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第12周-多继承、虚基类,继承和组合-课后实践·阅读程序

2015-05-24 20:34 405 查看
/* 
* Copyright (c) 2014, 烟台大学计算机学院 
* All rights reserved. 
* 文件名称:test.cpp 
* 作    者:刘畅 
* 完成日期:2015 年 5  月  24 日 
* 版 本 号:v1.0 
* 
* 问题描述:阅读程序~~
* 输入描述: ;
* 程序输出:。
(1)

#include <iostream>
using namespace std;
class A
{
public:
    A()
    {
        a=0;
    }
    A (int i)
    {
        a=i;
    }
    void print()
    {
        cout<<a<<"  ";
    }
private:
    int a;
};
class B: public A
{
public:
    B()
    {
        b=0;
    }
    B(int i, int j, int k): A(i),aa(j)
    {
        b=k;
    }
    void print()
    {
        A::print();
        aa.print();
        cout<<b<<endl;
    }
private:
    int b;
    A aa;
};
int main()
{
    B test[2];
    test[0]=B(1,4,7);
    test[1]=B(2,5,8);
    for(int i=0; i<2; i++)
        test[i].print();
    return 0;
}


运行结果:



(2)

#include <iostream>
using namespace std;
class A
{
public:
    A(string s)
    {
        cout<<s<<endl;
    }
};
class B:public A
{
public:
    B(string s1, string s2):A(s1)
    {
        cout<<s2<<endl;
    }
};
class C:public A
{
public:
    C(string s1,string s2):A(s1)
    {
        cout<<s2<<endl;
    }
};
class D:public B,C
{
public:
    D(string s1,string s2,string s3,string s4):B(s1,s2),C(s3,s4)
    {
        cout<<s4<<endl;
    }
};
int main()
{
    D d("class A","class B","class C","class D");
    return 0;
}


运行结果:



贺老给的阅读代码好像出了点问题,char 的没有输出,于是我给改成了string。。。看运行结果就能大致明白它的构造函数的执行顺序了。

(3)

#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:



这个还存有疑问。

(4)

#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:



#include<iostream>
using namespace std;
class A
{
public:
    int n;
};
class B:virtual 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;
}


修改后运行结果:



(5)

#include <iostream>
using namespace std;
class A
{
protected:
    int a,b;
public:
    A(int aa, int bb):a(aa), b(bb) {}
    void printA()
    {
        cout<<"a: "<<a<<"\tb: "<<b<<endl;
    }
};
class B: public A
{
    int c;
public:
    B(int aa, int bb, int cc):A(aa,bb),c(cc) {}
    void printB()
    {
        cout<<"a: "<<a<<"\tb: "<<b<<"\tc: "<<c<<endl;
    }
};
int main()
{
    A a(1,1);
    B b(2,3,4);
    //此处加入下面各小题中的代码
    return 0;
}


(a)
a=b;
    a.printA();
    b.printA();
    b.printB();
运行结果:



(b)

b=a;  
a.printA();  
b.printA();  
b.printB();
运行结果:



可以通过派生类对象向基类赋值,但反过来是行不通的

(c)

A &r1=a;
    A &r2=b;
    r1.printA();
    r2.printA();
    r2.printB();
运行结果:



不能通过派生类对象去访问基类对象的成员,

删除后:



(d)

A *p=&a;
    p->printA();
    p=&b;
    p->printA();
    p->printB();


运行结果:



指向基类的指针只能访问派生类中的基类成员,而不能访问派生类中增加的成员。

删除后:



(e)

在classA中增加成员函数:

int getA(){return a;}


在main函数前增加一段函数:

void f(A x)
{
    cout<<"aaaaah, my a: "<<x.getA()<<endl;
}


main函数中指定部分为:

f(a);
    f(b);


代码如下:

#include <iostream>
using namespace std;
class A
{
protected:
int a,b;
public:
A(int aa, int bb):a(aa), b(bb) {}
void printA()
{
cout<<"a: "<<a<<"\tb: "<<b<<endl;
}
int getA()
{
return a;
}
};
class B: public A
{
int c;
public:
B(int aa, int bb, int cc):A(aa,bb),c(cc) {}
void printB()
{
cout<<"a: "<<a<<"\tb: "<<b<<"\tc: "<<c<<endl;
}
};
void f(A x) { cout<<"aaaaah, my a: "<<x.getA()<<endl; }

int main()
{
A a(1,1);
B b(2,3,4);
f(a); f(b);
return 0;
}


运行结果:



(6)补充阅读:

#include <iostream>
using namespace std;
class B
{
public:
    B(int x=0)
    {
        X=x;
        cout<<"B("<<x<<")\n";
    }
    ~B()
    {
        cout<<"~B()\n";
    }
    void print()
    {
        cout <<X<< " ";
    }
private:
    int X;
};

class D: public B
{
public:
    D (int x, int y):B(x)
    {
        Y=y;
        cout<<"D("<<x<<","<<y<<")\n";
    }
    ~D()
    {
        cout<<"~D()\n";
    }
    void print()   //与基类中的成员函数同名!这个冲突如何解决的?!
    {
        B::print();
        cout <<Y<<endl;
    }
private:
    int Y;
};
int  main()
{
    D d(11,22);
    d.print();
    return 0;
}


是通过作用域限定符解决了基类和派生类成员函数同名的问题。
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