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beta 18 'rint' function with gcc is broken
- To: gnu-win32 at cygnus dot com
- Subject: beta 18 'rint' function with gcc is broken
- From: Jeff Deifik <jdeifik at weasel dot com>
- Date: Sun, 11 May 1997 22:44:48 -0700
I am running windows nt 4.0 server, sp2.
I have a library regression test that I run on the compilers I use.
It showed a bug with the 'rint' function.
This function is supposed to convert a double to an integral double
value. It isn't working, as the enclosed test program shows.
Compile with no options, i.e.
gcc yy.c
yy
The output should be nothing, but it is:
47.4 47.125000
47.6 47.625000
-47.4 -47.125000
-47.6 -47.625000
Jeff Deifik
/* y.c, demonstrates a bug with gnu-w32 beta 18 */
/* Written 11/06/96 by Jeff Deifik */
/* Modified 05/11/97 by Jeff Deifik */
#include <stdio.h>
#include <math.h>
typedef enum { False = 0, True = 1 } Boolean;
float FLOAT_TO_ROUNDED_FLOAT(float);
int flo_close(double value,double f,double clo);
int main(int argc, char *argv[])
{
Boolean fail = False;
double d;
float f;
d = rint(47);
if (!flo_close(47.0f,d,0.0001f))
{ (void)printf("47 %f\n",d); fail = True; }
f = FLOAT_TO_ROUNDED_FLOAT(47.0);
if (!flo_close(47.0f,f,0.0001f))
{ (void)printf("47 %f\n",f); fail = True; }
d = rint(47.4);
if (!flo_close(47.0f,d,0.0001f)) { (void)printf("47.4 %f\n",d); fail = True; }
d = rint(47.6);
if (!flo_close(48.0f,d,0.0001f)) { (void)printf("47.6 %f\n",d); fail = True; }
d = rint(-47);
if (!flo_close(-47.0f,d,0.0001f)) { (void)printf("-47 %f\n",d); fail = True; }
d = rint(-47.4);
if (!flo_close(-47.0f,d,0.0001f)) { (void)printf("-47.4 %f\n",d); fail = True; }
d = rint(-47.6);
if (!flo_close(-48.0f,d,0.0001f)) { (void)printf("-47.6 %f\n",d); fail = True; }
return 0;
}
/* Value is the ideal number, f is what it really is.*/
int flo_close(double value,double f,double clo)
{
return(((f+clo) >= value) && ((f-clo) <= value));
}
float FLOAT_TO_ROUNDED_FLOAT(float x) { return(floor(x+0.5)); }