對于下面的問題，是否是我如此接近零，但將零與容差進(jìn)行比較不起作用？數(shù)字越精確，我對直線上圓弧點(diǎn)的檢查就越失敗，而越不精確，它就越有效。CAD 繪圖確實(shí)有一個(gè)弧線，該弧線在線段上有一個(gè)點(diǎn)，這就是我在此測試中獲取輸入坐標(biāo)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;}檢查了公式和堆棧溢出，該算法在大多數(shù)情況下都有效，但發(fā)現(xiàn)它失敗的情況，我將其追溯到包含的示例，并確定我的檢查針對上述輸入返回 Sign = 1 而不是 Sign = 0，并且精度更高。