const applied to “universal reference” parameter

问题

I've stumbled upon Scott Mayers article on universal references, link.

From what I understood universal reference, that is some type T&& can mean an rvalue or lvalue type in different contexts.

For example:

template<typename T>
void f(T&& param);               // deduced parameter type ⇒ type deduction;
                                 // && ≡ universal reference

In the above example depending on template parameter the T&& can be either an lvalue or rvalue, that is, it depends on how we call f

int x = 10;
f(x); // T&& is lvalue (reference)
f(10); // T&& is rvalue

However according to Scott, if we apply const to to above example the type T&& is always an rvalue:

template<typename T>
void f(const T&& param);               // “&&” means rvalue reference

Quote from the article:

Even the simple addition of a const qualifier is enough to disable the interpretation of “&&” as a universal reference:

Question: Why does const make an "universal" reference rvalue?

I think this is impossible because following code makes confusion then:

template<typename T>
void f(const T&& param);               // “&&” means rvalue reference

int x = 10;
f(x); // T&& is lvalue (reference) // how does 'x' suddenly become an rvalue because of const?
f(10); // T&& is rvalue // OK

来源：https://stackoverflow.com/questions/34556808/const-applied-to-universal-reference-parameter