如果我将浮点数复制到另一个变量，它们是否相等? [英] If I copy a float to another variable, will they be equal?

问题描述

我知道使用==检查浮点变量的相等性不是一个好方法.但是我只想通过以下语句知道这一点:

 float x = ...

float y = x;

assert(y == x)
 

由于从x复制了y，所以断言为真吗?

解决方案

除了kmdreko指出的assert(NaN==NaN);情况外，您还可能遇到x87-math的情况，即将80位浮点数临时存储到内存中，然后与值进行比较仍然存储在寄存器中.

可能的最小示例，当使用-O2 -m32编译时，它在gcc9.2上失败:

 #include <cassert>

int main(int argc, char**){
    float x = 1.f/(argc+2);
    volatile float y = x;
    assert(x==y);
}
 

Godbolt演示: https://godbolt.org/z/X-Xt4R

如果您设法创建足够的寄存器压力来存储y并从内存中重新加载，则可以省略volatile(但要使编译器感到困惑，不要完全省略比较).

>

请参阅GCC常见问题解答参考:

• 为什么浮点结果随优化级别或不同的编译器版本或不同的目标体系结构而改变?

I know that using == to check equality of floating-point variables is not a good way. But I just want to know that with the following statements:

float x = ...

float y = x;

assert(y == x)

Since y is copied from x, will the assertion be true?

解决方案

Besides the assert(NaN==NaN); case pointed out by kmdreko, you can have situations with x87-math, when 80bit floats are temporarily stored to memory and later compared to values which are still stored inside a register.

Possible minimal example, which fails with gcc9.2 when compiled with -O2 -m32:

#include <cassert>

int main(int argc, char**){
    float x = 1.f/(argc+2);
    volatile float y = x;
    assert(x==y);
}

Godbolt Demo: https://godbolt.org/z/X-Xt4R

The volatile can probably be omitted, if you manage to create sufficient register-pressure to have y stored and reloaded from memory (but confuse the compiler enough, not to omit the comparison all-together).

See GCC FAQ reference:

• Why floating-point results change with optimization levels or different compiler versions or different target architectures?

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