Understanding virtual base classes and constructor

问题:

I'm a bit confused about how virtual base classes work. In particular, I was wondering how the constructor of the base class gets called. I wrote an example to understand it:

#include <cstdio>
#include <string>
using std::string;

struct A{
    string s;
    A() {}
    A(string t): s(t) {}
};

struct B: virtual public A{
    B(): A("B"){}
};

struct C: virtual public A {};

struct D: public B, public C {};

struct E: public C, public B {};

struct F: public B {};

int main(){
    D d;
    printf("\"%s\"\n",d.s.c_str());
    E e;
    printf("\"%s\"\n",e.s.c_str());
    F f;
    printf("\"%s\"\n",f.s.c_str());
    B b;
    printf("\"%s\"\n",b.s.c_str());
}

Which outputs

""
""
""
"B"

I wasn't sure what would happen in the first two cases, but for the third one at least I was expecting the output to be "B". So now I'm just confused. What are the rules for understanding how the constructor of A gets called?

回答1:

There is always just one constructor call, and always of the actual, concrete class that you instantiate. It is your responsibility to endow each derived class with a constructor which calls the base classes' constructors if and as necessary, as you did in B's constructor.

Update: Sorry for missing your main point! Thanks to ildjarn.

However, your B inherits virtually from A. According to the standard (10.1.4 in the FIDS), "for each distinct baseclass that is specified virtual, the most derived object shall contain a single base class subobject of that type". In your case this means that when constructing the base, your class F immediately calls A's default constructor, not B's.

回答2:

Virtual base classes are always constructed by the most derived class.