Re: comparing doubles for equality

Bjørn Augestad wrote:
Here's how I do it (more or less):

#define TOLERANCE 0.0000001

int dbl_isequal(double a, double b)
{
return (fabs(a - b) < TOLERANCE) ? 1 : 0;
}

Works for me.

That works if your numbers stay within a certain range (which may indeed
be the case in some programs), but it eventually fails once you start to
get relatively big numbers. Here's a program:

#include <stdio.h>
#include <math.h>

#define TOLERANCE 0.0000001

int dbl_isequal(double a, double b)
{
return (fabs(a - b) < TOLERANCE) ? 1 : 0;
}

int main()
{
double u = 1.0;
double v = u + 1e-14;
double big = 1.0;

big *= 1<<16;
big *= 1<<16;
big *= 1<<16;
big *= 1<<16;

printf ("%g == %g? %s.\n", u, v, dbl_isequal(u, v) ? "yes" : "no");

u *= big;
v *= big;

printf ("%g == %g? %s.\n", u, v, dbl_isequal(u, v) ? "yes" : "no");

return 0;
}

And here's its output:

1 == 1? yes.
1.84467e+19 == 1.84467e+19? no.

It seems odd that multiplying two "equal" numbers by the same value,
especially when that value is a power of two and can be represented
exactly as a 64-bit IEEE-754 floating point number, causes them to
afterwards be unequal.

- Logan
.

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