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问题描述:
下面是给出的基类Animal声明和main()函数。
<strong><span style="font-family:KaiTi_GB2312;font-size:18px;color:#ff6666;">class Animal { public: virtual void cry() { cout<<"不知哪种动物,让我如何学叫?"<<endl; } }; int main( ){ Animal *p; p = new Animal(); p->cry(); Mouse m1("Jerry",'m'); p=&m1; p->cry(); Mouse m2("Jemmy",'f'); p=&m2; p->cry(); Cat c1("Tom"); p=&c1; p->cry(); Dog d1("Droopy"); p=&d1; p->cry(); Giraffe g1("Gill",'m'); p=&g1; p->cry(); return 0; }</span></strong>
程序的运行结果将是:
3、每一个Animal的派生类都有一个“名字”数据成员,这一共有的成员完全可以由基类提供改造上面的程序,将这一数据成员作为抽象类Animal数据成员被各派生类使用。
代码:
#include <iostream> #include <cstring> using namespace std; class Animal{ protected: string name; public: Animal (string s):name(s){} virtual void cry()=0; }; class Mouse:public Animal{ private: char sex; public: Mouse(string s,char x):Animal(s),sex(x){} void cry(){ cout<<"我叫"<<name<<",我是一只"<<(sex=='m'?"公":"母")<<"老鼠,我的叫声是:吱吱吱!\n"; } }; class Cat:public Animal{ public: Cat(string s):Animal(s){} void cry(){ cout<<"我叫"<<name<<",我是一只猫,我的叫声是:喵喵喵!\n"; } }; class Dog:public Animal{ public: Dog(string s):Animal(s){} void cry(){ cout<<"我叫"<<name<<",我是一只狗,我的叫声是:汪汪汪!\n"; } }; class Giraffe:public Animal{ private: char sex; public: Giraffe(string s,char x):Animal(s),sex(x){} void cry(){ cout<<"我叫"<<name<<",我是一只"<<(sex=='m'?"公":"母")<<"长颈鹿,我的脖子太长,发不出声音来!\n"; } }; int main( ){ Animal *p; Mouse m1("Jerry",'m'); p=&m1; p->cry(); Mouse m2("Jemmy",'f'); p=&m2; p->cry(); Cat c1("Tom"); p=&c1; p->cry(); Dog d1("Droopy"); p=&d1; p->cry(); Giraffe g1("Gill",'m'); p=&g1; p->cry(); return 0; }
运行结果:
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原文地址:http://blog.csdn.net/zp___waj/article/details/46049911