C++多继承（多重继承）详解（二）命名冲突

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。

#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BaseA类的 show()函数
    cout<<"m_e = "<<m_e<<endl;
}

int main(){
    Derived obj(1, 2, 3, 4, 5);
    obj.display();
    return 0;
}

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。