---- 整理自狄泰软件唐佐林老师课程

1. 问题

C++是否允许一个类继承自多个父类？

1.1 C++中的 多重继承

• C++支持编写多重继承的代码
  • 一个子类可以拥有多个父类
  • 子类拥有所有父类的成员变量
  • 子类继承所有父类的成员函数
  • 子类对象可以当作任意父类对象使用

1.2 多重继承的语法规则

• 多重继承的本质和单继承相同

2. 多重继承问题一

• 编程实验：多重继承问题一

#include <iostream>
#include <string>

using namespace std;

class BaseA
{
    int ma;
public:
    BaseA(int a)
    {
        ma = a;
    }
    int getA()
    {
        return ma;
    }
};

class BaseB
{
    int mb;
public:
    BaseB(int b)
    {
        mb = b;
    }
    int getB()
    {
        return mb;
    }
};

class Derived : public BaseA, public BaseB
{
    int mc;
public:
    Derived(int a, int b, int c) : BaseA(a), BaseB(b)
    {
        mc = c;
    }
    int getC()
    {
        return mc;
    }
    void print()
    {
        cout << "ma = " << getA() << ", "
             << "mb = " << getB() << ", "
             << "mc = " << mc << endl;
    }
};

int main()
{
    cout << "sizeof(Derived) = " << sizeof(Derived) << endl;    // 12
    
    Derived d(1, 2, 3);
    
    d.print();
    
    cout << "d.getA() = " << d.getA() << endl;
    cout << "d.getB() = " << d.getB() << endl;
    cout << "d.getC() = " << d.getC() << endl;
    
    cout << endl;
    
    BaseA* pa = &d;
    BaseB* pb = &d;
    
    cout << "pa->getA() = " << pa->getA() << endl;
    cout << "pb->getB() = " << pb->getB() << endl;
    
    cout << endl;
    
    void* paa = pa;
    void* pbb = pb;
    
    
    if( paa == pbb )
    {
        cout << "Pointer to the same object!" << endl; 
    }
    else
    {
        cout << "Error" << endl;
    }
    
    cout << "pa = " << pa << endl;
    cout << "pb = " << pb << endl;
    cout << "paa = " << paa << endl;
    cout << "pbb = " << pbb << endl; 
    
    return 0;
}

• 通过 多重继承 得到的对象可能拥有 “不同的地址”

  • 解决方案：无

3. 多重继承问题二

• 多重继承可能产生冗余的成员

Doctor中有两个m_name、m_age

• 编程实验：多重继承问题二

#include <iostream>
#include <string>

using namespace std;

class People
{
    string m_name;
    int m_age;
public:
    People(string name, int age)
    {
        m_name = name;
        m_age = age;
    }
    void print()
    {
        cout << "Name = " << m_name << ", "
             << "Age = " << m_age << endl;
    }
};

class Teacher : virtual public People
{
public:
    Teacher(string name, int age) : People(name, age)
    {
    }
};

class Student : virtual public People
{
public:
    Student(string name, int age) : People(name, age)
    {
    }
};

class Doctor : public Teacher, public Student
{
public:
    Doctor(string name, int age) : Teacher(name, age), Student(name, age), People(name, age)
    {
    }
};

int main()
{
    Doctor d("Delphi", 33);
    
    d.print();
    
    return 0;
}

• 当多重继承关系出现闭合时将产生数据冗余的问题

  • 解决方案：虚继承
    

• 虚继承能够解决 数据冗余 问题

• 中间层父类 不再关心顶层父类的初始化（中间层父类不调用父类构造函数）

• 最终子类 必须直接调用 顶层父类的构造函数

问题：当架构设计中需要继承时，无法确定使用直接继承还是虚继承

4. 多重继承问题三

• 多重继承可能产生 多个虚函数表
  

• 编程实验：多重继承问题三

#include <iostream>
#include <string>

using namespace std;

class BaseA
{
public:
    virtual void funcA()
    {
        cout << "BaseA::funcA()" << endl;
    }
};

class BaseB
{
public:
    virtual void funcB()
    {
        cout << "BaseB::funcB()" << endl;
    }
};

class Derived : public BaseA, public BaseB
{

};

int main()
{
    Derived d;
    BaseA* pa = &d;
    BaseB* pb = &d;
    BaseB* pbe = (BaseB*)pa;    // oops!!
    BaseB* pbc = dynamic_cast<BaseB*>(pa);
    
    cout << "sizeof(d) = " << sizeof(d) << endl;
    
    cout << "Using pa to call funcA()..." << endl;
    
    pa->funcA();
    
    cout << "Using pb to call funcB()..." << endl;
    
    pb->funcB();
    
    cout << "Using pbc to call funcB()..." << endl;
    
    pbc->funcB();
    
    cout << endl;
    
    cout << "pa = " << pa << endl;
    cout << "pb = " << pb << endl;
    cout << "pbe = " << pbe << endl;
    cout << "pbc = " << pbc << endl;
    
    return 0;
}

需要进行强制类型转换时，C++中推荐使用新式类型转换关键字
解决方案：dynamic_cast

5. 正确的使用多重继承

• 工程开发中的“多重继承”方式：
  • 单继承某个类 + 实现（多个）接口

• 编程实验：正确的多继承方式

#include <iostream>
#include <string>

using namespace std;

class Base
{
protected:
    int mi;
public:
    Base(int i)
    {
        mi = i;
    }
    int getI()
    {
        return mi;
    }
    bool equal(Base* obj)
    {
        return (this == obj);
    }
};

class Interface1
{
public:
    virtual void add(int i) = 0;
    virtual void minus(int i) = 0;
};

class Interface2
{
public:
    virtual void multiply(int i) = 0;
    virtual void divide(int i) = 0;
};

class Derived : public Base, public Interface1, public Interface2
{
public:
    Derived(int i) : Base(i)
    {
    }
    void add(int i)
    {
        mi += i;
    }
    void minus(int i)
    {
        mi -= i;
    }
    void multiply(int i)
    {
        mi *= i;
    }
    void divide(int i)
    {
        if( i != 0 )
        {
            mi /= i;
        }
    }
};

int main()
{
    Derived d(100);
    Derived* p = &d;
    Interface1* pInt1 = &d;
    Interface2* pInt2 = &d;
    
    cout << "p->getI() = " << p->getI() << endl;    // 100
    
    pInt1->add(10);
    pInt2->divide(11);
    pInt1->minus(5);
    pInt2->multiply(8);
    
    cout << "p->getI() = " << p->getI() << endl;    // 40
    
    cout << endl;
    
    cout << "pInt1 == p : " << p->equal(dynamic_cast<Base*>(pInt1)) << endl;
    cout << "pInt2 == p : " << p->equal(dynamic_cast<Base*>(pInt2)) << endl;
    
    return 0;
}

• 一些有用的工程建议
  • 先继承自一个父类，然后实现多个接口
  • 父类中提供equal()成员函数
  • equal()成员函数用于判断指针是否指向当前对象
  • 与多重继承相关的强制类型转换使用dynamic_cast完成

6. 小结

• C++支持多重继承的编程方式
• 多重继承容易带来问题：
  • 可能出现“同一个对象的地址不同”的情况
  • 虚继承可以解决数据冗余的问题
  • 虚继承使得架构设计可能出现问题
• 多重继承中可能出现多个虚函数表指针VPTR
• 与多重继承相关的强制类型转换用dynamic_cast完成
• 工程开发中采用“单继承多接口”的方式使用多继承
• 父类提供成员函数用于判断是否指向当前对象