Diamond Inheritance Lowest Base Class Constructor

问题

The Code is as follow :

The Code :

#include <iostream>

using namespace std;

class Animal{
   int a;

    public:
    Animal(int a) : a(a){}
    int geta(){return a;}
};

class Bird : virtual public Animal{
    string b;
    public:
    Bird(int a , string b) : Animal(a) , b(b){}
};

class Fish : virtual public Animal{
    int f;
    public:
    Fish(int a , int f) : Animal(a) , f(f){}
};

class Unknown : public Bird, public Fish{
    char u;
    public:
    Unknown(int a , int f , string b , char u )
     : Bird(a , b) , Fish(a , f) , u(u){}  //Problem
};

The Question :

1.)How am I going to initialize all the superclass if the Unknown class is instantiated?Since there's only one instance of Animal will be created , how can I avoid mysef from having to call its constructor twice ?

Thank you

回答1:

The most derived class initializes any virtual base classes. In your class hierarchy, Unknown must construct the virtual Animal base class (e.g. by adding Animal(a) to its initialization list).

When constructing an Unknown object, neither Fish nor Bird will call the Animal constructor. Unknown will call the constructor for the Animal virtual base.

来源：https://stackoverflow.com/questions/12500230/diamond-inheritance-lowest-base-class-constructor