Re: Bizarre output.

[corey taylor <corey dot taylor at gmail dot com>]

This is one method of fuzzy floating point equates using a
magnitude-based epsilon.

const double fuzzyEpsilon = 0.000001;

/**
 Computes an appropriate epsilon for comparing a and b.
 */
inline double eps(double a, double b) {
    // For a and b to be nearly equal, they must have nearly
    // the same magnitude.  This means that we can ignore b
    // since it either has the same magnitude or the comparison
    // will fail anyway.
    (void)b;
    const double aa = abs(a) + 1;
    if (aa == inf()) {
        return fuzzyEpsilon;
    } else {
        return fuzzyEpsilon * aa;
    }
}

inline bool fuzzyEq(double a, double b) {
    return (a == b) || (abs(a - b) <= eps(a, b));
}

corey

On Wed, 30 Mar 2005 00:12:58 +0200, Antonio Coralles
<noche.suapie@reflex.at> wrote:
> Srinath M wrote:
> 
> > Can someone tell me why the text, "I see it" is not displayed when this
> > program is run?
> >
> > #include <stdio.h>
> >
> >
> > int main(void)
> > {
> >   float a, b;
> >
> >
> >   double d;
> >
> >
> >   a = 0.;
> >
> >
> >   b = 1.;
> >
> >
> >   for (d  = 0.0; d < 0.011; d = d + 0.001)
> >     {
> >       printf("d = %f\n",d);
> >
> >
> >       if (d == 0.009)
> >         {
> >           printf ("I see it\n");
> >         }
> >     }
> >   return 0;
> > }
> >
> Well, in my opinion you should normally not use "==" in conection with
> float or double, because floating-point arithmetic is not exact (not
> only with gcc). Besides 0.001 has not an exact representation as
> floating point number ... Try
> ...
> if(d > 0.008999 && d < 0.009001)
>     printf(...);
> ...
> antonio
> 
>