fmod不正确吗? [英] Is fmod incorrect?

问题描述

给出以下双打， fmod 返回正确的值吗?

Given the following doubles, does fmod return the correct value?

    double x = .090;
    double y = .003;
    double r = fmod(x, y);
    // r = 0.0029999999999999949

为什么r = 0?

推荐答案

由于与大多数小数部分一样，这些数字不能用二进制浮点格式精确表示.要查看实际使用的值:

Because, like most decimal fractions, these numbers can't be represented exactly by a binary floating point format. To see what values are actually being used:

#include <cstdio>

int main() {
    printf("%.20f %.20f\n", .090, .003);
}

Output: 0.08999999999999999667 0.00300000000000000006

这些小的舍入误差意味着第一个值略小于第二个值的精确倍数，因此您可以看到结果.

These small rounding errors mean that the first value is slightly less than an exact multiple of the second, hence the result you see.

您可能希望允许较小的错误容忍度:

You may want to allow a small tolerance for error:

if (r > y - some_small_value) {
    r = 0;
}

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