最小浮点数(最接近零)

问题描述

我试图找到可以存储在单个精度浮点数中的最小值(最接近零).使用<limits>标头可以获取该值，但是如果我将其设置得更小，则浮点数仍然可以容纳它，并且给出正确的结果.这是一个使用g ++ 5.3.0编译的测试程序.

I'm trying to find the minimum value (closest to zero) that I can store in a single precission floating point number. Using the <limits> header I can get the value, but if I make it much smaller, the float can still hold it and it gives the right result. Here is a test program, compiled with g++ 5.3.0.

#include <limits>
#include <iostream>
#include <math.h>

using namespace std;

int main()
{
    float a = numeric_limits<float>::max();
    float b = numeric_limits<float>::min();

    a = a*2;
    b = b/pow(2,23);


    cout << a << endl;
    cout << b << endl;
}

正如我预期的那样，"a"给出无穷大，但是"b"即使将最小值除以2 ^ 23后仍然保持良好的结果，之后它给出0.

As I expected, "a" gives infinity, but "b" keeps holding the good result even after dividing the minimum value by 2^23, after that it gives 0.

给出numeric_limits<float>::min()的值是2 ^(-126)，我相信这是正确的答案，但是为什么我的程序中的浮动内容持有如此小的数字呢?

The value that gives numeric_limits<float>::min() is 2^(-126) which I belive is the correct answer, but why is the float on my progam holding such small numbers?

推荐答案

std::numeric_limits::min对于浮点类型给出最小的非零值，该值可以表示而不会损失精度. std::numeric_limits::lowest给出最小的可表示值.使用IEEE表示形式，该值是次标准值(以前称为非标准化).

std::numeric_limits::min for floating-point types gives the smallest non-zero value that can be represented without loss of precision. std::numeric_limits::lowest gives the smallest representable value. With IEEE representations that's a subnormal value (previously called denormalized).

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