Xref: news2.mv.net comp.lang.c:54599 comp.lang.c++:70107 comp.os.msdos.djgpp:690 gnu.g++.help:4446 gnu.gcc.help:5657
From: wfeldt AT physik DOT uni-leipzig DOT de (Steffen Winterfeldt)
Newsgroups: comp.lang.c,comp.lang.c++,gnu.gcc.help,gnu.g++.help,comp.os.msdos.djgpp
Subject: Re: float != float and floats as return types
Followup-To: comp.lang.c,comp.lang.c++,gnu.gcc.help,gnu.g++.help,comp.os.msdos.djgpp
Date: 30 Jan 1996 18:21:39 GMT
Organization: Uni Leipzig
Lines: 32
Message-ID: <4elnjj$er4@server2.rz.uni-leipzig.de>
References: <4ej9lb$mpc AT fu-berlin DOT de>
NNTP-Posting-Host: tph100.physik.uni-leipzig.de
To: djgpp AT delorie DOT com
DJ-Gateway: from newsgroup comp.os.msdos.djgpp

Axel Thimm (axl AT zedat DOT fu-berlin DOT de) wrote:
: Hello,
: I am getting confused, about how C/C++ manage float binary operations,
: in particular multiplication. The next C++ example gives me surprising
: results:
: 	*** cut here: begin file t_prec.cc
: 	#include <iostream.h>
: 	#include <iomanip.h>
: 	#include <math.h>
: 	float quad( float );
: 	int main() {
: 	  for( int i=0; i<10; ++i ) {
: 	    float a, b, c;
: 	    a = i/13.123123;
: 	    b = a*a;
: 	    c = quad(a);
: 	    cout << (b - c) << '\t';
: 	    cout << (b - a*a) << '\t';
: 	    cout << (c - quad(a)) << '\n';
: 	  }
: 	  return 0;
: 	}
: 	float quad( float x ) { return x*x; }
: [...]

(c - quad(a)) is not zero, because quad's return value is in a floating point
register and so has higher precision than c.

BTW, with higher optimization (say, -O3), even the second column becomes zero.


Steffen