該函數還用於返回其參數中提到的2個浮點數的餘數(模數)。
remainder = number – rquot * denom
其中rquot是以下結果:: numer /denom,四舍五入到最接近的整數值(中間情況四舍五入到偶數)。
用法:
double remainder(double a, double b) float remainder(float a, float b) long double remainder(long double a, long double b) 參數: a and b are the values of numerator and denominator. 返回: The remainder() function returns the floating point remainder of numerator/denominator rounded to nearest.
錯誤或異常:必須同時提供兩個參數,否則會產生錯誤-不能像這樣調用“ remainder()”的匹配函數。
#代碼1
// CPP program to demonstrate
// remainder() function
#include <cmath>
#include <iostream>
using namespace std;
int main()
{
double a, b;
double answer;
a = 50.35;
b = -4.1;
// here quotient is -12.2805 and rounded to nearest value then
// rquot = -12.
// remainder = 50.35 - (-12 * -4.1)
answer = remainder(a, b);
cout << "Remainder of " << a << "/" << b << " is " << answer << endl;
a = 16.80;
b = 3.5;
// here quotient is 4.8 and rounded to nearest value then
// rquot = -5.
// remainder = 16.80 - (5 * 3.5)
answer = remainder(a, b);
cout << "Remainder of " << a << "/" << b << " is " << answer << endl;
a = 16.80;
b = 0;
answer = remainder(a, b);
cout << "Remainder of " << a << "/" << b << " is " << answer << endl;
return 0;
}
輸出:
Remainder of 50.35/-4.1 is 1.15 Remainder of 16.8/3.5 is -0.7 Remainder of 16.8/0 is -nan
#代碼2
// CPP program to demonstrate
// remainder() function
#include <cmath>
#include <iostream>
using namespace std;
int main()
{
int a = 50;
double b = 41.35, answer;
answer = remainder(a, b);
cout << "Remainder of " << a << "/" << b << " = " << answer << endl;
return 0;
}
輸出:
Remainder of 50/41.35 = 8.65
相關用法
注:本文由純淨天空篩選整理自pawan_asipu大神的英文原創作品 remainder() in C++。非經特殊聲明,原始代碼版權歸原作者所有,本譯文未經允許或授權,請勿轉載或複製。