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Posting-Version: version B 2.10.1 6/24/83; site petsd.UUCP
Path: utzoo!watmath!clyde!burl!ulysses!mhuxj!houxm!vax135!petsd!joe
From: joe@petsd.UUCP (Joe Orost)
Newsgroups: net.lang.c
Subject: Re: Standard extensions (logical exclusive or)
Message-ID: <376@petsd.UUCP>
Date: Thu, 8-Nov-84 09:20:14 EST
Article-I.D.: petsd.376
Posted: Thu Nov  8 09:20:14 1984
Date-Received: Fri, 9-Nov-84 08:37:22 EST
References: <374@petsd.UUCP>, <3081@alice.UUCP>
Organization: Perkin-Elmer DSG, Tinton Falls, N.J.
Lines: 39

<>
>Joseph Orost suggests a new operator ^^ such that
>
>	a^^b
>
>would be equivalent to
>
>	a? (b?0:1): (b?1:0)
>
>What is wrong with writing
>
>	a!=b
>
>for this purpose?

Sorry, this is not the same.  Take, for instance, a=3 and b=4.

	a^b  = 7
	a!=b = 1
	a^^b = 0

Next take a=3 and b=0:

	a^b  = 3
	a!=b = 1
	a^^b = 1

Another expression for a^^b is (a==0)^(b==0).
Yet another one is (a!=0)^(b!=0).
And yes, you have to evaluate both a and b no matter what.

					regards,
					joe

--
Full-Name:  Joseph M. Orost
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