AND WORKS FOR ARRAYS!

#define swap(a, b) { \
	register int	i; \
	for (i = 0; i < sizeof(a); i++) { \
		((char *) &a)[i] ^= ((char *) &b)[i]; \
		((char *) &b)[i] ^= ((char *) &a)[i]; \
		((char *) &a)[i] ^= ((char *) &b)[i]; \
	} \
}

OK, so it's not as pretty or efficient as everyone would like, but it's
completely general purpose and doesn't need any new features.

Mike Lindner