Why only base class default constructor is called in virtual base multiple inheritance? [duplicate]

Question

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• c++ virtual inheritance 3 answers

In multiple inheritance, I have a virtual Base class which is inherited by class A and class B. A and B are base classes of AB. Please see the code below. In constructor of A and B, Base(string) constructor is called. I am expecting to get following output:

Base::Base(std::string)

A::A()

B::B()

But I am getting following output:

Base::Base()

A::A()

B::B()

Why default constructor of Base is being called?

#include<iostream>
#include<string>
using namespace std;

class Base{
public:
        Base(){
                cout<<__PRETTY_FUNCTION__<<endl;
        }
        Base(string n):name(n){
                cout<<__PRETTY_FUNCTION__<<endl;
        }
private:
string name;
};

class A : public virtual Base {
public:
        A():Base("A"){
                cout<<__PRETTY_FUNCTION__<<endl;
        }
private:
string name;
};

class B : public virtual  Base {
public:
        B():Base("B"){
                cout<<__PRETTY_FUNCTION__<<endl;
        }
private:
string name;
};

class AB : public A, public B{

};

int main(){
        AB a;
}

Answer 1

The virtual base is constructed by the most-derived object. So AB's constructor call the Base constructor, but since you didn't specify a constructor for AB, its default constructor just calls the default constructor of Base.

You could call the string-constructor from AB like this:

struct AB : A, B
{
    AB() : Base("hello"), A(), B() { }
};

Note that the constructors A::A() and B:B() do not call the Base constructor in this setup!