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

Question

Motivation:

Almost for fun, I am trying to write a function overload that can tell apart whether the argument is a fixed-size array or a pointer.

double const  d[] = {1.,2.,3.}; double a; double const* p = &a; f(d); // call array version f(p); // call pointer version 

I find this particularly difficult because of the well known fact that arrays decay to pointer sooner than later. A naive approach would be to write

void f(double const* c){...} template<size_t N> void f(double const(&a)[N]){...} 

Unfortunately this doesn't work. Because in the best case the compiler determines the array call f(d) above to be ambiguous.

Partial solution:

I tried many things and the closest I could get was the following concrete code. Note also that, in this example code I use char instead of double, but it is very similar at the end.

First, I have to use SFINAE to disable conversions (from array ref to ptr) in the pointer version of the function. Second I had to overload for all possible arrays sizes (manually).

[compilable code]

#include<type_traits> // for enable_if (use boost enable_if in C++98) #include<iostream> template<class Char, typename = typename std::enable_if<std::is_same<Char, char>::value>::type> void f(Char const* dptr){std::cout << "ptr" << std::endl;} // preferred it seems  void f(char const (&darr)[0] ){std::cout << "const arr" << std::endl;} void f(char const (&darr)[1] ){std::cout << "const arr" << std::endl;} void f(char const (&darr)[2] ){std::cout << "const arr" << std::endl;} void f(char const (&darr)[3] ){std::cout << "const arr" << std::endl;} void f(char const (&darr)[4] ){std::cout << "const arr" << std::endl;} void f(char const (&darr)[5] ){std::cout << "const arr" << std::endl;} void f(char const (&darr)[6] ){std::cout << "const arr" << std::endl;} // this is the one called in this particular example // ad infinitum ...  int main(){          f("hello"); // print ptr, ok because this is the fixed size array     f(std::string("hello").c_str()); // print arr, ok because `c_str()` is a pointer } 

This works, but the problem is that I have to repeat the function for all possible values of N and using template<size_t N> gets me back to square zero, because with the template parameter the two calls get back to equal footing. In other words, template<size_t N> void f(char const(&a)[N]){std::cout << "const arr" << std::endl;} doesn't help.

Is there any way to generalize the second overload without falling back to an ambiguous call? or is there some other approach?

A C++ or C++1XYZ answer is also welcome.

Two details: 1) I used clang for the experiments above, 2) the actual f will end up being an operator<<, I think know if that will matter for the solution.

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Summary of solutions (based on other people's below) and adapted to a the concrete type char of the example. Both seem to depend on making the char const* pointer less obvious for the compiler:

1. One weird (portable?), (from the comment of @dyp.) Adding a reference qualifier to the pointer version:

    template<class Char, typename = typename std::enable_if<std::is_same<Char, char>::value>::type>     void f(Char const* const& dptr){std::cout << "ptr" << std::endl;}          template<size_t N>     void f(char const (&darr)[N] ){std::cout << "const arr" << std::endl;} 

2. One elegant (special case from @user657267)

    template<class CharConstPtr, typename = typename std::enable_if<std::is_same<CharConstPtr, char const*>::value>::type>     void f(CharConstPtr dptr){std::cout << "ptr" << std::endl;}          template<size_t N>     void f(char const (&darr)[N] ){std::cout << "const arr" << std::endl;} 

Answer 1

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>> 

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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.

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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.

Answer 2

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