为什么通用引用必须对成员函数使用模板而不是类？

Question

For example: This is NOT universal reference:

template<typename T>
class C {
  public:
    void gogo(T&& c);
};

But this is:

template<typename A>
class C {
  public:
    template<typename B>
    void gogo(B&& par);
}

Answer 1

In the second snippet, B&& par is not necessarily an universal reference. It's considered to be universal only if you let the compiler deduce B when the call is made:

C<float> c;
c.gogo(42); // T is deduced as int, `par` acts as an universal reference.
c.gogo<int>(42); // T is fixed, `par` is a regular rvalue reference.

Since in the first snippet T is always fixed when the call is made, the reference is never universal.

C<int> c; // Must specify T here, there's no way for it to be deduced.
c.gogo(42); // T is fixed, `par` is a regular rvalue reference.

Answer 2

This is because a universal, or a forwarding reference is, by definition an rvalue reference to a template parameter of the function call itself .

Deducing template arguments from a function call [temp.deduct.call]

A forwarding reference is an rvalue reference to a cv-unqualified template parameter that does not represent a template parameter of a class template

The standard explicitly calls out the fact that the template parameter is not a class template parameter. You can actually figure it out yourself:

template<typename T>
class C {
  public:
    void gogo(T&& c);
};

Here, goto() is not a template function. The template here is the class. Once the template class gets instantiated with some specific type, gogo() is just an ordinary, non-template class method that takes a parameter of a defined, specific type, an rvalue reference to whatever type the class template parameter is.