遍历java中一条线/路径上的每个点

问题描述

我是使用迭代器的新手，想知道如何遍历线段上的每个点(Line2D.Double，准确地说)——我需要检查线上的每个点是否满足某些要求.

I'm new to using iterators and was wondering how one would iterate through each point on a line segment (Line2D.Double, to be precise) -- I need to check to see if each point on the line fulfills certain requirements.

另外，给定一个路径对象(如 GeneralPath)，你会如何做同样的事情(遍历形状轮廓上的每个点)?

Also, given a path object (like GeneralPath), how would you do the same thing (iterate through each point on the outline of the shape)?

理想情况下，我想要这样的东西(带有一条线或一条路径):

Ideally I'd like something like this (with either a line or a path):

Line2D line = new Line2D.Double(p1,p2);
for (Point2D point : line)
{
   point.callSomeMethod();
}

推荐答案

Java API 中似乎没有任何内容使 Bresenham 的算法对用户可见.所以我写了一个迭代一行的类.

There seems to be nothing in the Java API that makes Bresenham's algorithm user-visible. So I wrote a class that iterates over a line.

你可以这样使用它:

List<Point2D> points = new ArrayList<Point2D>();
Line2D line = new Line2D.Double(0, 0, 8, 4);
Point2D current;

for (Iterator<Point2D> it = new LineIterator(line); it.hasNext();) {
    current = it.next();
    points.add(current);
}

assertThat(points.toString(),
    is("[Point2D.Double[0.0, 0.0], Point2D.Double[1.0, 0.0], " +
        "Point2D.Double[2.0, 1.0], Point2D.Double[3.0, 1.0], " +
        "Point2D.Double[4.0, 2.0], Point2D.Double[5.0, 2.0], " +
        "Point2D.Double[6.0, 3.0], Point2D.Double[7.0, 3.0], " +
        "Point2D.Double[8.0, 4.0]]"));

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