Why rvalue reference pass as lvalue reference?

Question

pass() reference argument and pass it to reference, however a rvalue argument actually called the reference(int&) instead of

Answer 1

template 
void pass(T&& v) {
    reference(v);
}

You are using a Forwarding reference here quite alright, but the fact that there is now a name v, it's considered an lvalue to an rvalue reference.

Simply put, anything that has a name is an lvalue. This is why Perfect Forwarding is needed, to get full semantics, use std::forward

template 
void pass(T&& v) {
    reference(std::forward(v));
}

What std::forward does is simply to do something like this

template 
void pass(T&& v) {
    reference(static_cast(v));
}

See this;