Re: Any way to detect the absense of virtual destructor in base class?

Alf P. Steinbach <alf.p.steinbach+usenet@gmail.com> wrote:


  How exactly do they achieve that? If I do, for example, this:

    std::unique_ptr<Base*> ptr = new Derived;

then how exactly is 'ptr' able to deduce the derived class destructor in
order to directly call it when it disposes of the object?

  (And where would it store it anyways? I thought the whole idea with
std::unique_ptr is that its size is that of one pointer, hence making
it as efficient as a raw pointer.)

  I wonder if you are being confused by shared_ptr (and possible other
smart pointers) being able to destroy an object even if they only see
a class declaration (rather than a definition), as long as the class was
fully declared at the point of construction of the pointer. (IOW the class
doesn't need to be fully declared at the place of destruction.) That's not
the same thing as a base-class type smart pointer being able to deduce
the derived-class type destructor.