Override virtual function of base classes, which do not share common interface

Question

#include <iostream>
struct B1
{
    virtual void method()=0;
    virtual ~B1(){}
};

struct B2
{
    virtual void method()=0;
    virtual ~B2(){}
};

struct D: B1, B2
{
    virtual void method()
    {
        std::cout << "D::method\n";
    };
};

int main(int argc,char *argv[])
{
    D d;
    B1 &b1=d;
    B2 &b2=d;
    b1.method();
    b2.method();
    return 0;
}

Note, B1 and B2 do not share common interface.

Is it this legal? If yes - in which standard? C++98/03/11 ?

Both, msvc and gcc have compiled it OK.

Previously I thought, that I have to use some common interface for such case (possible virtual inheritence).

Does such situation have some special name?

How it works in details, please? Maybe some ISO references?

Answer 1

Your code is well-formed: void D::method() overrides both void B1::method() and void B2::method().

The specification states (C++11 §10.3/2):

If a virtual member function vf is declared in a class Base and in a class Derived, derived directly or indirectly from Base, a member function vf with the same name, parameter-type-list, cv-qualification, and ref-qualifier (or absence of same) as Base::vf is declared, then Derived::vf is also virtual (whether or not it is so declared) and it overrides Base::vf.

B1 declares a virtual member function void B1::method(). Class D is derived from B1 and it also declares a member function with the same name (method), the same parameter list (no parameters), the same cv-qualification (no qualification) and the same ref-qualifier (no qualification).

Therefore, void D::method() overrides void B1::method().

The same logic applies for void B2::method() (just substitute B2 for B1 in the above explanation), so void D::method() overrides both void B1::method() and void B2::method().