http://bugzilla.redhat.com/bugzilla/show_bug.cgi?id=221319
raises a question whether:
std::complex<double> (1.0, 1.0) / 0.0
and
std::complex<double> (1.0, 1.0) / std::complex<double> (0.0, 0.0)
can or can't be NaN. In C say:
#define I (__extension__ 1.0iF)
_Complex double d = 1.0 + 0.0 * I, e, f, z = 0.0 + 0.0 * I;
int
main (void)
{
e = d / 0.0;
f = d / z;
__builtin_printf ("%g %g %g %g\n", __real__ e, __imag__ e, __real__ f,
__imag__ f);
return 0;
}
in -std=c99 always prints inf (where complex inf is even when just one of the
parts is inf), no matter what optimization options, in C89 the first division
when optimizing is usually optimized into the simpler __real__ d / 0.0,
__imag__ d / 0.0 division and thus gives inf too, but the latter and when not
optimizing
even the first one gives NaN.
What do we want in C++? C++ doesn't use flag_complex_method = 2, so unless
that changes, only the division by double (rather than std::complex <double>)
could generally (when not optimizing) return inf - with a help in the
specialization, see the above URL.
--
Summary: <complex> division by 0
Product: gcc
Version: 4.1.2
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: libstdc++
AssignedTo: unassigned at gcc dot gnu dot org
ReportedBy: jakub at gcc dot gnu dot org
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=30482
