在java中操作類

Question

我對如何從自定義類中提取數據感到困惑。該代碼將笛卡爾座標組織在一個名爲linesegment的類中，其中幾個類CartesianCoordinate作爲其成員。我被困在試圖找到兩組笛卡爾座標之間的距離。

我該如何解碼linesegment類，進入cartesiancoordinate類，然後訪問單個的double值從主類打印到屏幕？

下面是我的程序中使用的三類：

主類：

public class lab3 
{ 

    public static void main(String [] args) 
    { 
     cartesiancoordinate one, two; //instantsiating one and two as type cartesiancoordiante 
     one = new cartesiancoordinate(5, 6); //putting the information for one and two into type cartesiancoordinate 
     two = new cartesiancoordinate(4.5, -6.5); 

     linesegment oneandtwo; 
     oneandtwo = new linesegment(one, two);  

     System.out.println(one.toString()); //dual X/Y statements using a toString method 
     System.out.println(two.toString());  

     System.out.println(oneandtwo.tostring()); 

     System.out.println("X for one is: " + one.getx()); //individual X/Y statements using getter methods 
     System.out.println("Y for one is: " + one.gety()); 
     System.out.println("X for two is: " + two.getx()); //individual X/Y statements using getter methods 
     System.out.println("Y for two is: " + two.gety()); 

     double tester; 
     oneandtwo.test();  
     System.out.println("The test method returned the distance between the two cartesian coordinates to be: " + tester); 
    } 
} 

的cartesiancoordinate類：

class cartesiancoordinate 
{ 
    private double xposition; 
    private double yposition; 

    public cartesiancoordinate(double x, double y) 
    { 
     this.xposition = x; 
     this.yposition = y; 
    } 

    public double getx() 
    { 
     return this.xposition; 
    } 

    public double gety() 
    { 
     return this.yposition; 
    } 

    public String toString() 
    { 
     return "(" + this.xposition + "/" + this.yposition + ")"; 
    } 
} 

的麻煩線段類：

class linesegment 
{ 
    private cartesiancoordinate startpoint, endpoint, s1, e1;  
    public cartesiancoordinate one, two; 


    public linesegment(cartesiancoordinate x, cartesiancoordinate y) 
    { 
     this.startpoint = x; 
     this.endpoint = y; 
    } 

    public cartesiancoordinate getstartpoint() 
    { 
     return this.startpoint; 
    } 

    public cartesiancoordinate getendpoint() 
    { 
     return this.endpoint; 
    } 

    public String tostring() 
    { 
     return ("The start point is " + this.startpoint + " and the end point is " + this.endpoint); 
    } 


    public double test() 
    { 

     double x1,x2,y1,y2; 
     cartesiancoordinate s1,e1; 

     getstartpoint() = s1; 
     getendpoint() = e1 ; 

     s1.getx() = x1; 
     s1.gety() = y1; 
     e1.getx() = x2; 
     e1.gety() = y2; 

     double tester; 
     tester = x1 + x2 + y1 + y2; 
     return tester; 
    } 
} 

Answer 1

相反的：

double tester; 
oneandtwo.test(); 

你想：

double tester = oneandtwo.test(); 

Answer 2

這是一個問題。

getstartpoint() = s1; 
getendpoint() = e1 ; 

你的方法是return -ing值，所以如果你想將它們分配到s1和e1，那麼你就可以做到這一點，像這樣

s1 = getstartpoint(); 
e1 = getendpoint(); 

這不會解決的邏輯你代碼，但它至少應該編譯。

Answer 3

你linesegment類被打破：

class linesegment { 
    // a linesegment is nothing more than defined by a starting and ending point 
    private cartesiancoordinate startpoint, endpoint;  

    public linesegment(cartesiancoordinate x, cartesiancoordinate y) { 
     this.startpoint = x; 
     this.endpoint = y; 
    } 

    public cartesiancoordinate getstartpoint() { 
     return this.startpoint; 
    } 

    public cartesiancoordinate getendpoint() { 
     return this.endpoint; 
    } 

    public String toString() { 
     return ("The start point is " + this.startpoint + " and the end point is " + this.endpoint); 
    }  

    public double test() 
    { 
     double dx = endpoint.getx()-startpoint.getx(); 
     double dy = endpoint.gety()-startpoint.gety(); 
     return Math.sqrt(dx*dx+dy*dy); 
    } 
} 

現在主：

double tester = oneandtwo.test(); // get the distance 
System.out.println("Distance is "+tester); 

請使用標準的命名約定。您的課程必須拼寫CartesianCoordinates,LineSegment,Lab3。您的變量：oneAndTwo,startPoint,endPoint ...