使用IEEE浮点，在最小的非零正数和最小的非零负数之间，存在两个值:正零和负零。测试一个值是否在最小的非零值之间等价于测试与零相等;然而，赋值可能会产生影响，因为它会将负0变为正0。

It would be conceivable that a floating-point format might have three values between the smallest finite positive and negative values: positive infinitesimal, unsigned zero, and negative infinitesimal. I am not familiar with any floating-point formats that in fact work that way, but such a behavior would be perfectly reasonable and arguably better than that of IEEE (perhaps not enough better to be worth adding extra hardware to support it, but mathematically 1/(1/INF), 1/(-1/INF), and 1/(1-1) should represent three distinct cases illustrating three different zeroes). I don't know whether any C standard would mandate that signed infinitesimals, if they exist, would have to compare equal to zero. If they do not, code like the above could usefully ensure that e.g. dividing a number repeatedly by two would eventually yield zero rather than being stuck on "infinitesimal".

假设系统无法区分1.000000000000000000000和1.00000000000000001。这是1.0和1.0 + 1e-20。你认为在-1e-20和+1e-20之间还有一些值可以表示吗?

X和X的下一个值之间的差值根据X而变化。 Epsilon()只是1和下一个1的值之间的差。 0和下一个0值之间的差不是()。

相反，你可以使用std::nextafter来比较双精度值和0，如下所示:

bool same(double a, double b)
{
  return std::nextafter(a, std::numeric_limits<double>::lowest()) <= b
    && std::nextafter(a, std::numeric_limits<double>::max()) >= b;
}

double someValue = ...
if (same (someValue, 0.0)) {
  someValue = 0.0;
}

我认为这取决于你电脑的精度。 看一下这张表:你可以看到，如果用double表示，但你的精度更高，比较并不等于

someValue == 0.0

不管怎样，这是个好问题!

你不能把这个应用到0，因为有尾数和指数部分。 由于指数可以存储很小的数，小于， 但是当你尝试做一些类似(1.0 -“非常小的数字”)的事情时，你会得到1.0。 Epsilon不是值的指示器，而是值精度的指示器，值精度是尾数。 它显示了我们可以存储多少个正确的十进制数字。

“1和大于1的最小值之间的差值”意味着1 +“机器零”，这大约是10^-8或10^-16，这取决于你是否使用双变量的浮点数。你可以用1除以2，直到计算机看到1 = 1+1/2^p，如下所示:

#include <iostream>
#include "math.h"
using namespace std;

int main() {
    float a = 1;
    int n = 0;
    while(1+a != 1){
        a = a/2;
        n +=1;
    }
    cout << n-1 << endl << pow(2,-n);
    return 0;
}