C++ Virtual Inheritance of virtual method

Question

struct A {
    virtual void foo() { std::cout << "a";};
};

struct B:public virtual A {
    void foo() { std::cout << "b";}
};

struct C:public virtual A {
    void foo() { std::cout << "c";}
};

struct D:public B, public C {

};
int main() {
    return 0;
}

So this one while compiling gives us the following error:

 \main.cpp:16:8: error: no unique final overrider for 'virtual void A::foo()' in 'D'
 struct D:public B, public C {

If we would make the inheritance of B and C structs non-virtual, the code is compiling just right without any errors (but of course the error happens if we're calling dd.foo()). So what's the difference? Why do we have an error when we inherit our class virtually and no error if we do it straight?

I think this is not a duplicate of http://stackoverflow.com/questions/21558/in-c-what-is-a-virtual-base-class — Othman Benchekroun, Jun 23 '15 at 15:30
I voted for reopening. This question is about a specific detail which is not covered thouroughly in the marked duplicate. — davidhigh, Jun 23 '15 at 15:30
@OthmanBenchekroun, The answer isn't so much directly in the definition of virtual inheritance, but it follows from the definition. — chris, Jun 23 '15 at 15:31
Which part of the error is unclear? There is no unique final overrider. — Kerrek SB, Jun 24 '15 at 15:18

chris · Accepted Answer · 2015-06-23T15:41:15.550

Making A a virtual base class of B and C ensures that D contains exactly one A subobject^[1]. For this, both B and C provide final overriders for foo^[2] and both are inherited by D^[2], so D has two final overriders for foo, making the program ill-formed^[2].

When A is not a virtual base class of B and C, D will contain two distinct A subobjects^[1]. Each of these subobjects will have their own inherited final overrider for foo^[2].

[1]: N4140 §10.1 [class.mi]/4:

A base class specifier that does not contain the keyword virtual, specifies a non-virtual base class. A base class specifier that contains the keyword virtual, specifies a virtual base class. For each distinct occurrence of a non-virtual base class in the class lattice of the most derived class, the most derived object shall contain a corresponding distinct base class subobject of that type. For each distinct base class that is specified virtual, the most derived object shall contain a single base class subobject of that type.

[2]: §10.3 [class.virtual]/2 (emphasis mine):

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. For convenience we say that any virtual function overrides itself. A virtual member function C::vf of a class object S is a final overrider unless the most derived class of which S is a base class subobject (if any) declares or inherits another member function that overrides vf. In a derived class, if a virtual member function of a base class subobject has more than one final overrider the program is ill-formed.

I've had the same idea, so this one looks right, thanks :) So when we have diamond inheritance we don't know which foo overrides A subobject — Utikeev, Jun 23 '15 at 15:33
@Utikeev, Since both `B` and `C` override it, `D` must provide its own, which will become the unique final overrider (and will cause the ones in `B` and `C` to lose their final overrider status). — chris, Jun 23 '15 at 15:42
@chris What exactly is the entity `S` mentioned in [class.virtual]/2 above? — Alexander, Aug 23 '20 at 12:46
@Alexander, It looks to be an arbitrary object of the class, so you can sub it out for whichever concrete object you're applying this to. — chris, Aug 23 '20 at 18:30

C++ Virtual Inheritance of virtual method

1 Answers1