lvalue binding to rvalue reference

Question

I am trying to understand how lvalues bind to rvalue references. Consider this code:

#include 

template
void f(T&&         


        

Answer 1

Since T is a template argument, T&& becomes a forwarding-reference. Due to reference collapsing rules, f(T& &&) becomes f(T&) for lvalues and f(T &&) becomes f(T&&) for rvalues.

Answer 2

0x499602D2 has already answered your question; nevertheless, the following changes to your code might give further insights.

I have added a static_assert to f to check the deduced type:

#include <type_traits>

template<typename T>
void f(T&& x) {
    static_assert(std::is_same<T&&, int&>::value,"");
    std::cout << x;
}

The assert does not fail, so the type of x in f is eventually int& (in this particular example).

I have changed how g is called in main:

g(std::move(x));

Now the code compiles and the program works as expected and prints 44.

Hope this helps a bit in understanding rvalue references.