Why is 'virtual' optional for overridden methods in derived classes?

Question

When a method is declared as virtual in a class, its overrides in derived classes are automatically considered virtual as well, and the C++ language ma

Answer 1

Yeah, it would really be nicer to make the compiler enforce the virtual in this case, and I agree that this is a error in design that is maintained for backwards compatibility.

However there's one trick that would be impossible without it:

class NonVirtualBase {
  void func() {};
};

class VirtualBase {
  virtual void func() = 0;
};

template
class CompileTimeVirtualityChoice : public VirtualChoice {
  void func() {}
};

With the above we have compile time choice wether we want virtuality of func or not:

CompileTimeVirtualityChoice -- func is virtual
CompileTimeVirtualityChoice -- func is not virtual

... but agreed, it's a minor benefit for the cost of seeking a function's virtuality, and myself, I always try to type virtual everywhere where applicable.