Is value returned by std::unique_ptr::get valid after moving unique_ptr?

Question

Consider the following code snippet:

class Owner {
public:
 Owner(std::unique_ptr<int> ptr) : owned_pointer<int>(std:move(ptr)) {}
private:
 std::unique_ptr<int> owned_pointer;
};


std::unique_ptr<int> ptr(new int);
int* ptr1 = ptr.get();
Owner new_owner(std::move(ptr));

Is it safe to assume that ptr1 is valid as long as new_owner stays in scope? It seems to work, but I can't find a specification that states that explicitly - is it undefined behavior/implementation specific and just happen to work for me, or the code posted above is valid (ptr1 is guaranteed to point to moved pointer as long as it stays alive)?

Answer 1

Yes, the C++11 specification guarantees that transferring ownership of an object from one unique_ptr to another unique_ptr does not change the location of the object itself, and that get() on the second unique_ptr returns the same as it would have on the first unique_ptr before the transfer.

Looking at N3337, section 20.7.1:

4. Additionally, u can, upon request, transfer ownership to another unique pointer u2. Upon completion of such a transfer, the following postconditions hold:

   • u2.p is equal to the pre-transfer u.p,
   • u.p is equal to nullptr, and
   • if the pre-transfer u.d maintained state, such state has been transferred to u2.d.

where u is a unique_ptr object that stores a pointer u.p.

The first bullet answers the question directly, since get() is specified as returning the u.p.