通过 ArrayList 的反向迭代给出 IndexOutOfBoundsException

2022-01-24 00:00:00 iteration reverse arraylist java

当我对 ArrayList 进行反向迭代时，我得到了 IndexOutOfBoundsException.我尝试进行前向迭代，没有问题.我期望并且知道列表中有五个元素.代码如下:

When I reverse iterate over an ArrayList I am getting a IndexOutOfBoundsException. I tried doing forward iteration and there is no problem. I expect and know that there are five elements in the list. The code is below:

Collection rtns = absRtnMap.values();
List list = new ArrayList(rtns);
Collections.sort(list);

for(int j=list.size();j>0;j=j-1){
  System.out.println(list.get(j));
}

<小时>

前向迭代 - 工作正常，但对我没用:

Forward iteration - which is working fine, but not useful for me:

for(int j=0;j<list.size();j++){
    System.out.println(list.isEmpty());
    System.out.println(list.get(j));
} // this worked fine

<小时>

错误:

Exception in thread "Timer-0" java.lang.IndexOutOfBoundsException: Index: 3, Size: 3
    at java.util.ArrayList.RangeCheck(Unknown Source)
    at java.util.ArrayList.get(Unknown Source)
    at model.Return.getReturnMap(Return.java:61)
    at controller.Poller$1.run(Poller.java:29)
    at java.util.TimerThread.mainLoop(Unknown Source)
    at java.util.TimerThread.run(Unknown Source)

另外，如果有人知道反向迭代的更好用语，我很乐意尝试一下.

Also if anyone knows of a better idiom for reverse iteration I would be happy to try that out.

推荐答案

从 list.size() - 1 开始迭代，因为数组(或 ArrayList)元素是从 0 到小于列表大小的 1 编号.这是一个相当标准的成语:

Start the iteration at list.size() - 1 because array (or ArrayList) elements are numbered from 0 up through 1 less than the size of the list. This is a fairly standard idiom:

for (int j = list.size() - 1; j >= 0; j--) {
    // whatever
}

请注意，您的前向迭代是有效的，因为它会在 到达 list.size() 之前停止.

Note that your forward iteration works because it stops before reaching list.size().

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