Question

我希望写一个带有 Point 和 Line 的构造函数的类，并且在Line类中我希望写一个查找两条线是否相交的方法。这是我的Point类：

public class Point {
static double x;
static double y;

public Point(double x, double y) {
    this.x = x;
    this.y = y;
}

}

这是我的Line Class

public class Line {

public Line(Point x, Point y) {
}

// create two points for a line
static Point x1y1;
static Point x2y2;

// create another two points for another line
static Point x3y3;
static Point x4y4;

public static void main(String[] args) {
    // create lines
    Line line1 = new Line(x1y1, x2y2);
    Line line2 = new Line(x3y3, x4y4);

    //initialize points
    x1y1 = new Point(1.0, 1.0);
    x2y2 = new Point(3.0, 3.0);
    x3y3 = new Point(1.0, 2.0);
    x4y4 = new Point(4.0, 2.0);

    //call method to find if lines intersect
    findIntersection(line1, line2);
    System.out.println(x1y1.x);
}

public static void findIntersection(Line line1, Line line2) {
    double denominator = (x1y1.x - x2y2.x) * (x3y3.y - x4y4.y)
            - (x1y1.y - x2y2.y) * (x3y3.x - x4y4.x);
    double px = 0;
    double py = 0;

    if (denominator == 0) {
        System.out.println("Lines are parallel, they do not intersect");
    } else {
        px = ((x1y1.x * x2y2.y - x1y1.y * x2y2.x) * (x3y3.x - x4y4.x) - (x1y1.x - x2y2.x)
                * (x3y3.x * x4y4.y - x3y3.y * x4y4.x))
                / denominator;
        py = ((x1y1.x * x2y2.y - x1y1.y * x2y2.x) * (x3y3.y - x4y4.y) - (x1y1.y - x2y2.y)
                * (x3y3.x * x4y4.y - x3y3.y * x4y4.x))
                / denominator;

        System.out.println(px + "," + py);

    }
}

}

问题是我初始化所有点以使线条相交，但是当我尝试打印出点的值时，第1点的x-es和ys的值等于第4点一个虽然我用不同的值初始化它们，因此该方法计算出线不相交。为什么前三个y和x的值等于第四个？

Answer 1

这是因为x类的y和Point是static。这意味着，Point类的每个实例都将使用相同的值。删除static以修复它。

public class Point {
double x;
double y;

你应该改进课程设计：两个类（Point，Lines）另一个用main方法。

Line也一样。或者你最终会得到每一行相同的积分。（你想用4个变量来避免这种情况。）

像这样的东西

public class Point {
    double x;
    double y;

    public Point(double x, double y) {
        this.x = x;
        this.y = y;
    }
}

public class Line {
    Point a;
    Point b;

    public Line(Point a, Point b) {
        this.a = a;
        this.b = b;
    }

    public Point itIntersect(Line line) {
        // here change the logic
        Point point = null;

        double denominator = (x1y1.x - x2y2.x) * (x3y3.y - x4y4.y)
                - (x1y1.y - x2y2.y) * (x3y3.x - x4y4.x);

        double px = 0;
        double py = 0;

        if (denominator != 0) {
            px = ((x1y1.x * x2y2.y - x1y1.y * x2y2.x) * (x3y3.x - x4y4.x) - (x1y1.x - x2y2.x)
                    * (x3y3.x * x4y4.y - x3y3.y * x4y4.x))
                    / denominator;
            py = ((x1y1.x * x2y2.y - x1y1.y * x2y2.x) * (x3y3.y - x4y4.y) - (x1y1.y - x2y2.y)
                    * (x3y3.x * x4y4.y - x3y3.y * x4y4.x))
                    / denominator;

//            System.out.println(px + "," + py);

            point = new Point(px, py);

        }

        return point;
    }
}

P.S我没有改变交叉的逻辑，因为它太长了，但你可以做这样的事情来改进设计。

然后你需要做Point point = line.itIntersect（anotherLine）;

如果它为null则不相交，否则它将返回该点。

P.S将x和y设为私有，我没有，因为它是一个例子。

如何找到两条线的交叉点

1 个答案: