Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!usc!cs.utexas.edu!rice!uupsi!grebyn!ckp
From: ckp@grebyn.com (Checkpoint Technologies)
Newsgroups: comp.lang.c
Subject: Re: The nonexistent operator (along = v. == lines)
Keywords: xor
Message-ID: <1991Apr2.193936.5774@grebyn.com>
Date: 2 Apr 91 19:39:36 GMT
References: <156@revcan.UUCP>
Organization: Grebyn Timesharing
Lines: 30

In article <156@revcan.UUCP> darren@revcan.UUCP (Darren Morbey) writes:
>I've noticed in my writing C code that there is no such operator
>as ^^ (which would be to ^ as || is to |).  I feel in this case
>I *have* to write a macro for this *nonexistent* operator (you may
>recall I wrote #define XOR(a,b) ((a)^(b)) ).  However...
>
>I would like to use one of the following three macros, but each has
>its own particular problems.  I would appreciate any advice on which
>one should be used.  I would like it to operate like && and ||.

An ^^ operator cannot operate just like && and ||.  This is because
there is no possible notion of a shortcut evaluation; both sides must be
evaluated in every case.

I'll assume you mean for ^^ to operate on operands which are only
treated as 0 or non-0, and returns either 0 or 1.

>1.    #define XOR(a,b) ( ( !(a) && (b) ) || ( (a) && !(b) ) )
>2.    #define XOR(a,b) ( !(a) != !(b) )
>3.    #define XOR(a,b) ( (a) ? !(b) : (b) )  /* my favourite. */

Here's mine:

#define XOR(a,b) (((a) != 0) ^ ((b) != 0))

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