输出fmod函数c ++ [英] output of fmod function c++

问题描述

给定：

#include <iostream>
#include <cmath>
#include <limits>
using namespace std;

int main() {
    // your code goes here
    double h = .1;
    double x = 1;
    int nSteps = abs(x / h);

    double rem = fmod(x, h);
    cout<<"fmod output is "<<rem<<endl;
    if(abs(rem)<std::numeric_limits<double>::epsilon())
        cout<<"fmod output is almost near 0"<<endl;

    rem = remainder(x,h);
    cout<<"remainder output is "<<rem<<endl;
    if(abs(rem)<std::numeric_limits<double>::epsilon())
        cout<<"remainder output is almost near 0"<<endl;

    return 0;
}

给定 int（x / h）== 10 ，我会期望 fmod（）结果接近0，但我得到的是.0999999999。这是一个显着的差异。 remainder（）的结果仍然可以接受。您可以在 http://ideone.com/9wBlva 尝试使用验证码

Given int(x/h) == 10, I would have expected the fmod() result to be near 0 but what i get is .0999999999. It is a significant difference. The result of remainder() still seem acceptable. Code could be tried at http://ideone.com/9wBlva

为什么fmod（）结果的显着差异？

Why this significant difference for fmod() result?

推荐答案

fmod 您正在使用似乎遵循在 cppreference ：

The problem you're seeing is that the version of fmod you're using appears to follow the implementation defined at cppreference:

double fmod(double x, double y)
{
    double result = std::remainder(std::fabs(x), (y = std::fabs(y)));
    if (std::signbit(result)) result += y;
    return std::copysign(result, x);
} 

std :: remainder 计算非常非常小的结果，几乎为零（当我使用1和0.1时，-5.55112e-17，对于2和0.2使用-1.11022e-16）。但是，重要的是结果是否定，这意味着 std :: signbit 返回true，导致 y 添加到结果中，结果等于 y 。

std::remainder computes a very very small result, nearly zero (-5.55112e-17 when using 1 and 0.1 for me, -1.11022e-16 for 2 and 0.2). However, what's important is that the result is negative, which means std::signbit returns true, causing y to get added to the result, effectively making the result equal to y.

请注意 std :: fmod 没有说明使用 std :: remainder ：

由此函数计算的除法运算x / y的浮点余数正好是值x-n * y，其中n是x / y，其小数部分被截断。

The floating-point remainder of the division operation x/y calculated by this function is exactly the value x - n*y, where n is x/y with its fractional part truncated.

所以如果你自己计算的值，你最终以零（即使你使用 std :: round 结果而不是纯整数截断）

So if you compute the value yourself, you do end up with zero (even if you use std::round on the result instead of pure integer truncation)

当 x 是2和 y 是0.2

double x = 2;
double y = .2;

int n = static_cast<int>(x/y);
double result = x - n*y;
std::cout << "Manual: " << result << std::endl;
std::cout << "fmod: " << std::fmod(x,y) << std::endl;

输出（ gcc demo ）

手动：0

fmod ：0.2

Manual: 0
fmod: 0.2

但是问题并没有降级为gcc;我也看到它在MSVC和铛。在clang中，如果使用 float 而不是 double ，有时会有不同的行为。

However the problem is not relegated to only gcc; I also see it in MSVC and clang. In clang there is sometimes different behavior if one uses float instead of double.

来自 std :: remainder 的真正小的负值来自这样的事实，即0.1和0.2都不能在浮点数学中精确表示。如果你改变x和y，比如说2和0.25，那么一切都很好。

This really small negative value from std::remainder comes from the fact that neither 0.1 nor 0.2 can be represented exactly in floating point math. If you change x and y to, say 2 and 0.25, then all is well.

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