Select function based on if template is pointer/reference or none

Question

I would like to provide different implementations of a function dependant on if it is a pointer, a reference or a regular type. This is my code so far:

template<class T,
         class = typename std::enable_if_t<std::is_reference<T>::value>>
void f(T && in)
{}

// This causes redefinition error 
template<class T,
    class = typename std::enable_if_t<std::is_pointer<T>::value>>
void f(T && in)
{}

template<class T,
    class = typename std::enable_if_t<!std::is_reference<T>::value>,
    class = typename std::enable_if_t<!std::is_pointer<T>::value>>
void f(T && in)
{}

The middle function causes:

12:13: error: redefinition of 'template void f(T&&)'

7:13: note: 'template void f(T&&)' previously declared here

Funnily only the first and last function together compile.

Any ideas how to fix it or simplify this code.

Answer 1

The usual way is to provide appropriate overloads:

#include <iostream>

template <class T> void f(T&&) {
    std::cout << "T&&\n";
}

template <class T> void f(T*) {
    std::cout << "T*\n";
}

template <class T> void f(T&) {
    std::cout << "T&\n";
}

int main() {
    int i;
    f(std::move(i));
    f(&i);
    f(i);
}

This produces the following output:

[temp]$ clang++ -std=c++11 test.cpp
[temp]$ ./a.out
T&&
T*
T&
[temp]$