求一道关于类和对象的题

Question

自定义如下形式的 Point 类，其对象表示平面上的一个点(x, y)，通过类成员方式对该类重载二目运算符“十”和“^”，用来求出两个对象的和以及两个对象(平面点)的距离。各运算符的使用含义(运算结果）如下所示：
(1.2,-3.5)+(-1.5,6)=(-0.3,2.5)
(1.2,-3.5)^（-1.5,6)＝9.87623
在类外定义类中已声明的函数，并编写主函数，验证所定义函数的正确性。
class Point {
friend ostream & operator<((ostream& os, const Point & p):
friend istream & operator>>(istream & is, Point&p);
public:
Point (double x0 = 0, double y0 =0){x = x0:y = y0;)
Point(const Point & p);
Point operator＋(const Point & pt);
double operator^(const Point & pt):
private:
double x,y;
};

Answer 1

代码及运行结果如下：

#include <iostream>
using namespace std;

class Point {
    friend ostream& operator<<(ostream& os, const Point& p);
    friend istream& operator>>(istream& is, Point& p);
public:
    Point(double x0 = 0, double y0 = 0) {
        x = x0; y=y0;
    }
    Point(const Point& p) {
        x = p.x;
        y = p.y;
    }
    Point operator-(const Point& pt) {
        Point pp;
        pp.x = x - pt.x;
        pp.y = y - pt.y;
        return pp;
    }
    Point operator +(const Point& pt) {
        Point pp;
        pp.x = x + pt.x;
        pp.y = y + pt.y;
        return pp;
    }
    double operator ^(const Point& pt) {
        double t = sqrt((x-pt.x) * (x-pt.x) + (y-pt.y) * (y-pt.y));
        return t;
    }
private:
    double x, y;
};

ostream& operator<<(ostream& os, const Point& p) {
    os <<"(" << p.x << "," << p.y<<")";
    return os;
}
istream& operator>>(istream& is, Point& p) {
    is >> p.x >> p.y;
    return is;
}


int main()
{
    Point p1(1.2, -3.5);
    Point p2(-1.5, 6);
    Point p3 = p1 + p2;
    cout <<p1 <<"+" << p2 <<"=" << p3 << endl;
    double t = p1 ^ p2;
    cout <<p1 <<"^"<<p2 <<"=" << t << endl;
    return 0;
}