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)。用如下的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,c2,参数为圆半径 Rectangle r1(4.5,8.4),r2(5.0,2.5);//建立Rectangle类对象r1,r2,参数为矩形长、宽 Triangle t1(4.5,8.4),t2(3.4,2.8); //建立Triangle类对象t1,t2,参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,使它每一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; }
原文地址:http://blog.csdn.net/i__am__legend/article/details/46340405