Protected non-virtual destructor in the base class

Question

I am trying to understand virtual destructors. The following is a copy paste from this page When to use virtual destructors?

Here, you'll notice that I didn't declare Base's destructor to be virtual. Now, let's have a look at the following snippet:

Base *b = new Derived(); // use b 
delete b; // Here's the problem!

[...] If you want to prevent the deletion of an instance through a base class pointer, you can make the base class destructor protected and non-virtual; by doing so, the compiler won't let you call delete on a base class pointer.

I don't understand why the deletion is prevented by having a protected non-virtual base class destructor. Doesn't the compiler think that we're trying to call delete from a base class object? What does protected have to do with that?

Answer 1

The C++ Standard has this to say about delete (section 5.3.5p10):

Access and ambiguity control are done for both the deallocation function and the destructor (12.4, 12.5).

Therefore, only code that has access to the destructor is able to use delete. Since the destructor is protected, that means that no one can call delete on a pointer of type Base*. Only subclasses can use the destructor at all (and the only thing that will is the subclass's own destructor, as part of the subobject destruction process).

Of course, the subclass should make its own destructor public, allowing you to delete objects through the subclass type (assuming that is the correct actual type).

NOTE: Actually, other members of Base can do delete (Base*)p; since they have access. But C++ assumes that someone using this construct will not be doing that -- C++ access control only provides guidance to code outside your class.