为什么使用英特尔C ++编译器的NaN-NaN == 0.0? [英] Why does NaN - NaN == 0.0 with the Intel C++ Compiler?

问题描述

众所周知，NaN是以算术方式传播的，但是我找不到任何示范，所以我写了一个小测试:

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;
}

该示例(在此处运行)基本上产生了我所期望的(否定有点怪异，但是很有道理):

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产生类似的结果.但是，英特尔C ++ 15会产生:

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

具体是qNaN - qNaN == 0.0.

这...不对，对吗?有关标准(ISO C，ISO C ++，IEEE 754)对此有何评论?为什么编译器之间的行为有所不同?

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?

推荐答案

Intel C ++编译器中的默认浮点处理是/fp:fast，它不安全地处理NaN(这也会导致NaN == NaN被).尝试指定/fp:strict或/fp:precise，看看是否有帮助.

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.

这篇关于为什么使用英特尔C ++编译器的NaN-NaN == 0.0?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！