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The C++11 FDIS it says

If a virtual function is marked with the virt-specifier override and does not override a member function of a base class, the program is ill-formed. [ Example:

struct B {
    virtual void f(int);
};
struct D : B {
    void f(long) override; // error: wrong signature overriding B::f
    void f(int) override; // OK
};

What if B::f would not have been marked virtual? Is the program ill-formed, then? Or is override then to be ignored`. I can not find any handling of this case in the std text.

Update 1/2 (merged) I forwarded a request to the C++ Editors to look into things. Thanks Johannes to pointing that out to me.

  • "void f(long) override" does not override a function, esp. no virtual one,
  • therefore it is not virtual
  • therefore the text "If a virtual function is marked with..." does not apply
  • therefore the example does not match the text.

But by realizing this I found, that the intention of the "override" contextual keyword can not be met: if a typo in the function name or the wrong argument type does make the function itself non-virtual, then the standard's text never applies -- and "override" is rendered useless.

The best possible solution may be

  • putting "virtual" in front of the example's functions
See Question&Answers more detail:os

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What if B::f would not have been marked virtual? Is the program ill-formed, then?

Yes, it is. Because in order to override something, that something has to be virtual. Otherwise it's not overriding, it's hiding. So, the positive answer follows from the quote in your question.


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