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Mail Archives: cygwin/2000/11/07/15:48:16

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From: "Keyser Soze" <keysers AT terra DOT com DOT br>
To: <cygwin AT sourceware DOT cygnus DOT com>
Subject: Newbie half off topic: '*' substituted by 'c'
Date: Tue, 7 Nov 2000 18:47:29 -0200
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I made a simple code to examplify, a calculation program, but when you =
do: "conta 2 * 4" it doesn't do the multiplication, after making a =
printf call to see what argc[2] contains, it appears 'c'.
Anyone knows what is happening?

Keyser Soze

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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; =
charset=3Diso-8859-1">
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<BODY bgColor=3D#ffffff>
<DIV><FONT size=3D2>I made a simple code to examplify, a calculation =
program, but=20
when you do: "conta 2 * 4" it doesn't do the multiplication, after =
making a=20
printf call to see what argc[2] contains, it appears 'c'.</FONT></DIV>
<DIV><FONT size=3D2>Anyone knows what is happening?</FONT></DIV>
<DIV><FONT size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT size=3D2>Keyser Soze</FONT></DIV></BODY></HTML>

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#include <stdio.h>

main(int argv, char *argc[]){

int num1, num2;
char oper;

if(argv < 4)	{
	printf("\nWRONG!!!\n%s num1 oper num2", argc[0]);
	return(-1);
		}

num1 = atoi(argc[1]);
num2 = atoi(argc[3]);
oper = *argc[2];

switch(oper){
	case '+': printf("\n %d\n", num1 + num2); break;
	case '-': printf("\n %d\n", num1 - num2); break;
	case '*': printf("\n %d\n", num1 * num2); break;
	case '/': printf("\n %d\n", num1 / num2); break;
	default: printf("\n %c nao e operador",oper); break;
		}
return 0;
}

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