15第十二周实践项目——程序阅读

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/*
 * Copyright (c) 2014, 烟台大学计算机学院
 * All rights reserved.
 * 文件名称:test.cpp
 * 作    者:李晓凯
 * 完成日期:2015年 5 月 24 日
 * 版 本 号:v1.0
 *
 * 问题描述:
 * 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;    }    //思考:这3处出现的print,有何区别    void print()      //(1)    {        A::print();   //(2)        aa.print();   //(3)        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;}


(2)

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


(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;}


(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;}


问题:

将程序中有注释的两条语句修改为注释的内容,即将“class B:public A {};”修改为“class B:virtual public A{};”,“class C:public A {}; ”修改为“class C:virtual public A{};”,重新编译运行程序,程序的运行结果又是多少?

答案:20

原因:在D类中只有一个n,而没有B类和C类之分,所以输出20.

(5)阅读下面类的定义,请说出在测试函数中不同情况的调用产生的结果

#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=b;    a.printA();    b.printA();    b.printB();

    b=a;    a.printA();    b.printA();    b.printB();

程序会发生编译错误,原因是:____基类不能对派生类赋值______

    A &r1=a;    A &r2=b;    r1.printA();    r2.printA();    r2.printB();
将会发生错误的一行删除;r2.printB();

那一行的错误原因是:_______r2是类A的对象,不能调用类B的成员函数___________

    A *p=&a;    p->printA();    p=&b;    p->printA();    p->printB();
将会发生错误的一行删除;p->printB();

那一行的错误原因是:_______指针对象p属于A,不能指向类B中的成员函数___________

类A中加入以下代码:

    int getA(){return a;}
main函数前加以下代码:

void f(A x){    cout<<"aaaaah, my a: "<<x.getA()<<endl;}
main中指定部分代码:

    f(a);    f(b);





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