From:	Bryan Anderson <ba AT jb DOT man DOT ac DOT uk>
Newsgroups:	comp.os.msdos.djgpp
Subject:	Re: Problems with DJGPP V2.01 - atof() function
Date:	Sun, 1 Dec 1996 08:57:06 +0000
Organization:	Sirius Cybernetics Corporation
Lines:	60
Message-ID:	<Pine.SOL.3.91.961201085145.18913B-100000@aten>
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To:	djgpp AT delorie DOT com
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On Sat, 30 Nov 1996, Francois Charton wrote:

> Morten Welinder wrote:
> > 
> > afonso AT inesca DOT inesca DOT pt writes:
> > 
> > >                char string[]="1.13";
> > >                int  result;
> > >                ...
> > >                result = (int)(atof(string)*100);
> > >                ...
> > 
> > >        I've got result = 112!!! not 113 as I wished, because
> > >the function atof() return is 1.29999... not 1.13 (and I only have
> > >an old i386).
> > 
> > Getting 112 is well within the C standard.  If your program does
> > not work in this situation then you have a bug.
> > 
> 
> Sorry to disagree but this *is* a bug : to be sure try the following 
> program : 
> 
> int main(void)
> {
> char ch[8]="1.13";
> int result, otherresult;
> float f;
> result=(int)(atof(ch)*100.0);
> f=atof(ch)*100.0;
> otherresult=(int) f;
> printf("result: %d otherresult:%d\n", result, otherresult);
> return 0;
> }
> 
> On my machine I get result: 112 and otherresult: 113...
> 
> Francois
> 

It is not a bug. It is a fundamental problem: decimals are held 
imprecisely because many decimals have an infinite number of binary 
digits after the binary point. You cannot beat this.

Try running this test.

#include <stdio.h>

main()
{
  double f = 1.13;
  printf(" 113 - 100*1.13 = %16.13e\n",113-100*f);
}

Using GCC under Linux I get

 113 - 100*1.13 = 1.0658141036402e-14

    Bryan

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