C++ - 为什么钻石 inheritance 的这种结构不会引起歧义？

Question

I have scanned through dozens of multiple inheritance question but wasn't able to find an answer for this question.

I understand why this won't compile:

struct B{
    virtual void f(){
        printf("B");
    }
};

struct C1 : virtual B{
   void f() override{
        printf("C1");
    }
};

struct C2: virtual B{
    void f() override{
        printf("C2")
    }
};

struct D: C1,C2{
    
};

Because there is not way to determine which f should be in B 's vtable.

However, I was quite surprised to see that this code will compile:

struct B{
    virtual void f(){
        printf("B");
    }
};

struct C1 : virtual B{
   void f() override{
        printf("C1");
    }
};

struct C2: virtual B{

};

struct D: C1,C2{
    
};

My (wrong) intuition was that we still have two path from D to B and in each one the last place f is overridden is different - in one path it's in C1 and in one path it's in B , but this does compile.

Why doesn't this cause an ambiguity?

Answer 1

Because C1::f1 overrides B::f1, it will be used in preference on any object that has a dynamic type of C1 or any subclass.

In the ambiguous case, While both C1::f1 and C2::f1 override B::f1, they do not override each other, so the ambiguity is between them -- the fact that they both override a common base function is irrelevant.