Is the return expression treated as a temporary object in c++?

Question

class elem
{
public:
    elem()
    {
    }
    elem(elem &)
    {
    cout << 222222 << endl;
    }  
    elem(elem &&)
    {
    cout << 111111 << endl;
    }
};

elem fun()
{
    elem e;

    elem f(e);

    return e;
}

void main()
{
    fun();
}

Output:

22222

11111

My confusion:

As I return an lvalue in fun , why is fun not calling the copy constructor but move constructor? When does 'e' change to a rvalue in the return expression?

Note: I have turned off the copy elision.

Answer 1

If you declare a local variable within a function and return it out of the function, it is an rvalue. Also there is a binding rule which says... “The r-value reference in move constructor is preferred over const l-value reference in copy constructor — these are the rules for binding expression values to references. If move constructor is not available, the second preference — copy constructor — is chosen”. As from @Olef's comment, see also, Automatic move from local variables and parameters .