------- Comment #1 from pinskia at gmail dot com 2010-07-01 05:04 -------
Subject: Re: New: Complex division with NaN produces unexpected result
I think the issue is we don't implement imagainy types so 1 + nan I
turns into nan.
On Jun 30, 2010, at 9:51 PM, "ian at airs dot com" <gcc-bugzi...@gcc.gnu.org
> wrote:
> Annex G of the ISO C99 standard says that a complex value with one
> part being
> infinity is considered an infinity, even if the other part is a
> NaN. It's not
> clearly stated, but presumably if neither part of the number is an
> infinity,
> but one part is a NaN, then the number is a NaN. And presumably if
> a complex
> NaN is involved in a math operation, the result should be a complex
> NaN.
>
> So, I would expect that dividing a complex NaN by a complex 0 would
> give me a
> complex NaN. However, when I run this program:
>
>
> #include <stdio.h>
> #include <math.h>
> #include <complex.h>
>
> __complex float
> div (__complex float f1, __complex float f2)
> {
> return f1 / f2;
> }
>
> int
> main ()
> {
> __complex float f;
>
> f = div (NAN + NAN * I, 0);
> printf ("%g+%g*i\n", creal (f), cimag (f));
> f = div (1.0 + NAN * I, 0);
> printf ("%g+%g*i\n", creal (f), cimag (f));
> f = div (NAN + 1.0 * I, 0);
> printf ("%g+%g*i\n", creal (f), cimag (f));
> }
>
> with current mainline, it prints
>
> nan+nan*i
> nan+nan*i
> nan+inf*i
>
> That last answer seems incorrect according to the rules of Annex G.
> It is an
> infinity when it should be a NaN.
>
>
> --
> Summary: Complex division with NaN produces unexpected
> result
> Product: gcc
> Version: 4.6.0
> Status: UNCONFIRMED
> Severity: normal
> Priority: P3
> Component: c
> AssignedTo: unassigned at gcc dot gnu dot org
> ReportedBy: ian at airs dot com
>
>
> http://gcc.gnu.org/bugzilla/show_bug.cgi?id=44741
>
--
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=44741
