命名冲突

一起看海

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()：
纯文本复制
#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(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;
}
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。