https://gcc.gnu.org/bugzilla/show_bug.cgi?id=78249
Richard Biener <rguenth at gcc dot gnu.org> changed:
What |Removed |Added
----------------------------------------------------------------------------
Status|UNCONFIRMED |NEW
Last reconfirmed| |2016-11-08
Ever confirmed|0 |1
--- Comment #2 from Richard Biener <rguenth at gcc dot gnu.org> ---
#include <limits>
double x = 0.0 * std::numeric_limits<double>::infinity();
get's me
x:
.long 0
.long 2146959360
independent of optimization level. But I can confirm the findings for
#include <iostream>
#include <limits>
double y = 0.0 * std::numeric_limits<double>::infinity();
int main() {
double x = 0.0 * std::numeric_limits<double>::infinity();
std::cout << "CPP runtime: " << x << std::endl;
std::cout << "CPP compile-time: " << y << std::endl;
}
which evaluates 0.0 * std::numeric_limits<double>::infinity() at runtime via
4008fe: e8 9f 00 00 00 callq 4009a2
<_ZNSt14numeric_limitsIdE8
infinityEv>
400903: 66 0f 28 c8 movapd %xmm0,%xmm1
400907: 66 0f ef c0 pxor %xmm0,%xmm0
40090b: f2 0f 59 c1 mulsd %xmm1,%xmm0
where with optimization we simplify it to +NaN (at compile-time).
