Is it useless to declare a local variable as rvalue-reference, e.g. T&& r = move(v)?

by

No, AnyTypeMovable&& r = move(v); here is not useful at all.

Consider the following code:

#include <iostream>
#include <vector>

class MyMovableType
{
        int i;
public:
        MyMovableType(int val): i(val){}
        MyMovableType(MyMovableType&& r) { this->i = r.i; r.i = -1; }
        MyMovableType(const MyMovableType& r){ this->i = r.i; }
        int getVal(){ return i; }
};

int main()
{
        std::vector<MyMovableType> vec;
        MyMovableType a(10);
        MyMovableType&& aa = std::move(a);

        vec.push_back(aa);

        std::cout << a.getVal() << std::endl;

        return 0;

}

As aa is an l-value (as noted by R. Martinho Fernandes, and also by Xeo – a named rvalue-reference is an lvalue), this will print 10 indicating that moving has not been performed (nor in the assignment, nor in the push_back call), so you still need to std::move it to the push_back method, as in this case:

#include <iostream>
#include <vector>

class MyMovableType
{
        int i;
public:
        MyMovableType(int val): i(val){}
        MyMovableType(MyMovableType&& r) { this->i = r.i; r.i = -1; }
        MyMovableType(const MyMovableType& r){ this->i = r.i; }
        int getVal(){ return i; }
};

int main()
{
        std::vector<MyMovableType> vec;
        MyMovableType a(10);
        MyMovableType&& aa = std::move(a);

        vec.push_back(std::move(aa));

        std::cout << a.getVal() << std::endl;

        return 0;

}

move will be performed, so the printout will be -1. So, despite the fact that you’re passing aa to the push_back, you still need to pass it via std::move.

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