Function template with explicit specification of type

Question

I'm trying to understand below sample code I have. I know we can explicitly specify data type but not sure what "int, double" and "int, int" means. And why do we write function template that way instead of T tripleit (T Val) { T temp = val * 3; }? Thanks in advance.

#include <iostream>

using namespace std;
// why do we write this way instead of T tripleit (T Val) { T temp = val * 3; }?
template <class T, class U>
T tripleit (U val)
{
    T temp = val * 3;

}


int main()
{
    int a=5;
    double d=3.3;

    cout << "Explicit int; int argument: " << tripleit<int>(a) << endl;
    cout << "Explicit int; double argument: " << tripleit<int>(d) << endl;

    // what does <int, double> and <int, int> mean?
    cout << "Explicit int, double; double argument: " << tripleit<int, double>(d) << endl;
    cout << "Explicit int, int; double argument: " << tripleit<int, int>(d) << endl;

    return 0;
}

By the way, the output is :

Explicit int; int argument: 15

Explicit int; double argument: 9

Explicit int, double; double argument: 9

Explicit int, int; double argument: 9

Answer 1

If the template parameters had more descriptive names, they could look like this:

template <class ReturnType, class ParameterType>
ReturnType tripleit (ParameterType val)
{
  ReturnType temp = val * 3;
  return temp;  // I assume this line is actually present in your code, it makes no sense otherwise
}

With these names, it should be a bit more clear. The function can be used to multiply a number by 3 and convert it to a desired type at the same time.

Invoking the template with both template arguments specified simply suppresses template argument deduction. I think the really interesting case is missing there:

cout << "Explicit double, int; double argument: " << tripleit<double, int>(d) << '\n';

This will pass in a double value 3.3. However, as ParameterType is explicitly specified to be int, the value will be converted to int (perhaps with a warning). Inside the function, temp will be of type double (the first template argument), but the return value will still be 9, so the expected output is 9 or perhaps 9.0 or 9.0e0 depending on the current cout settings for floating-point numbers.