Can the default destructor be generated as a virtual destructor automatically?

Question

Can the default destructor be generated as a virtual destructor automatically?

If I define a base class but no default destructor, is there a default virtual destructor generated automatically?

Answer 1

No. There is a cost associated with making a method virtual, and C++ has a philosophy of not making you pay for things that you don't explicitly state that you want to use. If a virtual destructor would have been generated automatically, you would have been paying the price automatically.

Why not just define an empty virtual destructor?

Answer 2

In C++ 11 you can use:

class MyClass
{
  // create a virtual, default destructor
  virtual ~MyClass() = default;
};

Answer 3

Yes, by inheriting from a base class with a virtual destructor. In this case, you already pay the price for a polymorphic class (e.g. vtable).

Answer 4

No, all destructor's are by default NOT virtual.

You will need to define a virtual destructor on all the base classes

In addition to that.

To quote Scott Meyers in his book "Effective C++":

The C++ language standard is unusually clear on this topic. When you try to delete a derived class object through a base class pointer and the base class has a non-virtual destructor (as EnemyTarget does), the results are undefined

In practice, it's usually a good idea to define a class with a virtual destructor if you think that someone might eventually create a derived class from it. I tend to just make all classes have virtual destructor's anyway. Yes, there is a cost associated with that, but the cost of not making it virtual more often that not out weighs a measly bit of run-time overhead.

I suggest, only make it non-virtual when you're absolutely certain that you want it that way rather than the rely on the default non-virtual that the compilers enforce. You may disagree, however (in summary) I recently had a horrid memory leak on some legacy code where all I did was add a std::vector into one of the classes that had existed for several years. It turns out that one of it's base classes didn't have a destructor defined (default destructor is empty, non-virtual!) and as no memory was being allocated like this before no memory leaked until that point. Many days of investigation and time wasted later...

Answer 5

Uri and Michael are right -- I'll just add that if what's bugging you is having to touch two files to declare and define the destructor, it's perfectly all right to define a minimal one inline in the header:

class MyClass
{
   // define basic destructor right here
   virtual ~MyClass(){}

   // but these functions can be defined in a different file
   void FuncA();
   int FuncB(int etc);
}

Answer 6

Currently, Uri is right. On the other hand, after you have declared a virtual method in your class, you are paying the price for the existence of the virtual table anyway. In fact, the compiler will warn you if your class has a virtual method, but no virtual destructor. This could become a candidate for automatic generation of the default virtual destructor instead of the pesky warning.

Answer 7

No. You need to declare it as virtual.

Answer 8

Some people stated that it is never defaulted to virtual. That's not exactly true.

Even though destructors are not inherited, if a base class declares its destructor virtual, the derived destructor always overrides it.

This means that if the base class has a virtual destructor you don't have to define a destructor of the derived class as virtual nor explicitly write:

/// All of these are unnecessary and won't change the destruction behavior
/// as long as Class inherits from a base class with a virtual destructor
virtual ~Class() = default;
// or
~Class() override = default;
// or
~Class() {}