Re: Is finalizing a virtual method possible in C++?
* Jeff Schwab:
Alf P. Steinbach wrote:
* Kira Yamato:
On 2008-02-22 07:35:34 -0500, Pavel Lepin <p.lepin@ctncorp.com> said:
Kira Yamato <kirakun@earthlink.net> wrote in
<2008022206570016807-kirakun@earthlinknet>:
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.
<http://www.parashift.com/c++-faq-lite/>
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;
private:
struct OverrideTag {};
public:
virtual void foo( OverrideTag = OverrideTag() ) = 0;
};
class Derived: public Base
{
public:
virtual void foo( OverrideTag = OverrideTag() ) {}
};
class FurtherDerived: public Derived
{
public:
// Nix njet.
// virtual void foo( OverrideTag = OverrideTag() ) {}
};
int main()
{
Derived o;
o.foo();
}
What's to stop somebody from making OverrideTag public in the derived
class?
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.
Cheers,
- Alf
--
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