Tom Lane wrote:
> Bruce Momjian <bruce@momjian.us> writes:
> > I have made some more progress on this patch.
>
> I'm not convinced that you're fixing things so much as doing your best
> to destroy IEEE-compliant float arithmetic behavior.
>
> I think what we should probably consider is removing CheckFloat4Val
> and CheckFloat8Val altogether, and just letting the float arithmetic
> have its head. Most modern hardware gets float arithmetic right per
> spec, and we shouldn't be second-guessing it.

Well, I am on an Xeon and can confirm that our computations of large
non-infinite doubles who's result greatly exceed the max double are
indeed returning infinity, as the poster reported, so something isn't
working, if it supposed to. What do people get for this computation?

#include <stdio.h>
#include <stdlib.h>

int
main(int argc, char *argv[])
{
double a = 1e300, b = 1e300;
double c;

c = a * b;

printf("%e\n", c);
return 0;
}

I get 'inf'. I am on BSD and just tested it on Fedora Core 2 and got
'inf' too.

> A slightly less radical proposal is to reject only the case where
> isinf(result) and neither input isinf(); and perhaps likewise with
> respect to NaNs.

Uh, that's what the patch does for 'Inf':

result = arg1 + arg2;
CheckFloat4Val(result, isinf(arg1) || isinf(arg2));

I didn't touch 'Nan' because that is passed around as a value just fine
--- it isn't created or tested as part of an overflow.

--
Bruce Momjian bruce@momjian.us
EnterpriseDB http://www.enterprisedb.com

+ If your life is a hard drive, Christ can be your backup. +

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> I get 'inf'. I am on BSD and just tested it on Fedora Core 2 and got
> 'inf' too.

Ubuntu Edgy 64bit on Athlon 64X2 returns inf.

Joshua D. Drake


>
> > A slightly less radical proposal is to reject only the case where
> > isinf(result) and neither input isinf(); and perhaps likewise with
> > respect to NaNs.
>
> Uh, that's what the patch does for 'Inf':
>
> result = arg1 + arg2;
> CheckFloat4Val(result, isinf(arg1) || isinf(arg2));
>
> I didn't touch 'Nan' because that is passed around as a value just fine
> --- it isn't created or tested as part of an overflow.
>
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