负零（-0.0）与正零（+0.0）相比的行为 [英] Behaviour of negative zero (-0.0) in comparison with positive zero (+0.0)

问题描述

在我的代码中，

  float f = -0.0; //负数
  

与负零比较

  f == -0.0f 
  

结果将会是 true 。

但是

  float f = 0.0; //正数
  

与负零相比较

  f == -0.0f 
  

结果也是 true 而不是 false

=http://coliru.stacked-crooked.com/a/4962a3b70d76a260 =nofollow noreferrer>一个MCVE来测试它（live coliru）：

  #include< iostream> 
 
 int main（）
 {
 float f = -0.0; 
 
 std :: cout<<====><< f< std :: endl< std :: endl; 
 
 if（f == -0.0f）
 {
 std :: cout<<true<< std :: endl; 
} 
 else 
 {
 std :: cout<<false<< std :: endl; 
 
 
 
 
 
 输出：  p> 
 
 
  ====> -0 //这里打印负零
 
 
 
 $ div class =h2_lin>解决方案

C ++中的浮点运算通常是 IEEE-754 。这个标准与实数集合的数学定义不同。 为零值定义了两个不同的表示：正零和负零。这也是定义，这两个表示必须比较等于，所以定义：



  +0.0 == -0.0 
  

至于为何如此，请在


在IEEE算术中，定义log 0 =-∞和log x当x <1时是一个NaN是自然的。假设x表示一个下溢到零的小负数。由于签名为零，x将是负数，所以日志可以返回一个NaN。但是，如果没有有符号零，日志函数不能区分下溢负数和0，因此必须返回-∞。




In my code,

float f = -0.0; // Negative 

and compared with negative zero

f == -0.0f

result will be true.

But

float f = 0.0; // Positive

and compared with negative zero

f == -0.0f

also, result will be true instead of false

Why in both cases result to be true?

---

Here is a MCVE to test it (live on coliru):

#include <iostream>

int main()
{
    float f = -0.0;

    std::cout<<"==== > " << f <<std::endl<<std::endl;

    if(f == -0.0f)
    {
        std::cout<<"true"<<std::endl;
    }
    else
    {
        std::cout<<"false"<<std::endl;
    }
}

Output:

==== > -0  // Here print negative zero

true

解决方案

Floating point arithmetic in C++ is often IEEE-754. This norm differs from the mathematical definition of the real number set.

This norm defines two different representations for the value zero: positive zero and negative zero. It is also defined that those two representations must compare equals, so by definition:

+0.0 == -0.0

As to why it is so, in its paper What Every Computer Scientist Should Know About Floating Point Arithmetic, David Goldberg, 1991-03 (linked in the IEEE-754 page on the IEEE website) writes:

In IEEE arithmetic, it is natural to define log 0 = -∞ and log x to be a NaN when x < 0. Suppose that x represents a small negative number that has underflowed to zero. Thanks to signed zero, x will be negative, so log can return a NaN. However, if there were no signed zero, the log function could not distinguish an underflowed negative number from 0, and would therefore have to return -∞.

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