• 项目2-形状类族中的纯虚函数


    项目2-形状类族的中的纯虚函数


    写一个程序,定义抽象基类Shape,由它派生出3个派生类,Circle(圆形)、Rectangle(矩形)、Triangle(三角形)。用例如以下的main()函数。求出定义的几个几何体的面积和。 

    int main()
    {
        Circle c1(12.6),c2(4.9);//建立Circle类对象c1,c2,參数为圆半径
        Rectangle r1(4.5,8.4),r2(5.0,2.5);//建立Rectangle类对象r1,r2,參数为矩形长、宽
        Triangle t1(4.5,8.4),t2(3.4,2.8); //建立Triangle类对象t1,t2,參数为三角形底边长与高
        Shape *pt[6]= {&c1,&c2,&r1,&r2,&t1,&t2}; //定义基类指针数组pt,使它每个元素指向一个派生类对象
        double areas=0.0; //areas为总面积
        for(int i=0; i<6; i++)
        {
            areas=areas + pt[i]->area();
        }
        cout<<"totol of all areas="<<areas<<endl;   //输出总面积
        return 0;
    }


     

    #include <iostream>
    using namespace std;
    class Shape
    {
    public:
        virtual double area()=0;
    };
    class Circle:public Shape
    {
    public:
        Circle(double r):radius(r) {}
        virtual double area()
        {
            return 3.14159*radius*radius;
        };
    protected:
        double radius;
    };
    class Rectangle:public Shape
    {
    public:
        Rectangle(double w,double h):width(w),height(h) {}
        virtual double area()
        {
            return width*height;
        }
    protected:
        double width,height;
    };
    
    class Triangle:public Shape
    {
    public:
        Triangle(double w,double h):width(w),height(h) {}
        virtual double area()
        {
            return 0.5*width*height;
        }
    protected:
        double width,height;
    };
    
    int main()
    {
        Circle c1(12.6),c2(4.9);
        Rectangle r1(4.5,8.4),r2(5.0,2.5);
        Triangle t1(4.5,8.4),t2(3.4,2.8);
        Shape *pt[6]= {&c1,&c2,&r1,&r2,&t1,&t2};
        double areas=0.0;
        for(int i=0; i<6; i++)
        {
            areas=areas + pt[i]->area();
        }
        cout<<"totol of all areas="<<areas<<endl;
        return 0;
    }
    


     

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  • 原文地址:https://www.cnblogs.com/llguanli/p/6745036.html
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