Why does NaN - NaN == 0.0 with the Intel C++ Compiler?

Question

It is well-known that NaNs propagate in arithmetic, but I couldn\'t find any demonstrations, so I wrote a small test:

#include 
#include 

        

Answer 1

Since I see an answer impugning the standards compliance of Intel's compiler, and no one else has mentioned this, I will point out that both GCC and Clang have a mode in which they do something quite similar. Their default behavior is IEEE-compliant —

$ g++ -O2 test.cc && ./a.out 
neg: -nan
sub: nan nan nan
add: nan nan
div: nan nan nan
mul: nan nan

$ clang++ -O2 test.cc && ./a.out 
neg: -nan
sub: -nan nan nan
add: nan nan
div: nan nan nan
mul: nan nan

— but if you ask for speed at the expense of correctness, you get what you ask for —

$ g++ -O2 -ffast-math test.cc && ./a.out 
neg: -nan
sub: nan nan 0.000000
add: nan nan
div: nan nan 1.000000
mul: nan nan

$ clang++ -O2 -ffast-math test.cc && ./a.out 
neg: -nan
sub: -nan nan 0.000000
add: nan nan
div: nan nan nan
mul: nan nan

I think it is entirely fair to criticize ICC's choice of default, but I would not read the entire Unix wars back into that decision.