What happens to unique_ptr after std::move()?

Question

This code is what I want to do:

Tony& Movie::addTony()
{
    Tony *newTony = new Tony;
    std::unique_ptr tony(newTony);
    attachActor(std::m         


        

Answer 1

The standard says (§ 20.8.1.2.1 ¶ 16, emphasis added) that the move constructor of std::unique_ptr

unique_ptr(unique_ptr&& u) noexcept;

Constructs a unique_ptr by transferring ownership from u to *this.

Therefore, after you move-construct the temporary object that gets passed as argument to attachActor form your tony, tony no longer owns the object and hence tony.get() == nullptr. (This is one of the few cases where the standard library actually makes assertions about the state of a moved-away-from object.)

However, the desire to return the reference can be fulfilled without resorting to naked new and raw pointers.

Tony&
Movie::addTony()
{
  auto tony = std::make_unique();
  auto p = tony.get();
  attachActor(std::move(tony));
  return *p;
}

This code assumes that attachActor will not drop its argument on the floor. Otherwise, the pointer p would dangle after attachActor has returned. If this cannot be relied upon, you'll have to re-design your interface and use shared pointers instead.

std::shared_ptr
Movie::addTony()
{
  auto tony = std::make_shared();
  attachActor(tony);
  return tony;
}