[Bug c/44741] New: Complex division with NaN produces unexpected result

[ian at airs dot com]

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