C++ 中的 For 循环使用双突破提前一步，未达到边

For-loop in C++ using double breaking out one step early, boundary value not reached(C++ 中的 For 循环使用双突破提前一步，未达到边界值)

问题描述

我有一个在 32 位 Ubuntu 8.04 上使用 gcc 4.2.4 编译的简单 C++ 程序.它有一个 for 循环，其中 double 变量以特定步长从 0 递增到 1.当步长为 0.1 时，行为正是我所期望的.但是当步长为'0.05'时，循环在0.95后退出.谁能告诉我为什么会这样?输出遵循下面的源代码.

I have a simple C++ program compiled using gcc 4.2.4 on 32-bit Ubuntu 8.04. It has a for-loop in which a double variable is incremented from zero to one with a certain step size. When the step size is 0.1, the behavior is what I expected. But when the step size is '0.05', the loop exits after 0.95. Can anyone tell me why this is happening? The output follows the source code below.

#include <iostream>

using namespace std;

int main()
{
    double rangeMin = 0.0;
    double rangeMax = 1.0;
    double stepSize = 0.1;

    for (double index = rangeMin; index <= rangeMax; index+= stepSize)
    {
        cout << index << endl;
    }
    cout << endl; 

    stepSize = 0.05;
    for (double index = rangeMin; index <= rangeMax; index+= stepSize)
    {
        cout << index << endl;
    }

    return 0;
}

输出

sarva@savija-dev:~/code/scratch$ ./a.out 
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1

0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
sarva@savija-dev:~/code/scratch$

推荐答案

当使用浮点值时，并非每个值都可以精确表示，0.95+0.05 >1 因为 0.95 不能完全用 double 值表示.

When using floating point values not every value is exactly representable, 0.95+0.05 > 1 because 0.95 is not exactly representable by a double value.

看看维基百科对浮点精度的看法.

See what Wikipedia has to say about floating point accuracy.

如果您查看 IEEE 浮点转换器，您会看到0.95 在 64 位浮点数 (double) 中的值是 0-01111111110-1110011001100110011001100110011001100110011000 输入这个http://www.ajdesigner.com/fl_ieee_754_word/ieee_32_bit_word.php" rel="nofollow noreferrer">浮点计算器 你得到的值为 0.95000016 并添加 0.05 将带您超越 1.0 标记.