Re: I wish c++ did interfaces better.

From:
Pete Becker <pete@versatilecoding.com>
Newsgroups:
comp.lang.c++
Date:
Wed, 6 Aug 2008 21:30:54 -0400
Message-ID:
<2008080621305416807-pete@versatilecodingcom>
On 2008-08-06 20:51:21 -0400, Stuart Golodetz
<sgolodetz@NdOiSaPlA.pMiPpLeExA.ScEom> said:

Ok, I've recreated the problem:

struct I1
{
    virtual void f() = 0;
};

struct I2 : virtual I1
{
    virtual void g() = 0;
};

struct C1 : virtual I1
{
    void f()
    {
        std::cout << "f()" << std::endl;
    }
};

struct C2 : C1, virtual I2
{
    void g()
    {
        std::cout << "g()" << std::endl;
    }
};


[example simplified]

When I do this, I get compiler warnings:

Warning 1 warning C4250: 'C2' : inherits 'C1::C1::f' via dominance


[additional warnings elided]

Please can someone explain what's going on?


The compiler is warning you that it's doing what the language
definition says it should do, and that you might not be smart enough to
understand what you've done. But you are: the code is exactly right.

Here's what's going on: C2 sees two definitons of f, one from C1 and
one from i2 (which inherits it from i1). That would be ambiguous if it
weren't for the dominance rule. Both C1 and i2 have i1 as a virtual
base, so they both see the declaration of i1::f. C1 overrides i1::f,
and i2 does not override it. C2 inherits from both, and the dominance
rule says that a call to f on a C2 object is not ambiguous and calls
C1::f. The definition of f that's nearer to C2 in the hierarchy
dominates the one that's farther away. This rule only applies when all
the inheritance paths lead to a common virtual base type that declares
the function and only one of those paths has an overriding declaration.
So if i2 also defined f, the call from C2 would be ambiguous. Got it?
<g>

--
  Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

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