如何在 JavaScript 中处理线段、检查点

发布于 2024-10-15 11:36:39 字数 477 浏览 0 评论 0原文

我有以下代码：

Object.defineProperty(LineSegment.prototype,'hasPoint',{ value:
    function (point) {
        if (typeof point != 'object' || !(point instanceof Point)) {
            throw new TypeError('LineSegment.prototype.hasPoint requires a point value.'); }
        var m = (this.y1 - this.y2) / (this.x1 - this.x2);
        return (this.y1 - point.y) / (this.x1 - point.x) == m; }
    });

它适用于线条，但不适用于线段。我如何检查该点是否超出范围，并将其应用于 JavaScript？

原文

I have the following code:

Object.defineProperty(LineSegment.prototype,'hasPoint',{ value:
    function (point) {
        if (typeof point != 'object' || !(point instanceof Point)) {
            throw new TypeError('LineSegment.prototype.hasPoint requires a point value.'); }
        var m = (this.y1 - this.y2) / (this.x1 - this.x2);
        return (this.y1 - point.y) / (this.x1 - point.x) == m; }
    });

It works fine for lines, but not line segments. How would I check if the point is out of bounds, and apply it to JavaScript?

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我不是你的备胎 2024-10-22 11:36:39

仅当 x1 < 时才能使用最简单且最快的答案。 x2。在这种情况下，返回语句如下：

return (this.y1 - point.y) / (this.x1 - point.x) == m && point.x >= this.x1 && point.x <= this.x2;

第一部分检查点是否在线上，接下来的两个条件检查 x 坐标是否在 x1 和 x2（含）之间。

注：若x2< x1，您可以简单地在构造函数中添加一个条件来交换两个点。您还可以以 p+q*v 的形式更改点（其中 p 是直线的第一个点，q 是第二个点，va 标量值，只有当点在线上时才可能）。如果 v 介于 0 和 1（含）之间，则该点在线段上。

编辑：只是警告，由于计算机处理双打的方式，您的方法不会非常可靠。请参阅维基百科页面以获取一些示例：http://en.wikipedia.org/wiki/Double_ precision

The easiest and fastest answer can only be used if x1 < x2. In that case, the following return statement:

return (this.y1 - point.y) / (this.x1 - point.x) == m && point.x >= this.x1 && point.x <= this.x2;

The first part checks if the point is on the line, the two next conditions check if the x coordinate is between x1 and x2 (inclusive).

Note: if x2 < x1, you can simply add a condition in the constructor to swap the two points. You could also change the point in the form p+q*v (where p is the first point of the line, q the second and v a scalar value, this is only possible if the point is on the line). If v is between 0 and 1 (inclusive), the point is on the line segment.

EDIT: Just a warning, your method won't be very reliable, because of the way computers handle doubles. See the wikipedia page for some examples: http://en.wikipedia.org/wiki/Double_precision

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