Multiple inheritance ambiguous base class

Question

Consider the code

struct Base{};
struct Derived: public Base{};

struct A: public Base{};

struct B: public A, public Base{};

struct C: public A, public Derived{}; // why no ambiguity here?

int main() {}

The compiler (g++5.1) warns that

warning: direct base 'Base' inaccessible in 'B' due to ambiguity struct B: public A, public Base{};

I understand this, Base is duplicated in B.

Why is there no warning for C? Doesn't C inherit from both A and Derived, which both inherit from Base?
Why adding virtual
```
struct Derived: virtual Base{};
```

results now in both B and C emitting warnings, live on Wandbox

warning: direct base 'Base' inaccessible in 'B' due to ambiguity struct B: public A, public Base{};

warning: direct base 'Base' inaccessible in 'C' due to ambiguity struct C: public A, public Derived{};

@ThomasMatthews I know what the diamond problem is, and that's closely related to why B gives a warning. However what I don't understand is why C is OK. — vsoftco
– vsoftco, Commented May 15, 2015 at 19:46
I think it's just a case of gcc not detecting the ambiguity in C because Base is ambiguous in that case too. When you virtually derive from Base the most derived class (C) is responsible for calling the Base constructor, and so gcc starts detecting the ambiguity again. — Praetorian
– Praetorian, Commented May 15, 2015 at 19:57
@Praetorian it makes sense. clang++/vc++ don't spit any warnings at all, unless I make struct Derived: public virtual Base — vsoftco
– vsoftco, Commented May 15, 2015 at 20:00
It could also be because with the original C I can address Base's members by explicitly qualifying them. Say Base had a member function void foo(){}. Then I can say C{}.A::foo(); C{}.Derived::foo();, but as far as I can tell, there's no way to do that with B. — Praetorian
– Praetorian, Commented May 15, 2015 at 20:05

Brian Bi · Accepted Answer · 2015-05-15 21:01:24Z

2

In B, it's impossible to refer to members of the Base subobject inherited directly. Consider:

struct Base {
    int x;
};

struct B: public A, public Base {
    void foo() {
        int& x1 = A::x; // OK
        int& x2 = x; // ambiguous
        // no way to refer to the x in the direct base
    }
};

In C this is not a problem. Both x's can be referred to using qualified names:

struct C: public A, public Derived {
    void foo() {
        int& x1 = A::x; // OK
        int& x2 = Derived::x; // OK
    }
};

So the warning you get is one that only makes sense when a direct base is also inherited through another path.

For your second question, I couldn't reproduce the warning with C on Coliru with g++-5.1.

answered May 15, 2015 at 21:01

Brian Bi

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Rexxar · Accepted Answer · 2015-05-15 21:07:12Z

There are no ways to access unambiguously to Base members in "B" whereas it's possible in "C", as illustrated in the following code:

#include <iostream>

using namespace std;

struct Base
{
    void print()
    {
        cout << "Base" << endl;
    }
};

struct Derived : public Base {};

struct A : public Base
{
    void print()
    {
        cout << "A" << endl;
    }
};

struct B : public A, public Base
{
    void print()
    {
        A::print();

        //error (ambiguous), no way to access to Base::print => warning
        //Base::print();
    }
};

struct C : public A, public Derived
{
    void print()
    {
        A::print();
        Derived::print(); // Not Ambiguous, it's the Base inherited by 'Derived' which is used.
        // Still an error but you can access print indirectly through "Derived" => no warning needed
        //Base::print();
    }
};

int main() 
{
    B b;
    b.print();

    C c;
    c.print();

    return 0; 
}

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