#include <iostream>
using namespace std;
class Base1 {   
public:
    int var;
    void fun() { cout << "Member of Base1" << endl; }
};
class Base2 {   
public:
    int var;
    void fun() { cout << "Member of Base2" << endl; }
};
class Derived: public Base1, public Base2 {
public:
    int var;    
    void fun() { cout << "Member of Derived" << endl; }
};

int main() {
    Derived d;
    Derived *p = &d;

  //訪問Derived類成員
    d.var = 1;
    d.fun();    

    //訪問Base1基類成員
    d.Base1::var = 2;
    d.Base1::fun(); 

    //訪問Base2基類成員
    p->Base2::var = 3;
    p->Base2::fun();    

    return 0;
}
 

//7_7.cpp
#include <iostream>
using namespace std;
class Base0 {   //定義基類Base0
public:
    int var0;
    void fun0() { cout << "Member of Base0" << endl; }
};
class Base1: public Base0 { //定義派生類Base1 
public: //新增外部介面
    int var1;
};
class Base2: public Base0 { //定義派生類Base2 
public: //新增外部介面
    int var2;
};

例7-7  多繼承時的二義性和冗餘問題
class Derived: public Base1, public Base2 {
public: 
    int var;
    void fun() 
  { cout << "Member of Derived" << endl; }
};

int main() {    //程式主函式
    Derived d;
    d.Base1::var0 = 2;
    d.Base1::fun0();
    d.Base2::var0 = 3;
    d.Base2::fun0();
    return 0;
}