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width="85%" class="body18black">C++习题与解析(继承和派生-01)</TD>
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6.1 分析以下程序的执行结果</FONT><BR>#include<iostream.h><BR>class base<BR>{<BR> public:<BR> base(){cout<<"constructing base class"<<endl;}<BR> ~base(){cout<<"destructing base class"<<endl; }<BR>};<BR>class subs:public base<BR>{<BR> public:<BR> subs(){cout<<"constructing sub class"<<endl;}<BR> ~subs(){cout<<"destructing sub class"<<endl;}<BR>};<BR>void main()<BR>{<BR> subs s;<BR>}<BR>解:<BR> 本题说明单继承情况下构造函数和析构函数的调用顺序。这里base为基类,subs为派生类。<BR>所以输出为:<BR> constructing base class<BR> constructing sub class<BR> destructing sub class<BR> destrcuting base class<BR>注意:在单继承情况下,首先调用基类的构造函数 ,随后调用派生类的构造函数,析构函数的调用顺序则正好相反。<BR><BR>---------------------------------------------------<BR><BR><FONT color=#0000ff>6.2 分析以下程序的执行结果:</FONT><BR>#include<iostream.h><BR>class base<BR>{<BR> int n;<BR> public:<BR> base(int a)<BR> {<BR> cout<<"constructing base class"<<endl;<BR> n=a;<BR> cout<<"n="<<n<<endl;<BR> }<BR> ~base(){cout<<"destructing base class"<<endl;}<BR>};<BR>class subs:public base<BR>{<BR> base bobj;<BR> int m;<BR> public:<BR> subs(int a,int b,int c):base(a),bobj(c)<BR> {<BR> cout<<"constructing sub cass"<<endl;<BR> m=b;<BR> cout<<"m="<<m<<endl;<BR> }<BR> ~subs(){cout<<"destructing sub class"<<endl;}<BR>};<BR>void main()<BR>{<BR> subs s(1,2,3);<BR>}<BR>解:<BR> 本题说明 派生类中含有对象成员情况下构造函数和析构函数的调用顺序。这里base为基类,subs为派生类,subs类的构造函数中含有对象成员。<BR>所以输出为:<BR> constrcuting base class<BR> n=1<BR> constructing base class<BR> n=3<BR> constructing sub class<BR> m=2<BR> destructing sub class<BR> destructing base class<BR> destructing base class<BR>注意:当派生类中含有对象成员时,构造函数的调用顺序如下:<BR>1)基类的构造函数<BR>2)对象成员的构造函数<BR>3)派生类的构造函数<BR><BR>析构函数的调用顺序与之相反<BR><BR>-------------------------------------------------<BR><BR><FONT color=#0000ff>6.3 分析以下程序的执行结果</FONT><BR>#include<iostream.h><BR>class A<BR>{<BR> public:<BR> int n;<BR>};<BR>class B:public A{};<BR>class C:public A{};<BR>class D:public B,public C<BR>{<BR> int getn(){return B::n;}<BR>};<BR>void main()<BR>{<BR> D d;<BR> d.B::n=10;<BR> d.C::n=20;<BR> cout<<d.B::n<<","<<d.C::n<<endl;<BR>}<BR>解:<BR> D类是从类和类派生的而类和类又都是从类派生的,但各有自己的副本。所以对于对象d,d.B::n与d.C::n是两个不同的数据成员它们互无联系。<BR>所以输出为:<BR>10,20</P>
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<P><FONT color=#0000ff>6.4 分析以下程序的执行结果</FONT><BR>#include<iostream.h><BR>class A<BR>{<BR> public:<BR> int n;<BR>};<BR>class B:virtual public A{};<BR>class C:virtual public A{};<BR>class D:public B,public C<BR>{<BR> int getn(){return B::n;}<BR>};<BR>void main()<BR>{<BR> D d;<BR> d.B::n=10;<BR> d.C::n=20;<BR> cout<<d.B::n<<","<<d.C::n<<endl;<BR>}<BR>解:<BR> D类是从类和类派生的而类和类又都是从类派生,但这是虚继承关系即是虚基类因此和共用一个的副本所以对于对象d,d.B::n与d.C::n是一个成员。<BR>所以输出为:<BR>20,20</P>
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<P><FONT color=#0000ff>6.5 假设图书馆的图书包含书名、编号作者属性读者饮包含姓名和借书证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。<BR></FONT>解:<BR> 设计一个类,从它派生出胃病书类book和读者类reader,在reader类中有一个rentbook()成员函数用于借阅图书。<BR>程序代码如下:<BR>本程序的执行结果如下:<BR>#include<iostream.h><BR>#include<string.h><BR>class object<BR>{<BR> char name[20];<BR> int no;<BR> public:<BR> object(){}<BR> object(char na[],int n)<BR> {<BR> strcpy(name,na);no=n;<BR> }<BR> void show()<BR> {<BR> cout<<name<<"("<<no<<")";<BR> }<BR>};<BR>class book:public object<BR>{<BR> char author[10];<BR> public:<BR> book(){}<BR> book(char na[],int n,char auth[]):object(na,n)<BR> {<BR> strcpy(author,auth);<BR> }<BR> void showbook()<BR> {<BR> show();<BR> cout<<"作者:"<<author;<BR> }<BR>};<BR>class reader:public object<BR>{<BR> book rent[5];<BR> int top;<BR> public:<BR> reader(char na[],int n):object(na,n){top=0;}<BR> void rentbook(book &b)<BR> {<BR> rent[top]=b;<BR> top++;<BR> }<BR> void showreader()<BR> {<BR> cout<<"读者:";show();<BR> cout<<endl<<"所借图书:"<<endl;<BR> for(int i=0;i<top;i++)<BR> {<BR> cout<<" "<<i+1<<":"; // 5个空格<BR> rent[i].show();<BR> cout<<endl;<BR> }<BR> }<BR>};<BR>void main()<BR>{<BR> book b1("C语言",100,"谭浩强"),b2("数据结构",110,"严蔚敏");<BR> reader r1("王华",1234);<BR> r1.rentbook(b1);<BR> r1.rentbook(b2);<BR> r1.showreader();<BR>}<BR><BR>读者王华(1234)<BR>所借图书:<BR> 1:C语言(100)<BR> 2:数据结构(110)</P>
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