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

As a related note, in C++0x you have the option of enforcing being explicit with your overrides via the new attribute syntax.

struct Base {
  virtual void Virtual();
  void NonVirtual();
};

struct Derived [[base_check]] : Base {
  //void Virtual(); //Error; didn't specify that you were overriding
  void Virtual [[override]](); //Not an error
  //void NonVirtual [[override]](); //Error; not virtual in Base
  //virtual void SomeRandomFunction [[override]](); //Error, doesn't exist in Base
};

You can also specify when you intend to hide a member via the [[hiding]] attribute. It makes your code somewhat more verbose, but it can catch a lot of annoying bugs at compile time, like if you did void Vritual() instead of void Virtual() and ended up introducing a whole new function when you meant to override an existing one.