第53课.被遗弃的多重继承(上)
1.c++中的多重继承
a.c++支持编写多重继承的代码
b.一个子类可以拥有多个父类
c.子类拥有父类的成员变量
d.子类继承所有父类的成员函数
e.子类对象可以当做任意父类对象使用(退化)
语法规则:
class Derived : public BaseA,
public BaseB,
public BaseC
{
// 多重继承的本质与单继承相同
}
2.多重继承的问题一
多重继承时得到的对象可能拥有不同的地址。
#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 = " << getC() << ", " << endl;
}
};
int main()
{
cout << "sizeof(Derived) = " << sizeof(Derived) << endl; // 12, 子类是父类的叠加
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; // 把b对象的地址给pa指针 (退化)
BaseB* pb = &d; // 把b对象的地址给pb指针 这里是同一个对象理论上来说,地址应该是一样的。
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; // 输出,这表示。pa和pb从d中获得到地址不同
}
cout << "pa = " << pa << endl;
cout << "pb = " << pb << endl;
cout << "paa = " << paa << endl;
cout << "pbb = " << pbb << endl;
return 0;
}
解析:
多重继承时得到的对象可能拥有不同的地址。如下图所示。
pa为BaseA指针。它得到的地址为叠加时,它的成员在子类中存储的位置。
同理:pb为BaseB指针。它的得到的地址为叠加时,它的成员在子类中的存储位置
解决办法:无
3.多重继承的问题二
多重继承可能产生冗余的成员
eg:
#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 : public People
{
public:
Teacher(string name, int age) : People(name, age)
{
}
};
class Student : 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)
{
}
};
int main()
{
Doctor d ("Huang", 24);
d.print();
// d.Teacher::print(); // 可行,分开引用
// d.Student::print();
return 0;
}
解析:这里把People这个类叠加了两次。在Tacher和Student中各自叠加两次后。在Doctor中叠加。所以会出现两个print函数。函数调用时,系统不知道到底要调用Tacher中叠加的print函数还是Student中的。
当多重继承关系出现闭合时,将产生数据冗余问题。
解决办法:虚继承
a.虚继承能够解决数据冗余问题
b.中间层父类不在关心顶层的初始化(但是还是得正常写)
c.最终子类必须直接调用顶层父类的构造函数
eg:
class People
{
};
// 因为继承关系在这里开始闭合(这里继承最顶层父类),所以在这里使用虚继承
class Tacher : virtual public People
{
};
// 因为继承关系在这里开始闭合(这里继承最顶层父类),所以在这里使用虚继承
class Student : virtual public People
{
};
class Doctor : public Tacher, public Student
{
public:
Doctor (string name, int age) : Teacher(name, age), Student(name, age), People(name, age)
{
}
};