Question

This code works as expected (online here).
At the end v is empty and w is not empty as it has pilfered the contents of v.

    vector<int> v;
    v.push_back(1);
    cout << "v.size(): " << v.size() << endl;
    auto vp = move(v);
    vector<int> w(vp);
    cout << "w.size(): " << w.size() << endl;
    cout << "v.size(): " << v.size() << endl;

But if I replace auto vp=move(v) with

    vector<int> && vp = move (v);

Then it doesn’t move. Instead it copies and both vectors are non-empty at the end. As shown here.

Clarification: More specifically, what is the auto-derived type of vp? If it’s not vector<int> &&, then what else could it be? Why do the two examples give different results despite being so similar?

Extra: I also tried this, and it still copied instead of moving

    std :: remove_reference< vector<int> > :: type && vp = move(v);

Answer 1

Edit for OP’s clarification: the auto-derived type of move(v) is vector<int>. See C++11 "auto" semantics.

The 1st example does this:

move 'v' into 'vp'
copy 'vp' into 'w'

and the 2nd example does this:

set 'vp' as rvalue-reference of 'v'
copy 'vp' (which is 'v') into 'w'

---

What std:move does is simply casting a type to an rvalue (See What is std::move(), and when should it be used?). Therefore, in

vector<int>&& vp = move(v);

it’s just setting the rvalue-reference vp to v and do nothing else. Also, an rvalue-reference is an lvalue (it has a name), so

vector<int> w(vp);

will call the copy constructor to copy vp (which is v) into w.

It will call the move constructor if you make vp an rvalue (Example):

vector<int> w(move(vp))

You may want to read this: C++ Rvalue References Explained.