Is it possible to overload a function that can tell a fixed array from a pointer?

You may use the following:

namespace detail
{
    template <typename T> struct helper;

    template <typename T> struct helper<T*> { void operator() () const {std::cout << "pointer\n";} };
    template <typename T, std::size_t N> struct helper<T[N]> { void operator() ()const {std::cout << "array\n";} };
}


template <typename T>
void f(const T& )
{
    detail::helper<T>{}();
}

Live example

This seems to work for me

#include <iostream>

template<typename T>
std::enable_if_t<std::is_pointer<T>::value>
foo(T) { std::cout << "pointer\n"; }

template<typename T, std::size_t sz>
void foo(T(&)[sz]) { std::cout << "array\n"; }

int main()
{
  char const* c;
  foo(c);
  foo("hello");
}

Bonus std::experimental::type_traits:

using std::experimental::is_pointer_v;
std::enable_if_t<is_pointer_v<T>>

---

Your comment made me try something even simpler

template<typename T> void foo(T) { std::cout << "pointer\n"; }
template<typename T, unsigned sz> void foo(T(&)[sz]) { std::cout << "array\n"; }

Of course the problem here is that foo is now callable for any type, depends on how lax you want your parameter checking to be.

---

One other way is to (ab)use rvalue references

void foo(char const*&) { std::cout << "pointer\n"; }
void foo(char const*&&) { std::cout << "array\n"; }

Obviously it's not foolproof.