Perfect forwarding a member of object

问题

Suppose I have two structs:

struct X {};
struct Y { X x; }

I have functions:

void f(X&);
void f(X&&);

How do I write a function g() that takes Y& or Y&& but perfect forwarding X& or X&& to f(), respectively:

template <typename T>
void g(T&& t) {
  if (is_lvalue_reference<T>::value) {
    f(t.x);
  } else {
    f(move(t.x));
  }
}

The above code illustrate my intention but is not very scalable as the number of parameters grows. Is there a way make it work for perfect forwarding and make it scalable?

回答1:

template <typename T>
void g(T&& t) {
  f(std::forward<T>(t).x);
}

回答2:

I think this will work, although I'm not sure:

template<class T, class M>
struct mforward {
  using type = M&&; 
};
template<class T, class M>
struct mforward<T&, M> {
  using type = M&; 
};

template <typename T>
void g(T&& t) {
  f(std::forward<typename mforward<T, decltype(t.x)>::type>(t.x));
}

来源：https://stackoverflow.com/questions/8570655/perfect-forwarding-a-member-of-object