如何计算c ++中两个数字的差异？

Question

If I enable double and integer only, then it is 4 functions. But I want to enable all data types (int long float double unsigned numbers etc.) How is it possible?

#include <iostream>

using namespace std;

double diff(int num1, int num2) {
    return double(num1-num2);
}

double diff(int num1, double num2) {
    return double(num1)-num2;
}

double diff(double num1, int num2) {
    return num1-double(num2);
}

double diff(double num1, double num2) {
    return num1-num2;
}

int main() {
    int a = 10;
    double b = 4.4;
    cout << diff(a, b) << endl;
    return 0;
}

Answer 1

template <typename T, typename U>
double diff(T a, U b) {
    return a - b;
}

You don't need the cast to double -- this is done for you if either argument is a double , and during return when both are integers. However,

double diff(double a, double b);

can be called with int arguments as well.

Answer 2

Use a template function:

template <typename T1, typename T2>
double diff(const T1& lhs, const T2& rhs)
{
  return lhs - rhs;
}

Answer 3

您不必“启用”操作，只需写：

cout << (a - b) << endl;

Answer 4

Unlike all of previous answers I would add about C++11. In C++11 you can use decltype .

#include <iostream>

template <typename T1, typename T2>
auto diff(T1 a, T2 b) -> decltype(a)
{
   return (a - b);
}

int main() {
   std::cout << diff(3.5, 1) << std::endl;
   std::cout << diff(3, 1.5) << std::endl;
}

diff function will always return value of type like first argument. Note in first case it is float number, but in second it is integer.

Answer 5

You could define a template for the same

template <typename T, typename U>
T diff(const T& a, const U& b) {
    return a - b;
}

This code makes a lot of assumptions, like operator - is defined for T, and return type will be always of type T and so on...

Answer 6

You could always calculate the difference using absolute values, for instance

cout << abs(a - b) << endl;

you might want to use templates like previous answers said though.