Why does this template function not want to be reloaded?

Question

C++ / C#

#include<iostream>
using namespace std;

template < typename T > void Function1(T a, T b);
void main() {
  int i;
  double d;
  char c;
  int b;
  cout << "Enter int number: ";
  cin >> i;
  Function1(i, b);
  cout << endl;
  cout << "Enter double number: ";
  cin >> d;
  Function1(d, b);
  cout << endl;
  cout << "Enter char number: ";
  cin >> c;
  Function1(c, b);
  cout << endl;
}
template < typename T >
void Function1(T a, T b = 2) {
  T c;
  for (int i = 0; i <= b; ++i) {
    c = a*a;
  }
  cout << c;
}

Answer 1

I will not paint all the problems of this code, I will focus on the main thing:
Function1 is declared as taking 2 arguments of the same type . In the second and third cases, you call it by slipping arguments of different types. Accordingly, the compiler cannot decide what T will be in this situation - double or int, char or int? Either make 2 template types, or make the 2nd argument non-template, or pass arguments of the same type, or explicitly tell the function when calling what T
is . , in the second - no.