Re: Is finalizing a virtual method possible in C++?

"Alf P. Steinbach" <>
Fri, 22 Feb 2008 20:57:41 +0100
* Jeff Schwab:

Alf P. Steinbach wrote:

* Kira Yamato:

On 2008-02-22 07:35:34 -0500, Pavel Lepin <> said:

Kira Yamato <> wrote in

So, I don't want to say too hestantly that it is
impossible to simulate the finalizing of virtual methods
and classes in C++.

Hence, I rather turn to the experts and ask if anyone
knows if it is possible to simulate them in any way?

This is a FAQ.


Yes. Thanks. But the FAQ does not say that finalizing method is
impossible. If in fact it did, then the answer is found. Instead,
it just proposes a "solution."

So let me try asking differently, has anyone gotten further with a
better solution than merely stating a comment?

There should be no reason to finalize a member function in C++.

But if you absolutely must it's not that difficult.

    class Base
    friend class Derived;
        struct OverrideTag {};
        virtual void foo( OverrideTag = OverrideTag() ) = 0;

    class Derived: public Base
        virtual void foo( OverrideTag = OverrideTag() ) {}

    class FurtherDerived: public Derived
        // Nix njet.
    // virtual void foo( OverrideTag = OverrideTag() ) {}

    int main()
        Derived o;;

What's to stop somebody from making OverrideTag public in the derived

struct Derived: Base {
    using Base::OverrideTag;
    virtual void foo( OverrideTag = OverrideTag() ) { }

Nothing. So Base just provides a service to Derived: "you can make this
function final if you want, and that's what you get by default". The
whole thing about support in Base is just because C++ doesn't have
'final', which would only appear in Derived, none of Base's business.

But I gather, if someone want's it bad enough, they can add support in
the base class...

Or introduce an artifical base class.


- Alf

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