Re: determine if a type is a free function pointer

I am trying to experiment with typetraits such that I can determine if a
 type is a free function pointer type. The following code example works
but it's not generic. As you can see, I have to enumerate the function
signatures. Is there a generic solution? (... doesn't work btw).

Fei

#include <iostream>

using namespace std;

template <typename T>
class typetraits{
    template <typename U>
    struct is_free_func_ptr {
        enum { result = false };
    };

    template <typename U>
    struct is_free_func_ptr<U (*)()>{
        enum { result = true };
    };

    template <typename U, typename V>
    struct is_free_func_ptr<U (*)(V)>{
        enum { result = true };
    };

public:
    enum { result = is_free_func_ptr<T>::result };
};

void foo() { }
void foo(int) { }

struct f{
    void operator ()() const {}
};

int main(){
    typedef void (*foo_fp)();
    typedef int (*foo_fp2)(int);
    cout << "result: " << typetraits<foo_fp>::result << endl;
    cout << "result: " << typetraits<foo_fp2>::result << endl;
    cout << "result: " << typetraits<void *>::result << endl;
    cout << "result: " << typetraits<void>::result << endl;
    cout << "result: " << typetraits<f>::result << endl;
}