检查一个点是否在线段上，除了某些情况下有效（舍入错误？）

Question

对于下面的问题，是否是我如此接近零，但将零与容差进行比较不起作用？数字越精确，我对直线上圆弧点的检查就越失败，而越不精确，它就越有效。CAD 绘图确实有一个弧线，该弧线在线段上有一个点，这就是我在此测试中获取输入坐标的地方。class Line{   public Point Point1 {get;set;}   public Point Point2 {get;set;}   public Line(double x1, double y1, double x2, double y2)   {      Point1 = new Point(x1,y1); Point2 = new Point(x2,y2);   }}class Point{    public double X {get;set;}   public double Y {get;set;}   public Point (double x, double y)   {       X = x; Y = y;  }}//4 decimal place numbers, worksPoint arcEnd = new Point(3.8421, 16.9538); // these numbers don't //3.84212141717697, //16.9538136440052Point arcStart = new Point(4.0921, 17.2038);//test an arc point on/off the lineLine line = new Line(3.9336, 16.9538, 3.7171, 16.9538); //these numbers don't 3.93362776812308, 16.9538136440053, //3.71712141717697, 16.9538136440054bool on_line = Sign(line.Point1, line.Point2, arcEnd) //true//more precise numbers, from CAD / dxf drawing for line and arc, arc end //point touches somewhere on the line (included in comments above, fail)//so on_line = true for the above inputs and the Sign function gives zero, //but when using the commented precise numbers sign gives back 1 and the //value computed in sign is 3.0639866299190109E-14.public static bool Sign(Point Point1, Point Point2, Point point){    double value = (Point2.X - Point1.X) * (p.Y - Point1.Y) - (Point2.Y - Point1.Y) * (p.X - Point1.X);    return Equals(Math.Sign(value), 0);}public static bool Equals(double d1, double d2, double tolerance=0.000001){    double a = Math.Abs(d1 - d2);    double b = Math.Abs(d1 * tolerance);    if (a <= b)    {       return true;    }    return false;}检查了公式和堆栈溢出，该算法在大多数情况下都有效，但发现它失败的情况，我将其追溯到包含的示例，并确定我的检查针对上述输入返回 Sign = 1 而不是 Sign = 0，并且精度更高。

Answer 1

你犯了两个错误。首先，您在“Return Equals(Math.Sign(value), 0);”中使用了 Sign 函数，它将为任何正数提供 1 的值，为任何负数提供 -1 的值。这会破坏你使用宽容的尝试。其次，您尝试将差异与第一个数字“b = Math.Abs(d1 * 容差)”的比率进行比较，这将始终返回 False。我建议你将它与宽容本身进行比较，就像这样。

public static bool Sign(Point Point1, Point Point2, Point point)

{

    double value = (Point2.X - Point1.X) * (point.Y - Point1.Y) - (Point2.Y - Point1.Y) * (point.X - Point1.X);

    return Equals(value, 0);

}

public static bool Equals(double d1, double d2, double tolerance = 0.000001)

{

    double a = Math.Abs(d1 - d2);

    if (a <= tolerance)

        return true;

    return false;

}