使用epsilon将double比较为零 [英] Compare double to zero using epsilon

问题描述

今天，我在寻找一些C ++代码（由别人写），并找到这一部分：

  double someValue = ... 
 if（someValue< std :: numeric_limits< double> :: epsilon（）&& 
 someValue> -std :: numeric_limits& 
 someValue = 0.0; 
} 
  

我试图找出这是否有意义。

epsilon（）的说明文件：

函数返回1和大于1的最小值之间的差值，可以用[double]表示。

这也适用于0，即 epsilon（）是大于0的最小值吗？或者是 0 和 0 + epsilon 之间的数字，可以用 double ？

如果没有，那么不等于 someValue == 0.0 ？

解决方案

假设64位IEEE双精度，有一个52位尾数和11位指数。查看以下数字：

  1.0000 00000000 00000000 00000000 00000000 00000000 00000000×2 ^ 0 = 1 
  

大于1的最小可表示数字：

  1.0000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000001×2 ^ 0 = 1 + 2 ^ -52 
  

因此：

  epsilon =（1 + 2 ^ -52） -  1 = 2 ^ -52 
  

0和epsilon之间是否有任何数字？很多...最小正表示（正常）数是：

  1.0000 00000000 00000000 00000000 00000000 00000000 00000000×2 ^ -1022 = 2 ^ 1022 
  

其实有（1022 - 52 + 1）×2 ^ 52 = 4372995238176751616 0和epsilon之间的数字，约为所有可表示数字的正数的47％...

Today, I was looking through some C++ code (written by somebody else) and found this section:

double someValue = ...
if (someValue <  std::numeric_limits<double>::epsilon() && 
    someValue > -std::numeric_limits<double>::epsilon()) {
  someValue = 0.0;
}

I'm trying to figure out whether this even makes sense.

The documentation for epsilon() says:

The function returns the difference between 1 and the smallest value greater than 1 that is representable [by a double].

Does this apply to 0 as well, i.e. epsilon() is the smallest value greater than 0? Or are there numbers between 0 and 0 + epsilon that can be represented by a double?

If not, then isn't the comparison equivalent to someValue == 0.0?

解决方案

Assuming 64-bit IEEE double, there is a 52-bit mantissa and 11-bit exponent. Look at the following numbers:

1.0000 00000000 00000000 00000000 00000000 00000000 00000000 × 2^0 = 1

The smallest representable number greater than 1:

1.0000 00000000 00000000 00000000 00000000 00000000 00000001 × 2^0 = 1 + 2^-52

Therefore:

epsilon = (1 + 2^-52) - 1 = 2^-52

Are there any numbers between 0 and epsilon? Plenty... E.g. the minimal positive representable (normal) number is:

1.0000 00000000 00000000 00000000 00000000 00000000 00000000 × 2^-1022 = 2^-1022

In fact there are about (1022 - 52 + 1)×2^52 = 4372995238176751616 numbers between 0 and epsilon, which is about 47% of all the positive representable numbers...

这篇关于使用epsilon将double比较为零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！