Why are rvalues references variables not rvalue?

Question

Let\'s say I have two overloads of a function f. f(T&&) and f(T&). Then in the body of g: g(T&&

Answer 1

In g() t is a named variable. All named variables are lvalues. If T is a template type then you can forward the variable to f() using std::forward. This will call f() with the same type as was passed to g()

template
g(T&& t) { f(std::forward(t));}

If T is not a template type but just a type then you can use std::move

g(T&& t) { f(std:move(t)); }