覆盖非虚拟功能

Question

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

Answer 1

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.

Answer 2

如果B:f是非虚拟的，则两个 D:f函数都将格式错误。

Answer 3

Yes, the program is ill formed when override is added to any non-virtual function.

Generally, functions with differing signatures (overloaded), are as different as functions with different names. The example given in the Spec is not meant to imply that the function name effects override . It's meant to show the common error that override is designed to prevent.