课程首页在:http://blog.csdn.net/sxhelijian/article/details/11890759,内有完整教学方案及资源链接


【项目2 - 圆的比较】
   在项目1的基础上,在圆类上重载关系运算符(6种),使之能够按圆的面积比较两个圆的大小。自编main函数完成测试。


参考解答:

#include <iostream>
#include<Cmath>
using namespace std;
class Point
{
public:
    Point(double a=0,double b=0):x(a),y(b) {}		//构造函数
protected:										 //受保护成员
    double x,y;
};

class Circle:public Point //circle是Point类的公用派生类
{
public:
    Circle(double a=0,double b=0,double r=0): Point(a,b),radius(r) { }//构造函数
    double area ( ) const; //计算圆面积
    friend ostream &operator<<(ostream &,const Circle &);//重载运算符“<<”
    //重载关系运算符运算符,使之能够按圆的面积比较两个圆的大小;
    bool operator>(const Circle &);
    bool operator<(const Circle &);
    bool operator>=(const Circle &);
    bool operator<=(const Circle &);
    bool operator==(const Circle &);
    bool operator!=(const Circle &);
protected:
    double radius;
};

//计算圆面积
double Circle::area( ) const
{
    return 3.14159*radius*radius;
}

//重载运算符“<<”,使之按规定的形式输出圆的信息
ostream &operator<<(ostream &output,const Circle &c)
{
    output<<"Center=["<<c.x<<", "<<c.y<<"], r="<<c.radius;
    return output;
}

//重载关系运算符(种)运算符,使之能够按圆的面积比较两个圆的大小;
bool Circle::operator>(const Circle &c)
{
    return (this->radius - c.radius) > 1e-7;
}

bool Circle::operator<(const Circle &c)
{
    return (c.radius - this->radius) > 1e-7;
}

bool Circle::operator>=(const Circle &c)
{
    return !(*this < c);
}

bool Circle::operator<=(const Circle &c)
{
    return !(*this > c);
}

bool Circle::operator==(const Circle &c)
{
    return abs(this->radius - c.radius) < 1e-7;
}

bool Circle::operator!=(const Circle &c)
{
    return abs(this->radius - c.radius) > 1e-7;
}

int main( )
{
    Circle c1(3,2,4),c2(4,5,5);      //c2应该大于c1
    cout<<"圆c1( "<<c1<<" )的面积是 "<<c1.area()<<endl;
    cout<<"圆c2( "<<c2<<" )的面积是 "<<c2.area()<<endl;
    cout<<"圆c1 ";
    if(c1>c2) cout<<"大于, ";
    if(c1<c2) cout<<"小于, ";
    if(c1>=c2) cout<<"大于等于, ";
    if(c1<=c2) cout<<"小于等于, ";
    if(c1==c2) cout<<"等于, ";
    if(c1!=c2) cout<<"不等于, ";
    cout<<"圆c2"<<endl;

    return 0;
}



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