What does T::* mean in template's parameters?

Question

Following the article written in here :

I came across this code (shortened and changed for clarity):

template <class T> struct hasSerialize
{
    // This helper struct permits us to check that serialize is truly a method.
    // The second argument must be of the type of the first.
    // For instance reallyHas<int, 10> would be substituted by reallyHas<int, int 10> and works!
    // reallyHas<int, &C::serialize> would be substituted by reallyHas<int, int &C::serialize> and fail!
    // Note: It only works with integral constants and pointers (so function pointers work).
    // In our case we check that &C::serialize has the same signature as the first argument!
    // reallyHas<std::string (C::*)(), &C::serialize> should be substituted by 
    // reallyHas<std::string (C::*)(), std::string (C::*)() &C::serialize> and work!
    template <typename U, U u> struct reallyHas;

    // We accept a pointer to our helper struct, in order to avoid to instantiate a real instance of this type.
    // std::string (C::*)() is function pointer declaration.
    template <typename C>
    static char&
    test(reallyHas<std::string (C::*)(), &C::serialize>* /*unused*/) { }
};

So this

std::string (C::*)()

caught my attention.

Can anyone explain me the C::* part? That is a function pointer that returns std::string but what more?

Answer 1

A member function pointer to a member in class C that returns a std::string .

Check isocpp.org for more info on pointers to member functions.