問(wèn)題描述

我是使用迭代器的新手，想知道如何遍歷線段上的每個(gè)點(diǎn)(準(zhǔn)確地說(shuō)是 Line2D.Double)——我需要檢查線上的每個(gè)點(diǎn)是否滿足某些要求.

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，給定一個(gè)路徑對(duì)象(如 GeneralPath)，你會(huì)如何做同樣的事情(遍歷形狀輪廓上的每個(gè)點(diǎn))?

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

理想情況下，我想要這樣的東西(帶有一條線或一條路徑):

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();
}

推薦答案

Java API 中似乎沒(méi)有任何東西可以讓 Bresenham 的算法對(duì)用戶可見(jiàn).所以我寫(xiě)了一個(gè)遍歷一行的類.

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.

你可以這樣使用它:

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]]"));

這篇關(guān)于遍歷java中一行/路徑上的每個(gè)點(diǎn)的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！