第十二周课后实践：程序阅读（3）（4）

最新推荐文章于 2015-06-15 11:38:08 发布

原创最新推荐文章于 2015-06-15 11:38:08 发布 · 389 阅读

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本文深入探讨了C++中多重继承与虚拟基类的构造顺序及其运行结果，详细解释了如何通过实例代码展示构造函数调用的过程，并分析了输出结果的原因。

(3)问题及代码：

#include <iostream>
using namespace std;
class Base
{
public:
    Base(char i) { cout<<"Base constructor. --"<<i<<endl; }
};
class Derived1:virtual public Base
{
public:
    Derived1(char i,char j):Base(i)
    {
        cout<<"Derived1 constructor. --"<<j<<endl;
    }
};
class Derived2:virtual public Base
{
public:
    Derived2(char i,char j):Base(i)
    {
        cout<<"Derived2 constructor. --"<<j<<endl;
    }
};
class MyDerived:public Derived1,public Derived2
{
public:
    MyDerived(char i,char j,char k,char l,char m,char n,char x): Derived2(i,j), Derived1(k,l), Base(m), d(n)
    {
        cout<<"MyDerived constructor. --"<<x<<endl;
    }
private:
    Base d;
};
int main()
{
    MyDerived obj('A','B','C','D','E','F','G');
    return 0;
}

运行结果：

(4)问题及代码：

#include<iostream>  
using namespace std; 
class A  
{  
public:  
   int n;  
};  
class B:public A {};   // class B:virtual public A{};  
class C:public A {};   // class C:virtual public A{};  
class D:public B,public C  
{  
public:
    int getn() {return B::n;}  
};  
int main()  
{  
   D d;  
   d.B::n=10;  
   d.C::n=20;
   cout<<d.getn()<<endl; 
   return 0;
}

运行结果：