多重继承、虚基类继承时，显式地调用基类构造函数，对象是如何初始化的问题

//多重继承、虚基类继承时，显式地调用基类构造函数，对象是如何初始化的问题
#include <iostream>
using std::cout;
using std::endl;

class Base
{
    public:
        Base(){cout << "hahahha"
                    << endl;n = 78;}
        Base(int n0){cout << "Base\t" 
                         << n0
                         << endl;n = n0;}
        int n;
};

class C_A0: virtual public Base
{
    public:
        C_A0(){}
        C_A0(int n0,int n1):Base(n0){cout << "C_A0\t"
                                          << n1
                                          << endl;}
};

class C_A1: virtual public Base
{
    public:
        C_A1(){}
        C_A1(int n0,int n1):Base(n0){cout << "C_A1\t"
                                          << n1
                                          << endl;}
};

class C_B0: public C_A0,public C_A1 
{
    public:
        C_B0(){};
        C_B0(int n,int n0,int n1,int n2):Base(n),C_A0(n0,n0),C_A1(n1,n1)//显式地调用基类构造函数
        //在基类是虚拟的时，禁止信息通过中间类自动传递给基类，因此，C_A0(n0,n0),C_A1(n1,n1)中的第二个
        //参数将传递给对应的类，由于显式地调用了Base构造函数，子对象调用的是Base(int n0)构造函数。如果没
        //有默认构造函数，将报错:，error: no matching function for call to 'Base::Base()'
        {
          cout << "C_B0\t" << n2 << endl;
        }
};

int main()
{ 
    C_B0 a(1,2,3,4); //显式地调用Base(int n0)构造函数，因此，Base::n = 1；
    cout << &(a.C_A1::n) << '\t' << (a.C_A1::n) << endl;
    cout << &(a.C_A0::n) << '\t' << (a.C_A0::n) << endl; //由运行结果可知，派生类对象a从Base类中继承了两个n成员数据，这两个n成员数据是相同的
    return 0;
}

运行结果