Re: Pure virtual functions and multiple inheritance

From:
Kaz Kylheku <kkylheku@gmail.com>
Newsgroups:
comp.lang.c,comp.lang.c++
Date:
Sun, 1 Feb 2009 08:09:11 +0000 (UTC)
Message-ID:
<20090207032204.139@gmail.com>
["Followup-To:" header set to comp.lang.c.]
On 2009-01-30, Jeff Schwab <jeff@schwabcenter.com> wrote:

Kevin Smith wrote:

Can I provide the implementation of a pure virtual function by inheriting
from another class? eg.:

class A{
    public:
        virtual void f() = 0;

};

class B{
    public:
        void f(){};

};

class C: public A, public B{};

My compiler (VC++) tells me that C is an abstract class, even though there
is a public implementation of f() in C. Is this Standard?


Yes.


Yes, what? It is standard?

What you want is called a "mixin." Of course, the "easy" way is:

struct A { virtual void f() =0; };
struct B: virtual A { void f() { } };


Virtual base classes are a complicated way to solve this. And also,
you are editing one of the original classes. B::f is now virtual,
and overrides the pure virtual A::f.

The virtual inheritance of A ensures that the C object gets only one A. Since
the object has only one A, and since it contains a B which seals the pure
virtual A::f with B::f, the C object has no pure virtuals.

This is like duct-tape.

(In some 13 years of commercial C++ development, I've yet to find use for a
virtual base class, and I'm not sure you understand what they are for).

struct C: virtual A, B { };


See there is no reason for C now to inherit A directly, since it's
getting it from B. You can change this whole thing to:

  struct A { virtual void f() = 0; };
  struct B: public A { void f() { } };
  struct C: public B { };

Look ma, no virtual inheritance.

If that's not an option for you, try this:

struct A { virtual void f() =0; };
struct B { void f() { } };
struct B_mixin: virtual A, private B { void f() { B::f(); } };
struct C: virtual A, B_mixin { };


Yuck. Five years in the Java slammer for you!

But I know a good language lawyer who may be able to get you three.

But at least this doesn't touch A and B. But note that instead
of this B_mixin implementing the override, C can just do it.

The above is a complicated way of doing this:

  struct A { virtual void f() =0; };
  struct B { void f() { } };
  struct C: public A, public B { void f() { B::f() } };

No mixin, no virtual bases. Keep it simple and stupid.

Also note that C doesn't actually solve any problem in your
above solution. So what we can do is simply drop C,
and rename B_mixin to C:

  struct A { virtual void f() =0; };
  struct B { void f() { } };
  // C was called B_mixin; original C gone
  struct C: virtual A, private B { void f() { B::f(); } };

Drop the virtual from virtual A, and you get the same thing
again.

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