Question

I am just a bit curious about the new function std::move() that has just been added into the newest C++ standard.

Finished reading an article about it and found out that the function’s definition is

 namespace std {
      template <class T>
      inline typename remove_reference<T>::type&& move(T&& x)
      {
           return x;
      }
 }

This seems like it doesn’t have any difference between calling std::move and using casting.

For example here,

class NPC{
    int m_number1;
    int m_number2;
public:
    NPC() : m_number1(1), m_number2(2) {
        cout << "NPC Constructor called!" << endl;
    }

    NPC(NPC&& _npc) : m_number1(_npc.m_number1), m_number2(_npc.m_number2) {
        _npc.m_number1 = 0;
        _npc.m_number2 = 0;

        cout << "NPC Move Constructor called!" << endl;
    }
};


int main() {
    NPC n1;
    NPC n2 = std::move(n1);

    cout << "----------------" << endl;
    NPC n3;
    NPC n4 = (NPC&&)n3;
    return 0;
}

Is it right to think that there is basically no difference?
Well, I am pretty much sure that I am right but also know that being too confident always backfires.

Thanks in advance!

Answer 1

std::move is defined as a specialized cast:

static_cast<typename remove_reference<T>::type&&>(t)

You could do this if you really want. But it would be much shorter and obvious to everyone what you’re doing if you just use std::move. The rules around && casting are weird, so it’s best to just use move and forward when that’s what you mean.