C++多继承(多重继承)
派生类都只有一个基类,称为单继承(Single Inheritance)。除此之外,C++也支持多继承(Multiple Inheritance),即一个派生类可以有两个或多个基类。
下面是一个多继承的实例:
运行结果:
BaseA constructor
BaseB constructor
Derived constructor
1, 2, 3, 4, 5
Derived destructor
BaseB destructor
BaseA destructor
从运行结果中还可以发现,多继承形式下析构函数的执行顺序和构造函数的执行顺序相反。
修改上面的代码,为 BaseA 和 BaseB 类添加 show() 函数,并将 Derived 类的 show() 函数更名为 display():
请读者注意第 64、65 行代码,我们显式的指明了要调用哪个基类的 show() 函数。
多继承容易让代码逻辑复杂、思路混乱,一直备受争议,中小型项目中较少使用,后来的 Java、C#、PHP 等干脆取消了多继承。多继承的语法也很简单,将多个基类用逗号隔开即可。例如已声明了类A、类B和类C,那么可以这样来声明派生类D:
class D: public A, private B, protected C{
//类D新增加的成员
}
多继承下的构造函数
多继承形式下的构造函数和单继承形式基本相同,只是要在派生类的构造函数中调用多个基类的构造函数。以上面的 A、B、C、D 类为例,D 类构造函数的写法为:D(形参列表): A(实参列表), B(实参列表), C(实参列表){
//其他操作
}
D(形参列表): B(实参列表), C(实参列表), A(实参列表){
//其他操作
}
下面是一个多继承的实例:
- #include <iostream>
- using namespace std;
- //基类
- class BaseA{
- public:
- BaseA(int a, int b);
- ~BaseA();
- protected:
- int m_a;
- int m_b;
- };
- BaseA::BaseA(int a, int b): m_a(a), m_b(b){
- cout<<"BaseA constructor"<<endl;
- }
- BaseA::~BaseA(){
- cout<<"BaseA destructor"<<endl;
- }
- //基类
- class BaseB{
- public:
- BaseB(int c, int d);
- ~BaseB();
- protected:
- int m_c;
- int m_d;
- };
- BaseB::BaseB(int c, int d): m_c(c), m_d(d){
- cout<<"BaseB constructor"<<endl;
- }
- BaseB::~BaseB(){
- cout<<"BaseB destructor"<<endl;
- }
- //派生类
- class Derived: public BaseA, public BaseB{
- public:
- Derived(int a, int b, int c, int d, int e);
- ~Derived();
- public:
- void show();
- private:
- int m_e;
- };
- Derived::Derived(int a, int b, int c, int d, int e): BaseA(a, b), BaseB(c, d), m_e(e){
- cout<<"Derived constructor"<<endl;
- }
- Derived::~Derived(){
- cout<<"Derived destructor"<<endl;
- }
- void Derived::show(){
- cout<<m_a<<", "<<m_b<<", "<<m_c<<", "<<m_d<<", "<<m_e<<endl;
- }
- int main(){
- Derived obj(1, 2, 3, 4, 5);
- obj.show();
- return 0;
- }
BaseA constructor
BaseB constructor
Derived constructor
1, 2, 3, 4, 5
Derived destructor
BaseB destructor
BaseA destructor
从运行结果中还可以发现,多继承形式下析构函数的执行顺序和构造函数的执行顺序相反。
命名冲突
当两个或多个基类中有同名的成员时,如果直接访问该成员,就会产生命名冲突,编译器不知道使用哪个基类的成员。这个时候需要在成员名字前面加上类名和域解析符::
,以显式地指明到底使用哪个类的成员,消除二义性。修改上面的代码,为 BaseA 和 BaseB 类添加 show() 函数,并将 Derived 类的 show() 函数更名为 display():
- #include <iostream>
- using namespace std;
- //基类
- class BaseA{
- public:
- BaseA(int a, int b);
- ~BaseA();
- public:
- void show();
- protected:
- int m_a;
- int m_b;
- };
- BaseA::BaseA(int a, int b): m_a(a), m_b(b){
- cout<<"BaseA constructor"<<endl;
- }
- BaseA::~BaseA(){
- cout<<"BaseA destructor"<<endl;
- }
- void BaseA::show(){
- cout<<"m_a = "<<m_a<<endl;
- cout<<"m_b = "<<m_b<<endl;
- }
- //基类
- class BaseB{
- public:
- BaseB(int c, int d);
- ~BaseB();
- void show();
- protected:
- int m_c;
- int m_d;
- };
- BaseB::BaseB(int c, int d): m_c(c), m_d(d){
- cout<<"BaseB constructor"<<endl;
- }
- BaseB::~BaseB(){
- cout<<"BaseB destructor"<<endl;
- }
- void BaseB::show(){
- cout<<"m_c = "<<m_c<<endl;
- cout<<"m_d = "<<m_d<<endl;
- }
- //派生类
- class Derived: public BaseA, public BaseB{
- public:
- Derived(int a, int b, int c, int d, int e);
- ~Derived();
- public:
- void display();
- private:
- int m_e;
- };
- Derived::Derived(int a, int b, int c, int d, int e): BaseA(a, b), BaseB(c, d), m_e(e){
- cout<<"Derived constructor"<<endl;
- }
- Derived::~Derived(){
- cout<<"Derived destructor"<<endl;
- }
- void Derived::display(){
- BaseA::show(); //调用BaseA类的show()函数
- BaseB::show(); //调用BaseB类的show()函数
- cout<<"m_e = "<<m_e<<endl;
- }
- int main(){
- Derived obj(1, 2, 3, 4, 5);
- obj.display();
- return 0;
- }
posted on 2022-11-29 10:11 mbaicizhan 阅读(76) 评论(0) 编辑 收藏 举报