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.
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-bugzilla@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 >
Note: a very similar issue probably affects C++.
1.0 + NAN * I is always (NaN, NaN) as we don't implement imaginary types. So this is the correct value since we don't do those. See http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1464.htm for ways of the C standard committee fixing this issue with respect of creating complex values.