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

Question

Could you guys give me an illustrative example under certain circumstance to prove the following statements are useful and necessary?

AnyTypeMovable   v; AnyTypeMovable&& r = move(v); 

Answer 1

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.