Is std::unique_ptr required to know the full definition of T?

Question

I have some code in a header that looks like this:

#include 

class Thing;

class MyClass
{
    std::unique_ptr< Thing > my_thing;
};
         


        

Answer 1

This isn't implementation-dependent. The reason that it works is because shared_ptr determines the correct destructor to call at run-time - it isn't part of the type signature. However, unique_ptr's destructor is part of its type, and it must be known at compile-time.