C++ 集成和派生练习题解答

桃扇骨 2022-07-15 01:58 158阅读 0赞

**综合题1：考察单继承方式**

综合题1.编写一个程序设计一个汽车类vehicle，包含的数据成员有车轮个数wheels和车重weight。小车类car是它的私有派生类，其中包含载人数pasenger\_load。卡车类truck是vehicle的私有派生类，其中包含载人数passenger\_load和载重量payload，每个类都有相关数据的输出方法。

#include<iostream>
    using namespace std;
    class vehicle 
    {
      int wheels,weight;
        public:
      vehicle() {wheels=4,weight=100;}
      void show1(){cout<<"wheels="<<wheels<<" ,wight="<<weight<<endl;}
    };
    class car:public vehicle
    {
    	int pasenger_load;
      public:
      car(){ pasenger_load=5;}
      void show2(){show1();cout<<"pasenger_load:"<<pasenger_load<<endl;}
    };
    class truck:public vehicle
    {
    	int pasenger_load,payload;
      public:
      truck(){ pasenger_load=5;payload=7;}
      void show3(){show1();cout<<"pasenger_load:"<<pasenger_load<<", payload="<<payload<<endl;}
    };
    
    int main()
    {
      vehicle a;
      car b;
      truck c;
      a.show1();
      b.show2();
      c.show3();
    }

**综合题2 考察构造函数和析构函数**

#include <iostream>
    using namespace std;
    class base
    {
    public:
    base(){cout<<"constructing base class"<<endl;}
    ~base(){cout<<"destructing base class"<<endl;}
    };
    class subs:public base
    {
    public:
    subs(){cout<<"constructing base class"<<endl;}
    ~subs(){cout<<"destructing base class"<<endl;}
    };
    int main()
    {
    subs s;
    }

**综合题3 考察累的继承方式**

题目要求：假设图书馆的图书包含书名、编号，作者三个属性；读者包含姓名和借书证两个属性，每位读者最多可借5本书，编写程序列出某读者的借书情况。

#include "iostream"
    #include "string"
    using namespace std;
    class obj
    {
     public:
        string name;
        string num;
     public:
     obj(){name="0";num="0";}
      obj(string x,string y)
      {
        name=x;
        num=y;
      }
    };
    class book :public obj
    {
        public:
            string writer;
        public:
    book():obj(){writer="0";}
    
    };
    class reader:public obj
    {
        private:
         book b[5];
         const int MAX;
         static int num;
        public:
        reader(string x,string y):obj(x,y),MAX(5){}
      int rentbook();
       void print();
    };
    int reader:: num =0 ;
    int reader:: rentbook()
     {
         int i;
         cout<<"输入选项：0、退出，1、借书，2、还书 "<<endl;
    
            while(cin>>i)
            {
                switch(i)
                {
                    case 0:
                        cout<<"!!!"<<endl;
                        return 0;
                    case 1:
                    if(num == 4) cout<<"已经借满5本"<<endl;
                    else
                    {
                        string na ,nu,author;
                        cout<<"输入所借书的书名:";
                        cin>>na;
                        cout<<"输入所借图书的编号:";
                        cin>>nu;
                        cout<<"输入图书的作者:";
                        cin>>author;
    
                        b[num].name = na;
                        b[num].num = nu;
                        b[num].writer = author;
                        num++;
                    }
                        break;
                    case 2:
                    if(num == 0) cout<<"没有书籍可还!"<<endl;
                    else
                    {
                        string na ;
                        cout<<"输入所还书的书名";
                        cin>>na;
                        int i;
                        for(i=0;i<MAX;i++)
                            if(b[i].name == na)
                               { b[i].name ="0";b[i].num ="0";b[i].writer="0";cout<<"还书成功"<<endl; goto B ;}
                                cout<<"没有此书"<<endl;
                    }
    
                    B:break;
                    default :
                        cout<<"please input the right number!!"<<endl;
                    break;
                }
       cout<<"输入选项：0、退出，1、借书，2、还书 "<<endl;
     }
     }
    
     void reader:: print()
     {
         int i;
        cout<<"姓名："<<obj::name<<"学号"<<obj::num<<endl;
        cout<<"所借图书如下："<<endl;
        cout<<"书名"<<"\t"<<"编号"<<"\t"<<"作者"<<endl;
        for(i=0;i<MAX;i++)
        if(b[i].name !="0")
        cout<<b[i].name<<"\t"<<b[i].num<<"\t"<<b[i].writer<<endl;
     }
    int main()
    {reader a("wang","1234");
     a.rentbook();
     a.print();
    }

**综合题4 - 考查虚基类的实现**

分析以下程序的执行结果

#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
    {
    int get() {return B::n;}
    };
    int main()
    {
    D d;
    d.B::n=10;
    d.C::n=20;
    cout<<d.B::n<<","<<d.C::n<<endl;
    }

综合题5  考察重集成类的设计

设计一个圆类circle和一个桌子类table，另设计一个圆桌类roundtable，它是从前两个类派生的，要求输出一个圆桌的高度、面积和颜色等数据。

#include<iostream>
    #include<string.h>
    using namespace std;
    class cicle 
    {
      float radius;
      public:
      cicle(float a){radius=a;}
      float getaerea() {return 3.14*radius*radius;}
    };
    class table
    {
      float height;
      public:
      table(float a){height=a;}
      float getheight() {return height;}
    };
    class roundtable:public cicle,public table
    {
    	string color;
    	public:
      roundtable(float a,float b,string c): cicle(a),table(b){color=c;}
      string get() {return color;}
    };
    int main()
    {
      roundtable AA(10,2,"wanglei");
     
      cout<<AA.getaerea()<<","<<AA.getheight()<<","<<AA.get()<<endl;
      
    }