Why does NaN - NaN == 0.0 with the Intel C++ Compiler?
It is well-known that NaNs propagate in arithmetic, but I couldn't find any demonstrations, so I wrote a small test:
#include <limits>
#include <cstdio>
int main(int argc, char* argv[]) {
float qNaN = std::numeric_limits<float>::quiet_NaN();
float neg = -qNaN;
float sub1 = 6.0f - qNaN;
float sub2 = qNaN - 6.0f;
float sub3 = qNaN - qNaN;
float add1 = 6.0f + qNaN;
float add2 = qNaN + qNaN;
float div1 = 6.0f / qNaN;
float div2 = qNaN / 6.0f;
float div3 = qNaN / qNaN;
float mul1 = 6.0f * qNaN;
float mul2 = qNaN * qNaN;
printf(
"neg: %f\\nsub: %f %f %f\\nadd: %f %f\\ndiv: %f %f %f\\nmul: %f %f\\n",
neg, sub1,sub2,sub3, add1,add2, div1,div2,div3, mul1,mul2
);
return 0;
}
The example ([running live here](http://ideone.com/FjphJj)) produces basically what I would expect (the negative is a little weird, but it kind of makes sense):
neg: -nan
sub: nan nan nan
add: nan nan
div: nan nan nan
mul: nan nan
MSVC 2015 produces something similar. However, Intel C++ 15 produces:
neg: -nan(ind)
sub: nan nan 0.000000
add: nan nan
div: nan nan nan
mul: nan nan
Specifically, `qNaN - qNaN == 0.0`.
This... can't be right, right? What do the relevant standards (ISO C, ISO C++, IEEE 754) say about this, and why is there a difference in behavior between the compilers?
The default floating point handling in Intel C++ compiler is /fp:fast, which handles NaN's unsafely (which also results in NaN == NaN being true for example). Try specifying /fp:strict or /fp:precise and see if that helps.
