Why does std :: move () from unique_ptr from the <unique_ptr> list really move it?

Question

using Ptr = std::unique_ptr<int>;

Ptr f(bool arg) {
  std::list<Ptr> list;
  Ptr ptr(new int(1));
  list.push_back(std::move(ptr));

  if (arg) {
    Ptr&& obj1 = std::move(list.front());
    // Here |obj1| and |list.front()| still point to the same location!
    list.pop_front();
    return std::move(obj1);
  }
  else {
    Ptr obj2 = std::move(list.front());
    list.pop_front();
    return obj2;
  }
};

Ptr&& ptr1 = f(true);   // |ptr1| is empty.
Ptr&& ptr2 = f(false);  // |ptr2| is fine.

Full source here .

I do not understand - why obj1and list.front()still point to the same place after the call std::move()?

Answer 1

You have a link to the std::unique_ptrinside list.

Ptr&& obj1 = std::move(list.front());
// ^^ THIS

So when you do

list.pop_front();

The unique pointer that is listdestroyed, and you are left with a saggy reference to some object that is already destroyed, which is illegally used, returning it from the function.

UPDATE: std::move std::unique_ptr. std::move , , r- . , :

Ptr&& obj1 = std::move(list.front());
assert(obj1.get() == list.front().get());

else , .

Ptr obj2 = std::move(list.front());
list.pop_front();  // OK; destroys the moved-from unique_ptr
return obj2;       // OK; obj2 points to something valid and is not a dangling reference

@nosid , , .

Answer 2

move-semantics std::unique_ptr . move-semantics .

list.push_back(std::move(ptr)); ptr nullptr ( ).

, arg , list.front() , std::move r-value r-value obj1. , , r-value . r- - , , .

, cout,

Ptr& obj1 = list.front();

, rj- obj1, , ,

Ptr&& obj1 = std::move(list.front());
std::cout << obj1.get() << std::endl << list.front().get() << std::endl; // same address
obj1.reset(new int(2));
std::cout << obj1.get() << std::endl << list.front().get() << std::endl; // same other address

, ,

Ptr obj1 = std::move(list.front());

(, , nullptr-ify ) obj1.

, ,

list.pop_front();

unique_ptr r-value undefined. ( ).