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SubjectRe: FPU, i386
On Thu, 25 Apr 2002, Mark Mielke wrote:

> On Thu, Apr 25, 2002 at 10:22:49AM -0400, Richard B. Johnson wrote:
> > To use the math macros, the comparison should be something like:
> > if (isless(fabs(a-b), 1.0e-38))
> > break;
> I might be saying something stupid, but, I was under the impression
> that floating point '==', assuming it follows IEEE rules, does exactly
> this.

A floating-point '==' tests for the two oprands to be exact. The
only time they will be exact is:

float a, b;
a = b = 10.000;

At this instant a = b and (a==b) will return TRUE.
However, if by previous math, you expected a to equal 10.0, by

a = b = 10.00

a += 12345.678;
a -= 12345.678;

Now, 'a' is not equal to 'b' if the 'C' compiler actually performed
the indicated operations (the compiler may obtimize them away).

In this case (a==b) will return FALSE because they are not equal.

Script started on Thu Apr 25 11:52:46 2002

# cat >xxx.c
#include <stdio.h>
#include <math.h>

int main()
double a, b;
a = b = 10.0;

printf("%d\n", (a==b));
a += 10.234567890;
printf("%d\n", (a==b));
a -= 10.234567890;
printf("%d\n", (a==b));
return 0;


# gcc -o xxx xxx.c
# ./xxx
# exit

Script done on Thu Apr 25 11:53:10 2002

So, as you can see, when first initialized with the same value,
regardless of how inaccurately in can be represented, it does
show equality.

However, as soon as I muck with it, add, then subtract the same
number, it will no longer be egual.

This is why you NEVER use '==' when dealing with floating-point
numbers. The use of '==' in floating-point operations is a bug.
'C' allows you to write buggy code. It will probably call
a floating-point compare routine in the FPU (fcom, fcomp, fcompp)
which will promptly return the 'not-equal' in C1 and the CPU flags.

'C' will even allow you to use floating-point numbers in loops, i.e.,


Just because it's allowed, it doesn't mean that it's not a bug.

Dick Johnson

Penguin : Linux version 2.4.18 on an i686 machine (797.90 BogoMips).

Windows-2000/Professional isn't.

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