/*
* Copyright (c) 2014, 烟台大学计算机学院
* All rights reserved.
* 文件名称:test.cpp
* 作 者:呼亚萍
* 完成日期:2015年5月27日
* 版 本 号:v1.0
*
* 问题描述: (1)先建立一个Point(点)类,包含数据成员x,y(坐标点);
(2)以Point为基类,派生出一个Circle(圆)类,增加数据成员(半径),基类的成员表示圆心;
(3)编写上述两类中的构造、析构函数及必要运算符重载函数(本项目主要是输入输出);
(4)定义友元函数int locate,判断点p与圆的位置关系(返回值<0圆内,==0圆上,>0 圆外);
* 程序输入:相应的程序
* 程序输出:对应得结果
*/
#include<iostream>
#include<cmath>
using namespace std;
class Point
{
public:
Point()
{
x=0;
y=0;
}
Point(int a,int b):x(a),y(b) {}
double distant(const Point&p);
friend ostream& operator<<(ostream&,Point&);
protected:
double x;
double y;
};
double Point::distant(const Point&p)
{
double dx,dy;
dx=x-p.x;
dy=y-p.y;
return sqrt(dx*dx+dy*dy);
}
ostream& operator<<(ostream&output,Point&p)
{
output<<"("<<p.x<<","<<p.y<<")";
return output;
}
class Circle:public Point
{
public:
Circle(double a=0,double b=0,double r=0):Point(a,b),radius(r) {}
friend ostream& operator<<(ostream&,Circle&);
friend int locate(Point&,Circle&);
private:
double radius;
};
ostream& operator<<(ostream&output,Circle&c)
{
output<<"("<<c.x<<","<<c.y<<")"<<" ";
cout<<"radius="<<c.radius;
cout<<endl;
return output;
}
int locate(Point&p,Circle&c)
{
Point cp(c.x,c.y);
double d=cp.distant(p);
if(d==c.radius)
return 0;
if(d>c.radius)
return 1;
if(d<c.radius)
return -1;
}
int main( )
{
Circle c1(3,2,4),c2(4,5,5); //c2应该大于c1
Point p1(1,1),p2(3,-2),p3(7,3); //分别位于c1内、上、外
cout<<"圆c1: "<<c1;
cout<<"点p1: "<<p1;
cout<<"点p1在圆c1之"<<((locate(p1, c1)>0)?"外":((locate(p1, c1)<0)?"内":"上"))<<endl;
cout<<"点p2: "<<p2;
cout<<"点p2在圆c1之"<<((locate(p2, c1)>0)?"外":((locate(p2, c1)<0)?"内":"上"))<<endl;
cout<<"点p3: "<<p3;
cout<<"点p3在圆c1之"<<((locate(p3, c1)>0)?"外":((locate(p3, c1)<0)?"内":"上"))<<endl;
return 0;
}
运算结果:
知识点总结:
类的派生以及运算符的重载!
学习心得:
在编写完程序后,在对照讲义整理程序的框架,也会有新的收获!