第十二周任务四类的组合与继承

本文介绍了一个基于C++实现的点与圆类的设计,包括类的组合与继承、构造与析构函数、友元函数及重载运算符等。通过具体实例展示了如何判断点与圆的位置关系,并计算点与圆心连线与圆的交点。

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/* (程序头部注释开始)
* 程序的版权和版本声明部分
* Copyright (c) 2011, 烟台大学计算机学院学生 
* All rights reserved.
* 文件名称: 第十二周任务四类的组合与继承                             
* 作    者:石丽君                              
* 完成日期:      2012   年      5 月   9     日
* 版 本 号:          

  * 对任务及求解方法的描述部分
  * 输入描述:类的组合与继承
  (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、析构函数及必要的输入输出函数
  (4)定义友元函数int locate,判断点p在圆c上、圆c内或圆c外,返回值<0圆内,==0圆上,>0 圆外;
  (5)重载关系运算符(6种)运算符,使之能够按圆的面积比较两个圆的大小;
  (6)给定一点p,求出该点与圆心相连成的直线与圆的两个交点并输出
  
	* 问题描述: 
	* 程序输出: 
	* 程序头部的注释结束
*/

#include <iostream>
#include<cmath>  
using namespace std;  
#define pi 3.1415926  
class Point   
{  
protected:  
	double x,y;  
public:  
	Point () {x=0;y=0;}  
	Point (double x0,double y0)  
	{  
		x=x0;  
		y=y0;  
	}  
	~Point();  
	double getx() {return x;}  
	double gety() {return y;} 
	void setx(double n) {x=n;}
	void sety(double n) {y=n;}
	friend ostream& operator<<(ostream &output,Point c);  
};  
class Circle:public Point  
{  
protected:  
	double r;  
public:  
	Circle(double x0,double y0,double r0):Point(x0,y0), r(r0) {}  
	friend ostream&operator<< (ostream& output,Circle c); 
	bool operator >(Circle &c1);
	bool operator <(Circle &c1);
	bool operator >=(Circle &c1);
	bool operator <=(Circle &c1);
	bool operator ==(Circle &c1);
	bool operator !=(Circle &c1);
	friend void crossover_point1(Point &p1,Circle &c1, Point &p4, Point &p5);
	friend double locate(Point &p,Circle &c);
	double getr() {return r;}  
	~Circle();  
};  

ostream& operator<<(ostream &output,Point c)  
{  
	output<<'('<<c.getx()<<','<<c.gety()<<')'<<endl;  
	return output;  
}  
ostream& operator<< (ostream& output,Circle c)  
{  
	output<<"圆心坐标为:"<<"("<<c.getx()<<','<<c.gety()<<")"<<"半径为:"<<c.getr()<<endl;  
	return output;  
}  
Point::~Point()  
{  
}  
Circle::~Circle()  
{  
}
bool Circle::operator >(Circle &c1)
{
	if(r>c1.r)
		return true;
	return false;
}
bool Circle::operator <(Circle &c1)
{
	if(r>c1.r)
		return false;
	return true;
}
bool Circle::operator >=(Circle &c1)
{
	if(r>=c1.r)
		return true;
	return false;
}
bool Circle::operator <=(Circle &c1)
{
	if(r>=c1.r)
		return false;
	return true;
}
bool Circle::operator ==(Circle &c1)
{
	if(r==c1.r)
		return true;
	return false;
}
bool Circle::operator !=(Circle &c1)
{
	if(r!=c1.r)
		return true;
	return false;
}
void crossover_point1(Point &p1,Circle &c1, Point &p4, Point &p5)
{
	double n,m;
	n=(c1.getx()-sqrt(((c1.r)*(c1.r))/((c1.gety()-p1.gety())/(c1.getx()-p1.getx()))*((c1.gety()-p1.gety())/(c1.getx()-p1.getx()))));
	p4.setx(n);
	n=(c1.getx()+sqrt(((c1.r)*(c1.r))/((c1.gety()-p1.gety())/(c1.getx()-p1.getx()))*((c1.gety()-p1.gety())/(c1.getx()-p1.getx()))));
	p5.setx(n);
	m=(c1.gety()-p1.gety())/(c1.getx()-p1.getx())*(p4.getx()-p1.getx())+p1.gety();
	p4.sety(m);
	m=(c1.gety()-p1.gety())/(c1.getx()-p1.getx())*(p5.getx()-p1.getx())+p1.gety();
	p5.sety(m);
}
double locate(Point &p,Circle &c)
{
	double d,d1;
	d=sqrt((p.getx()-c.getx())*(p.getx()-c.getx())+(p.gety()-c.gety())*(p.gety()-c.gety()));
	d1=d-c.r;
	return d1;
	
}
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;
	cout<<endl; 
	cout<<"圆c1: "<<c1;
	if(c1>c2) cout<<"大于"<<endl;
	if(c1<c2) cout<<"小于"<<endl; 
	if(c1>=c2) cout<<"大于等于"<<endl;
	if(c1<=c2) cout<<"小于等于"<<endl; 
	if(c1==c2) cout<<"等于"<<endl; 
	if(c1!=c2) cout<<"不等于"<<endl; 
	cout<<"圆c2: "<<c2;
	cout<<endl; 
	
	Point p4,p5;
	crossover_point1(p1,c1, p4, p5);
	
	cout<<"点p1: "<<p1;
	cout<<"与圆c1: "<<c1;
	cout<<"的圆心相连,与圆交于两点,分别是:"<<endl;
	cout<<"交点: "<<p4;
	cout<<"交点: "<<p5;
	cout<<endl; 
	system("pause");
	return 0;
}

经验积累:在使用void setx()函数时,其无返回值,在使用void crossover_point1(Point &p1,Circle &c1, Point &p4, Point &p5)时,应为个double getx()



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