C++ 两个浮点或双精度数的余数

在本教程中，我们将学习如何确定两个浮点数的余数。

示例：

Input: a = 9.7, b = 2.3 
Output: 0.5
Input: a = 36.5, b = 5.0 
Output: 1.5

C程序：

#include 
#include 

int main()
{
    int x = 10;
    int y = 3;

    float p = 10.23f;
    float q = 3.1f;

    double a = 123456789.10;
    double b = 1233.1;

        printf("remainder(%lf,%lf) = %lf\n", a, b, remainder(a, b));
        printf("remainder(%d,%d) = %lf\n", x, y, remainder(x, y));
         printf("remainder(%f,%f) = %lf\n", p, q, remainder(p, q));


    return 0;
}

输出：

remainder(123456789.100000,1233.100000) = 50.200000
remainder(10,3) = 1.000000
remainder(10.230000,3.100000) = 0.930000

C++ 程序:

#include 
using namespace std;

double findingMod(double x, double y)
{
    double mod;
    if (x < 0)
        mod = -x;
    else
        mod = x;
    if (y < 0)
        y = -y;
    while (mod >= y)
        mod = mod - y;
    if (x < 0)
        return -mod;

    return mod;
}
int main()
{
    double x = 9.7, y = 2.3;
    cout << findingMod(x, y);
    return 0;
}

输出结果：

0.5

将两个浮点数的模数找出来，我们可以利用内建的 fmod 函数。

C++ 程序：

#include 
using namespace std;

int main()
{
    double x = 9.7, y = 2.3;
    cout << fmod(x, y);
    return 0;
}

输出：

0.5