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From: higgin@cbmvax.commodore.com (Paul Higginbottom - CATS)
Newsgroups: comp.sys.amiga.misc
Subject: Re: Compairing C values
Message-ID: <18595@cbmvax.commodore.com>
Date: 5 Feb 91 20:26:32 GMT
References: <rat.3380@hotcity.UUCP>
Reply-To: higgin@cbmvax.commodore.com (Paul Higginbottom - CATS)
Organization: Commodore, West Chester, PA
Lines: 32

In article <rat.3380@hotcity.UUCP> rat@hotcity.UUCP (P W) writes:
$        I have + as an input arguement via *argv[] and whenever I try to
$compare its value it comes out as a 0.  For instance...  If I compile a
$program like this:
$
$#include <stdio.h>
$#include <math.h>
$
$void main(int argc, char *argv[])
${
$        And I type "test +" (I would name the file test) it would just print a
$equals 0.  Any help would be appereciated...
$char a = argv[1];
$if (a = '+') printf("a equals +");
$if (a = '0') printf("a equals '0'");
$if (a = 0) printf("a equals 0");
$}

Nice to catch an easier one to answer once in a while...

The problem is that argv[1] is in fact a POINTER to the argument string
of characters, and not a character itself.

char a = argv[1][0];	/* might work */

Otherwise:

char *cp = argv[1];
char a = *cp;

	Good luck.
	Paul.