使用 epsilon 将双精度数与零进行比较 [英] Compare double to zero using epsilon

问题描述

今天，我翻阅了一些 C++ 代码(由其他人编写)，发现了这个部分:

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;
}

我正在尝试弄清楚这是否有意义.

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

epsilon() 的文档说:

该函数返回 1 和大于 1 的最小值之间的差值，该值可以 [用双精度数] 表示.

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

这是否也适用于 0，即 epsilon() 是大于 0 的最小值?或者 0 和 0 + epsilon 之间是否有可以用 double 表示的数字?

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?

如果不是，那么比较不等于 someValue == 0.0?

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

推荐答案

假设 64 位 IEEE 双精度，有 52 位尾数和 11 位指数.让我们分解一下:

Assuming 64-bit IEEE double, there is a 52-bit mantissa and 11-bit exponent. Let's break it to bits:

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

大于1的最小可表示数:

The smallest representable number greater than 1:

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

因此:

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

0 和 epsilon 之间有数字吗?很多...例如最小正可表示(正常)数是:

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

实际上在 0 和 epsilon 之间有 (1022 - 52 + 1)×2^52 = 4372995238176751616 个数，占所有可表示正数的 47%...

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

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