How does returning std::make_unique work?

Question

I have a base class and its subclass:

class Base {
    public:
    virtual void hi() {
        cout << \"hi\" << endl;
    } 
};

class Derived : pub         


        

Answer 1

In every case but (2) the returned value was treated as (some kind of) rvalue. In (2) it was not, because the types did not match the implicit move was blocked.

In a later iteration of the standard, (2) would also implicitly move.

All of them shortly engage in undefined behaviour after being called, as they try to delete a Derived object via pointer-to-Base. To fix this, record a deleter function.

template
using smart_unique=std::unique_ptr;

template
smart_unique make_smart_unique( Args&&... args ){
  return {
    new T(std::forward(args)...),
    [](void*ptr){ delete static_cast(ptr); }
  };
}
template
static const smart_unique empty_smart_unique{ nullptr, [](void*){} };

These are unique pointers smart enough to handle polymorphism like shared_ptr can.