金刚石继承最低基类构造函数 [英] Diamond Inheritance Lowest Base Class Constructor

问题描述

代码如下：

代码：

#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
};

问题：

1.）如果未知类被实例化，我将如何初始化所有的超类？因为只有一个实例将被创建，我如何避免mysef必须调用它的构造函数两次？

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 ?

谢谢

推荐答案

大多数派生类都初始化任何虚拟基类。在你的类层次结构中， Unknown 必须构造虚拟 Animal 基类（例如通过添加 Animal （a）到其初始化列表。）

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).

当构造一个 Unknown Fish 或 Bird 都会调用 Animal 构造函数。 未知将调用 Animal 虚拟基础的构造函数。

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

这篇关于金刚石继承最低基类构造函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！