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
