Consider the following code:

class Widget {
public:
    Widget() {
        std::cout << "nCtor";
    }
    Widget(const Widget& obj) {
        std::cout << "nCopy ctor";
    }
    Widget(Widget&& obj) noexcept {
        std::cout << "nMove Ctor";
    }
};

Widget foo(Widget obj) {
    return obj;
}

And the caller side:

foo(std::move(w));

In this case we will have

Ctor
Move Ctor
Move Ctor

Even though we receive obj in foo by value, the move constructor will be called due to the std::move. The second call of the move constructor will be done by the compiler as an optimization (is that correct?)

However, if we will change foo to the

Widget foo(Widget&& obj) {
    return obj;
}

the output will be

Ctor
Copy ctor

So my question is: why is it called copy constructor instead of the move constructor in the return statement, as before? We can only force the move constructor to be called by return std::move(obj)