Why does the following program not select the argument of the same type as the first template parameter?

Question

I am trying to write a function such that f<T>(args..) returns the first parameter of type T.

The following program seems to always select the first specialization thus printing 97 (ASCII code of 'a'). Though the second one wouldn't require converting char to int. Could someone please explain the behavior?

I am new to SFINAE and meta-programming.

  #include <iostream>
  using namespace std;

  template <typename T, typename ...Ts>
  T f(T a, Ts... args) {
    return a;
  }

  template <typename R, typename T, typename ...Ts>
  R f(typename enable_if<!is_same<R, T>::value, T>::type a, Ts... args) {
    return f<R>(args...);
  }

  int main() {
    cout << f<int>('a', 12);
  }

Answer 1

The second template argument of the std::enable_if should be the R, which is what you desire to have.

Following should work

 template < typename R, typename T, typename ...Ts>
 typename enable_if<!is_same<R, T>::value, R>::type f(T const& t, Ts&&... args) 
 //                                       ^^^         ^^^^^^^^^^^
 {
       return f<R>(std::forward<Ts>(args)...); // forward the args further
 }