Quick Q: May a destructor be final?

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Quick A: Yes, as any other virtual member function.

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May a destructor be final?

May a C++ destructor be declared as final?

Yes.

And if so, does that prevent declaration of a derived class:

Yes, because the derived class would have to declare a destructor (either explicitly by you or implicitly by the compiler), and that destructor would be overriding a function declared final, which is ill-formed.

The rule is [class.virtual]/4:

If a virtual function f in some class B is marked with the virt-specifier final and in a class D derived from B a function D​::​f overrides B​::​f, the program is ill-formed.

It's the derivation itself that is ill-formed, it doesn't have to be used.

Is declaring a destructor to be final a workable idiom for indicating that a class is not intended to be used as a base class?

Effectively, but you should just mark the class final. It's quite a bit more explicit.

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