题目要求:
使用纯虚函数来完成.
定义一个共用接口,计算正方体,长方体,球体的体积和表面积
提示:
class shape    //创建一个图形类,作为父类
{
public:
    virtual double tj()=0;        //方法1:计算体积
    virtual double bmj()=0;    //方法2:计算表面积
};

本人答案:



#include<iostream.h>
#define Pi 3.1415926
class shape    
{
public:    
    double x;
    double y;
    double z;
    
    double getx(double a)
    {
        x=a;
        return x;
    }

    double gety(double a,double b)
    {
        x=a;y=b;
        return x,y;
    }

    double getz(double a,double b,double c)
    {
        x=a;y=b;z=c;
        return x,y,z;
    }
    virtual double tj()=0;    
    virtual double bmj()=0;    
};

class zft:public shape
{    
public:

    double tj()
    {
        cout<<"正方体体积:"<<x*x*x<<endl;
        return x*x*x;
    }

    double bmj()
    {
        cout<<"正方体表面积:"<<6*x*x<<endl;
        return 6*x*x;
    }
};

class cft:public shape
{    
public:

    double tj()
    {
        cout<<"长方体体积:"<<x*y*z<<endl;
        return x*y*z;
    }

    double bmj()
    {
        cout<<"长方体表面积:"<<2*(x*y+y*z+z*x)<<endl;
        return 2*(x*y+y*z+z*x);
    }
};

class qt:public shape
{    
public:

    double tj()
    {
        cout<<"球体体积:"<<(4.0/3)*Pi*x*y*z<<endl;
        return (4.0/3)*Pi*x*y*z;
    }

    double bmj()
    {
        cout<<"球体表面积:"<<4*Pi*x*y<<endl;
        return 4*Pi*x*y;
    }
};

void main()
{
    shape *p;

    zft obj;
    p=&obj;
    p->getx(3);
    p->tj();
    p->bmj();

    cft obj2;
    p=&obj2;
    p->getz(3,4,5);
    p->tj();
    p->bmj();

    qt obj3;
    p=&obj3;
    p->getz(6,6,6);
    p->tj();
    p->gety(6,6);
    p->bmj();

}

[[it] 本帖最后由 happycplusplus 于 2008-10-10 19:05 编辑 [/it]]

忘了说 至于计算结果可以自己输入也可以直接显示