Java ArrayIndexOutOfBounds 异常 [英] Java ArrayIndexOutOfBounds Exception

问题描述

也许我已经看这个太久了，因为我找不到问题，但它应该很简单.我在线上收到一个 ArrayIndexOutOfBounds 异常:

Perhaps I have been looking at this for too long as I cannot find the problem, yet it should be something simple. I am receiving an ArrayIndexOutOfBounds exception on the line:

nextWord = MyArray[i + 1].toLowerCase();

有人能明白为什么吗?

  String currentWord = "";
  String nextWord = "";

  for (int i = 0; i <= MyArray.length; i++) {

   // If not at the end of the array
   if (MyArray.length > 0 && i < MyArray.length) {

    currentWord = MyArray[i].toLowerCase();
    nextWord = MyArray[i + 1].toLowerCase(); /* EXCEPTION */

    System.out.println("CURRENT WORD: " + currentWord);
    System.out.println("NEXT WORD: " + nextWord);
   } 
  }

谢谢！

推荐答案

数组索引从 0 到 array.length - 1.

数组的典型循环结构如下:

The typical loop construct for arrays is thus:

for (int i=0; i<array.length; i++) // do stuff

在你的情况下，你有一个单一的位置向前看，所以为了避免越界，你需要将该循环限制一个位置:

in your case, you've a got a single position look ahead, so to avoid out-of-bounds you need to restrict that loop by one position:

for (int i=0; i<array.length-1; i++) // do stuff

如果您将索引范围设置在循环之外，则在循环之后它将具有正确的值来分配最后一个currentWord:

if you scope the index outside of the loop, after the loop it will have the right value to assign the last currentWord:

int i=0;
for (; i<array.length-1; i++) // do stuff
// here i == array.length - 1, provided you don't mess with i in the "do stuff" part

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