Path: utzoo!utgpu!water!watmath!clyde!rutgers!ames!fxgrp!ljz
From: ljz@fxgrp.UUCP (Lloyd Zusman, Master Byte Software)
Newsgroups: comp.lang.c
Subject: Re: "logical" xor
Message-ID: <195@fxgrp.UUCP>
Date: 10 Jan 88 23:49:20 GMT
References: <2946@zeus.TEK.COM>
Reply-To: fxgrp!ljz@ames.arpa (Lloyd Zusman, Master Byte Software)
Followup-To: <2946@zeus.TEK.COM> dant@tekla.UUCP (Dan Tilque)
Organization: FX Development Group, Inc., Mountain View, CA
Lines: 55

In article <2946@zeus.TEK.COM> dant@tekla.UUCP (Dan Tilque) writes:
> ...
>Tell me, what is the ^^ supposed to add that is not already done by ^ ?
>The answer is nothing.  The "bitwise" xor does exactly what you would want
>a "logical" xor to do.  In fact, the terms "bitwise" and "logical" are
>misnomers.  They should be "unoptimized" and "optimized".  And there's no
>way to optimize an exclusive-or operation.
>
>The only reason to add ^^ is esthetic.  ...
> ...

I'm sorry, but I disagree.  Try running this program:

#include <stdio.h>

main()
{
    int a = 2;
    int b = 0;
    int c = 7;

    printf("a = %d, b = %d, c = %d\n", a, b, c);
    printf("a & b = %d\n", a & b);
    printf("a && b = %d\n", a && b);
    printf("a & c = %d\n", a & c);
    printf("a && c = %d\n", a && c);
    printf("a | b = %d\n", a | b);
    printf("a || b = %d\n", a || b);
    printf("a | c = %d\n", a | c);
    printf("a || c = %d\n", a || c);
}

The results that printed out on my machine are:
a = 2, b = 0, c = 7
a & b = 0
a && b = 0
a & c = 2
a && c = 1
a | b = 2
a || b = 1
a | c = 7
a || c = 1

In my opinion, this illustrates the *exact* reason for the two versions
of the 'and' and 'or' operators.  The 'bitwise' versions are needed for
bit testing, etc.  The 'logical' versions are guaranteed to result in
1 or 0.



-------------------------------------------------------------------------
 Lloyd Zusman
 Master Byte Software
 Los Gatos, California	    	    	Internet:   fxgrp!ljz@ames.arpa
 "We take things well in hand."	    	UUCP:	    ...!ames!fxgrp!ljz