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。
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