virtual desctructor on pure abstract base class

Question

I have

struct IMyInterface
{
   virtual method1() = 0;
   virtual method2() = 0;
};

GCC insists that I have

struct IMyInterface
{
   virtual method1() = 0;
   virtual method2() = 0;
   virtual ~IMyInterface(){};
};

I dont see why. A pure interface is all about the interface (duh). The destructor is part of the internal implementation details of a concrete implementer of the interface; it does not form part of the interface. I understand the whole slicing issue (or at least I think I do)

So my question is - is GCC right to insist on it and if so why?

Answer 1

According to the C++ spec, yes.

You need to declare the destructor virtual because otherwise, later

    IMyInterface * ptr = getARealOne();
    delete ptr;

won't call the destructor on the derived class (because the destructor isn't in the VTable)

It needs to be non-pure because base class destructors are always called by the sub-class destructor.

To further explain, C++ doesn't have a concept of an interface in the same way that Java or C# do. It's just a convention to use only pure-virtual methods, and think of that as an interface. The other rules about C++ destructors make it need to be non-pure, which breaks the similarity to interfaces in other languages, but those languages didn't exist at the time these rules were made.

Answer 2

If you don't declare the virtual d'tor in the base class, deleting objects of derived classes through a pointer to the base class leads to the wrong destructor being called, and thus to undefined behaviour and resource leaking.

struct A {

  virtual ~A() {}

};

struct B : A {

   std::string us_constitution;  
};


B* pb = new B();
A* pa = pb;

delete pa; // without the virtual d'tor in the base class, 'B::us_constitution' would never be freed.

Answer 3

The code you specified there is not a destructor. A destructor would be something "virtual ~IMyInterface();".