determine if a type is a free function pointer

Fei Liu <>
Wed, 12 Sep 2007 11:49:59 -0400
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).


#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 };

     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;

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