C程序设计与应用基础第五章多态性习题答案.doc

第五章 多态性1、填空题1）在一个成员函数内调用一个虚函数时,对该虚函数的调用进行_动态_联编。2）动态联编是在_虚函数_的支持下实现的,它通过_指针和引用_来调用该函数操作。3）下列程序的运行结果如下:Bases cons.Deriveds cons.Deriveds des.Bases des.根据结果将程序补充完整。#incude class Basepublic:Base()coutBases cons.endl;_varitual Base（）_coutBases des.endl;class Derived:public Basepublic:Derived()coutDeriveds cons.endl;Derived()coutDeriveds des.endl;void main()Base *Ptr=_new Derived；_delete ptr;4）C+中_不支持_虚构造函数,但_支持_虚析构函数5）带有_纯虚函数_的类称为抽象类,它只能作为_基类_来使用。6）下列程序的运行结果如下:Derive1s Print() called.Derive2s Print() caIIed.根据结果将程序补充完整。#include class Basepublic:Base(int I)b=I;_vitual void print （）_protected:int b;class Derive1:public Basepublic:_Derivel1（int I):Base（I)_void Print()coutDerive1s Print() called.endl; ;class Derive2:public Base_Derivel2（int I):Base（I)void Print()coutDerive1s Print() called.Print();void main()_Dervel1* d1=new Dervel1（1）； Dervel2* d1=new Dervel2（2）；_fun(d1);fun(d2);2、编程题1）设计一个小型公司的人员信息管理系统。该公司主要有四类人员：经理、兼职技术人员、销售经理和兼职推销员。要求存储这些人员的姓名、编号、级别、当月薪水总额并显示全部信息，具体要求如下所述。_人员编号基数为1000,每输入一个人员信息,编号顺序加1。_程序具有对所有人员提升级别的功能。经理为4级,兼职技术人员和销售经理为3级,推销员为1级。_月薪计算方法是:经理拿固定月薪8000元；兼职技术人员按每小时100元领取月薪；兼职推销员的月薪按该推销员当月销售额的4%提成；销售经理既拿固定月薪也领取销售提成,固定月薪为5000元,销售提成为所管辖部门当月销售额的5。答案：#include #include class employeepublic:employee();employee()delete name;virtual void pay()=0;virtual void promote(int increment=0)grade+=increment;virtual void display()=0;protected:char *name;int no;int grade;double Pay;static int MaxNo;class technician:public employeepublic:technician()hourlyRate=100;void promote(int)employee:promote(2);void pay();void display();private:double hourlyRate;int workHours;class salesman:virtual public employeepublic:salesman()CommRate=0.04;void promote(int)employee:promote(2);void pay();void display();protected:double CommRate;double sales;class manager:virtual public employeepublic:manager()monthlyPay=8000;void promote(int)employee:promote(3);void pay();void display();protected:double monthlyPay;class salesmanager:public manager,public salesmanpublic:salesmanager()monthlyPay=5000;CommRate=0.005;void promote(int)employee:promote(2);void pay();void display();int employee:MaxNo=1000;employee:employee()char namestr50;coutnamestr;name=new charstrlen(namestr)+1;strcpy(name,namestr);no=MaxNo+;grade=1;Pay=0.0;void technician:pay()cout请输入nameworkHours;Pay=hourlyRate*workHours;cout兼职技术人员name的编号为:not本月工资为:Payendl;void technician:display()cout兼职技术人员name的编号为:.not级别为:grade级,己付本月工资:Payendl;void salesman:pay()cout请输入.namesales;Pay=sales*CommRate;cout推销员name的编号为:not本月工资为:Payendl;void salesman:display()cout推销员name的编号为:not级别为:grade级,己付本月工资:Payendl;void manager:pay()Pay=monthlyPay;cout经理name的编号为:not本月工资为:Payendl;void manager:display()cout经理name的编号为:not级别为:grade级,己付本月工资:Payendl;void salesmanager:pay()cout请输入employee:namesales;Pay=monthlyPay+CommRate*sales;cout销售经理name的编号为:not本月工资为:Payendl;void salesmanager:display()cout销售经理name的编号为:not级别为:grade级,己付本月工资:Payendl;void main()manager m1;technician t1;salesmanager sm1;salesman s1;employee *emp4=&m1,&t1,&sm1,&s1;int i;for(i=0;ipromote();empi-pay();empi-display();2）编写一个程序,建立两种类型的表:队列与堆钱,使它们可以共用一个接口访问。答案：#include #include #include class Listpublic:List()head=tail=next=NULL;virtual void Store(int i)=0;virtual int Retrieve()=0;List *head, *tail, *next;int num;class Quene:public Listpublic:void Store(int i);int Retrieve();void Quene:Store(int i)List *item;item=new Quene;if(!item)coutAllocation error.num=i;if(tail)tail-next=item;tail=item;item-next=NULL;if(!head)head=tail;int Quene:Retrieve()int i;List *p;if(!head)coutQuene empty.num;p=head;head=head-next;delete p;return i;class Stack :public Listpublic:void Store(int i);int Retrieve();void Stack:Store(int i)List *item;item=new Stack;if(!item)coutAllocation error.num=i;if(head)item-next=head;head=item;if(!tail)tail=head;int Stack:Retrieve()int i;List *p;if(!head)coutStack empty.num;p=head;head=head-next;delete p;return i;void main()List *p;Quene qobj;p=&qobj;for(int i=0;iStore(i+1);coutQuene.;for(i=0;i3;i+)coutRetrieve();Stack sobj;p=&sobj;for(i=0;iStore(i+1);coutendlStack:;for(i=0;i3;i+)coutRetrieve();3）编写一个程序，先设计一个整数链表List类,然后从此链表派生出一个整数集合类set,在集合举中增加一个元素个数的数据项。集合类的插入操作与链表相似(只是不插入重复元素),并且插入后,表示元素个数的数据成员需增值。集合类的删除操作是在链表删除操作的基础上对元素个数做减1操作:而查找和显示操作是相同的。答案：#include enum boolean False,True;struct Nodeint val;Node *next;class Listpublic:List()nodes=0;List()Node *temp=nodes;for(Node *n=nodes;nodes!=0;)temp=nodes;nodes=nodes-next;delete temp;virtual boolean Insert(int val)Node *temp=new Node;if(temp!=0)temp-val=val;temp-next=nodes;nodes=temp;return True;elsereturn False;virtual boolean Delete(int val)if(nodes=0)return False;Node *temp=nodes;if(nodes-val=val)nodes=nodes-next;delete temp;return True;elsefor(Node *n=nodes;n-next!=0;n=n-next)if(n-next-val=val)temp=n-next;n-next=temp-next;delete temp;return True;return False;boolean IsContain(int val)if(nodes=0)return False;if(nodes-val=val)return True;elsefor(Node *n=nodes;n-next!=0;n=n-next)if(n-next-val=val)return True;return False;void Print()if(nodes=0)return;for(Node *n=nodes;n!=0;n=n-next)coutval ;coutInsert(10);p-Insert(20);p-Insert(30);p-Insert(40);p-Print();p-Delete(30);p-Print();p=&set1;p-Insert(12);p-Insert(22);p-Insert(32);p-Print();p-Delete(22);p-Print();4)某人喜欢饲养宠物。假定他拥有的放置宠物的窝的数目是固定的。请设计一个程序,使得某人可以饲养任意种类任意数目的宠物。要求能够知道现在饲养了多少只宠物, 每只宠物所在的位置及其种类和姓名。答案：#include #include #include #include class Animalpublic:Animal()name=NULL;Animal(char *n)name=strdup(n);virtual Animal()delete name;virtual void WhoAmI()coutGeneric Animal.endl;protected:char *name;class Cat: public Animal public:Cat()Cat(char *n):Animal(n)virtual void WhoAmI()coutI am a cat namednameendl;class Pig: public Animal public:Pig()Pig(char *n):Animal(n)virtual void WhoAmI()coutI am a pig namednameendl;class Dog: public Animalpublic:Dog()Dog(char *n):Animal(n)virtual void WhoAmI()coutI am a dog namednameendl;class Kennelpublic:Kennel(int max);virtual Kennel()delete Residents;int Accept(Animal *d);Animal *Release(int pen);void ListAnimals();private:int MaxAnimals,NumAnimals;Animal