C++ template for all pointers and template for all arrays

Question

I'm looking for a solution for the following problem: I have a class in which I want to overload an operator (in this example &) for all types of pointers and for all types of arrays. Inside the implementation for arrays I need to have access to the arraysize and inside the implementation for pointers I must be able to do something with the dereferenced object.

As pointed out here, the way for the arrays is quite clear:

template<typename T, unsigned int N>
void operator&(T (&arr)[N])
{
    cout << "general array operator: " << N << "\r\n";
}

But for the pointers neither of the following works:

// if I use this, the operator gets ambigous for arrays
template<typename T>
inline void operator&(T* p)
{
    cout << "general pointer operator: " << (*p) << "\r\n";
}
// this doesn't work because one cannot dereference void* 
void operator&(void* p)
{
    cout << "general pointer operator\r\n";
    (*this) & (*p);
}

Is there any good and clean solution to achieve different behaviour of an operator for arbitrary arrays and arbitrary pointers?

Here is a complete example code:

#include <iostream>

struct Class
{
    template<typename T>
    void operator&(T* p)
    {
        std::cout << "general pointer operator" << (*p) << std::endl;
    }

    template<typename T, unsigned int N>
    void operator&(T (&arr)[N])
    {
        std::cout << "general array operator" << N << std::endl;
    }
};

int main()
{
    int myarr[5];
    int* p = myarr;
    Class obj;

    obj & myarr; // error: operator is ambigous
    obj & p; // works

    return 0;
}

Answer 1

I have to admit that I have no idea why your snippet fails to compile properly. Anyway, a good old tag dispatching workaround seems to be working.

class cClass
{

public:
    template<class T, size_t N>
    void impl(T (&x)[N], std::true_type)
    {
        cout << "general array operator" << N << '\n';
    }

    template<typename T>
    void impl(T* p, std::false_type)
    {
        cout << "general pointer operator" << (*p) << '\n';
    }

    template<typename T>
    void operator&(T && x)
    {
        impl( std::forward<T>(x), std::is_array< typename std::remove_reference<T>::type >() );
    }

};

Answer 2

The solution that changes the least code is:

template<typename T>
void operator&(T*const& p)

which gets rid of the ambiguity. I'd go with tag dispatching myself.