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From: steve@umigw.MIAMI.EDU (steve emmerson)
Newsgroups: comp.lang.c
Subject: Re: typeof isn't enough to define swap correctly
Message-ID: <109@umigw.MIAMI.EDU>
Date: Fri, 4-Sep-87 19:22:39 EDT
Article-I.D.: umigw.109
Posted: Fri Sep  4 19:22:39 1987
Date-Received: Sat, 5-Sep-87 21:33:15 EDT
References: <557@rocky.STANFORD.EDU> <1880@sol.ARPA>
Reply-To: steve@umigw.UUCP (steve emmerson)
Organization: University of Miami
Lines: 23
Summary: close but no cigar

In article <1880@sol.ARPA> crowl@cs.rochester.edu (Lawrence Crowl) writes:
>The following swap macro is correct.  It relies on the pointer features of C
>to evaluate its arguments exactly once.
>
>#define swap(a,b) { \
>    typeof(a) t,*p1,*p2; /* temporary and pointers to sources */ \
>    p1 = &(a); p2 = &(b); /* evaluate arguments exactly once */ \
>    t = *p1; *p1 = *p2; *p2 = t; /* do actual swap */ \
>}
>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*.

Close but no cigar.
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