How get the class (object type) from pointer to method

Question

I have a pointer to the method:

struct A { int method() { return 0; } };
auto fn = &A::method;

I can get a return type by std::result_of, but how I can get from fn the class owner of the method?

Answer 1

Try this:

template<class T>
struct MethodInfo;

template<class C, class R, class... A>
struct MethodInfo<R(C::*)(A...)> //method pointer
{
    typedef C ClassType;
    typedef R ReturnType;
    typedef std::tuple<A...> ArgsTuple;
};

template<class C, class R, class... A>
struct MethodInfo<R(C::*)(A...) const> : MethodInfo<R(C::*)(A...)> {}; //const method pointer

template<class C, class R, class... A>
struct MethodInfo<R(C::*)(A...) volatile> : MethodInfo<R(C::*)(A...)> {}; //volatile method pointer

Answer 2

You can match it using class-template-specialization:

//Primary template
template<typename T> struct ClassOf {};

//Thanks T.C for suggesting leaving out the funtion /^argument
template<typename Return, typename Class>
struct ClassOf<Return (Class::*)>{   using type = Class;    };

//An alias
template< typename T> using ClassOf_t = typename ClassOf<T>::type;

Hence given:

struct A { int method() { return 0; } };
auto fn = &A::method;

We can retrieve the class like:

ClassOf_t<decltype(fn)> a;

Full example Here.