Path: utzoo!utgpu!water!watmath!clyde!cbosgd!ihnp4!chinet!dag
From: dag@chinet.UUCP (Daniel A. Glasser)
Newsgroups: comp.lang.c
Subject: Re: swap macro -- beware -- oh yeah ?
Message-ID: <2154@chinet.UUCP>
Date: 29 Jan 88 06:21:41 GMT
References: <8801200422.AA19420@decwrl.dec.com> <3061@killer.UUCP> <1182@ark.cs.vu.nl>
Reply-To: dag@chinet.UUCP (Daniel A. Glasser)
Organization: Chinet - Public Access Unix
Lines: 30

In article <3061@killer.UUCP> jfh@killer.UUCP (John Haugh) writes:
[a ^= b; b ^= a; a ^= b;]
>What if, A and B are aliases?
What do you mean by aliases?  Are they both the same storage or just
the same value?
>
>	let's say, A = B = 7.
okay, they are similar values but different storage.
>	after a ^= b, both a and b == 0.
Nope, a == 0, b still == 7
>	then, after both b ^= a and a ^= b (the second time), a and b
>	are STILL == 0.
No again, after b ^= a, a == 0 and b ==7, and now after a ^= b,
a == 7 and b == 7.  I think you are being too quick to attack.

This dosn't hold up, of course if &a == &b, but depending on how
your compiler handles constant expressions, and assuming you don't
use pointers, a macro like the following:
#define swap(a,b)	if(&(a)!=&(b)){(a)^=(b);(b)^=(a);(a)^=b}
might evaluate to the same code as without the "if", since &a and
&b will usually be constant expressions (even if they are autos or
paramters, there relative addresses will be known).  As I said,
it depends on your compiler.

-- 
Nobody at the place where I work	Daniel A. Glasser
knows anything about my opinions	...!ihnp4!chinet!dag
my postings, or me for that matter!	...!ihnp4!mwc!dag
					...!ihnp4!mwc!gorgon!dag
	One of those things that goes "BUMP!!! (ouch!)" in the night.