Re: Using SFINAE with member function types

Below is some code illustrating the wrong way to do what I'd like to
do. I have a class "A" with function "void f(int,int)" and class "B"
without that function. Is there a way to construct a traits-like
class, Q, that contains a constant expression (Q<T>::value) that
evaluates to true if some class T has the function "void f
(int,int)" ?

Pretend A and B cannot be changed. You only have Q to work with.

struct A {
  void f (int,int){}
};

struct B{};

template <class T,class F=void>
struct Q {
  enum{value=false};
};

template<class T>
struct Q<T,void(T::*)(int,int)> {
  enum{value=true};
};

int main() {
  if( Q<A>::value ) {cout << "A";}
  if( !Q<B>::value ){cout << "B";}
}

I'd like this to output "AB". I've tried playing around with
boost::enable_if and boost::function_traits but have not come up with
a solution.

The above code is a trivial case. The real case is, I'm trying to come
up with a "is_serializable" type trait (Q) so that an I/O "write"-like
function does something special for non-serializable classes (and just
does normal serialization for those that are). Every serializable
class has this function defined (as a mem func):
template <typename Archive>
void serialize(Archive &, const unsigned int);

If I could add a typedef to each serializable class, no problem. I
could even make a typelist of all serializable classes and use a
visitor/enable_if or something but I'd like to consider these last
resorts.