點派生出線2

/** Copyright (c) 2013, 煙台大學電腦學院* All rights reserved.* 檔案名稱：test.cpp* 作者：邱學偉* 完成日期：2013 年 5 月 11 日* 版本號碼：v1.0* 輸入描述：無* 問題描述： 定義點類，並以點類為基類，派生出直線類，從基類中繼承的點的資訊表示直線的中點* 程式輸出：* 問題分析：* 演算法設計：略*/#include <iostream>#include <cmath>using namespace std;class Point{  public:  Point():x(0),y(0){};  Point(double x1,double y1)  {     x=x1;     y=y1;  }  double getx(){return x;}  double gety(){return y;}  void display();  private:  double x,y;};void Point::display(){    cout<<"Point:("<<x<<","<<y<<")"<<endl;}class Line:public Point{    public:    Line(Point p1,Point p2);    double Lengh();    void PrintLine();    void PrintPoint();    private:    class Point pts,pte;};Line::Line(Point p1,Point p2){        pts=p1;        pte=p2;}double Line::Lengh(){    double x0=pts.getx()-pte.getx();    double y0=pts.gety()-pte.gety();    return sqrt(x0*x0+y0*y0);}void Line::PrintLine(){    cout<<"端點為："<<endl;    pts.display();    pte.display();    cout<<"長度："<<Lengh()<<endl;}void Line::PrintPoint(){    cout<<"("<<(pts.getx()+pte.getx())/2<<","<<(pts.gety()+pte.gety())/2<<")"<<endl;}int main(){    Point pt(-2,5),pe(7,9);    Line l(pt,pe);    l.PrintLine();    cout<<"中點："<<endl;    l.PrintPoint();    return 0;}

 

心得體會：這是自己敲出來的，跟老師的有不少不同之處；

/** Copyright (c) 2013, 煙台大學電腦學院* All rights reserved.* 檔案名稱：test.cpp* 作者：邱學偉* 完成日期：2013 年 5 月 11 日* 版本號碼：v1.0* 輸入描述：無* 問題描述： 定義點類，並以點類為基類，派生出直線類，從基類中繼承的點的資訊表示直線的中點* 程式輸出：* 問題分析：* 演算法設計：略*/#include <iostream>#include <cmath>using namespace std;class Point{  public:  Point():x(0),y(0){};  Point(double x1,double y1)  {     x=x1;     y=y1;  }  double getx(){return x;}  double gety(){return y;}  void display();  private:  double x,y;};void Point::display(){    cout<<"Point:("<<x<<","<<y<<")"<<endl;}class Line:public Point{    public:    Line(Point p1,Point p2);    double Lengh();    void PrintLine();    void PrintPoint();    private:    class Point pts,pte;};Line::Line(Point p1,Point p2):Point((p1.getx()+p2.getx())/2,(p1.gety()+p2.gety())/2){        pts=p1;        pte=p2;}double Line::Lengh(){    double x0=pts.getx()-pte.getx();    double y0=pts.gety()-pte.gety();    return sqrt(x0*x0+y0*y0);}void Line::PrintLine(){    cout<<"端點為："<<endl;    pts.display();    pte.display();    cout<<"長度："<<Lengh()<<endl;}void Line::PrintPoint(){    cout<<"("<<(pts.getx()+pte.getx())/2<<","<<(pts.gety()+pte.gety())/2<<")"<<endl;}int main(){    Point pt(-2,5),pe(7,9);    Line l(pt,pe);    l.PrintLine();    cout<<"\n The middle point of Line: ";    l.PrintPoint() ;//輸出直線l中點的資訊    return 0;}