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

问题

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) 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?

回答1:

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.

回答2:

This . . . can't be right, right? My question: what do the relevant standards (ISO C, ISO C++, IEEE 754) say about this?

Petr Abdulin already answered why the compiler gives a 0.0 answer.

Here is what IEEE-754:2008 says:

(6.2 Operations with NaNs) "[...] For an operation with quiet NaN inputs, other than maximum and minimum operations, if a floating-point result is to be delivered the result shall be a quiet NaN which should be one of the input NaNs."

So the only valid result for the subtraction of two quiet NaN operand is a quiet NaN; any other result is not valid.

The C Standard says:

(C11, F.9.2 Expression transformations p1) "[...]

x − x → 0. 0 "The expressions x − x and 0. 0 are not equivalent if x is a NaN or infinite"

(where here NaN denotes a quiet NaN as per F.2.1p1 "This specification does not define the behavior of signaling NaNs. It generally uses the term NaN to denote quiet NaNs")

回答3:

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.

来源：https://stackoverflow.com/questions/32195949/why-does-nan-nan-0-0-with-the-intel-c-compiler

标签

c++

floating-point

ieee-754

icc