Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!sdd.hp.com!samsung!munnari.oz.au!goanna!ok
From: ok@goanna.cs.rmit.oz.au (Richard A. O'Keefe)
Newsgroups: comp.lang.c
Subject: Re: print % in c
Message-ID: <4829@goanna.cs.rmit.oz.au>
Date: 26 Feb 91 09:50:19 GMT
References: <61516@eerie.acsu.Buffalo.EDU> <1991Feb25.180600.5004@ux1.cso.uiuc.edu>
Organization: Comp Sci, RMIT, Melbourne, Australia
Lines: 15

In article <1991Feb25.180600.5004@ux1.cso.uiuc.edu>, gordon@osiris.cso.uiuc.edu (John Gordon) writes:
> 	To print special characters with printf(), precede the character
> with a \ character.  Example:

> 	printf("This a percent sign: \%\n");
> 	printf("This is a backslash: \\\n");

Did you *try* this?  Backslash is handled by one of the compiler
phases.  The string "This is a percent sign \%\n" turns into the
characters <T,h,i,s, ,i,s, ,a, ,p,e,r,c,e,n,t, ,s,i,g,n, ,%,NEWLINE>
and by the time printf() sees it there is nothing special about the
percent sign.  To get a percent sign from printf(), use two % signs:
	printf("The original answer was 100%% wrong.\n");
-- 
The purpose of advertising is to destroy the freedom of the market.