blocks|key|174257|text|试着跑|type|unstyled|depth|inlineStyleRanges|entityRanges|data|174258|for+(int+i+=+0;+i+<+count;+i%2B%2B)+{
++++System.out.println(i+%2B+numbers.length+-+1);
}|code-block|syntax|javascript|174259|它打印：|174260|5+6+7|174261|数组有6个项，索引从0到5。当您尝试获取具有索引>=6的项时，它将引发ArrayIndexOutOfBoundsException。|174262|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Tried to run

<pre><code>for (int i = 0; i &lt; count; i++) {
 System.out.println(i + numbers.length - 1);
}
</code></pre>

It prints:

5
6
7

Your array has 6 items and is indexed from 0 to 5. When you try to get item that has index >=6 it throws ArrayIndexOutOfBoundsException

blocks|key|3330758|text|如果您想按降序打印它们，请使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3330759|System.out.println(numbers[numbers.length+-+1+-+i])|code-block|syntax|javascript|3330760|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you want to print them in descending order use 

<pre><code>System.out.println(numbers[numbers.length - 1 - i])
</code></pre>

blocks|key|174281|text|因此，在这个上下文中，让我们从数组访问的最小值和最大值的角度来看它|type|unstyled|depth|inlineStyleRanges|entityRanges|data|174282|因此，对于i+(0)的最小值，我们得到i+%2B+numbers.length+-+1+=+0+%2B+6+-+1+=+5|offset|length|style|CODE|174283|对于i的最大值，我们要么得到3，要么得到numbers.length：|174284|如为3：i+%2B+numbers.length+-+1+=+3+%2B+6+-+1+=+8|174285|在numbers.length的情况下，我们得到i+%2B+numbers.length+-+1+=+6+%2B+6+-+1+=+12|174286|您需要确保数组之间迭代的可能值总是会在数组的范围内给出一个值(显然不是这样)。|174287|entityMap^0|0|J|12|0|0|4|12|0|O|13|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|V|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|W|8|@$D|X|E|Y|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|Z|8|@$D|10|E|11|F|G]]|9|@]|A|$]]|$1|N|3|O|5|6|7|12|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|Q|$]]

So in this context let's look at it in terms of the minimum and maximum values that the array access can have 

So for the minimum value of i (0), we get <code>i + numbers.length - 1 = 0 + 6 - 1 = 5</code> 

and for the maximum of i we either get 3 or numbers.length: 

In the case of 3: <code>i + numbers.length - 1 = 3 + 6 - 1 = 8</code>

In the case of numbers.length we get <code>i + numbers.length - 1 = 6 + 6 - 1 = 12</code>

You need to make sure that the possible values that the array will iterate between is always going to give a value within the bounds of the array (as clearly this does not)

blocks|key|3330837|text|数组类似于对象链的图像。如果知道这些对象的索引，则始终可以访问其中的任何对象。因此，在这种情况下，您有一个长度为9的数组，其中包含前9个正整数，在0到8之间将有有效的索引。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3330838|+0++1++2++3++4++5++6++7++8
[1][2][3][4][5][6][7][8][9]|code-block|syntax|javascript|3330839|要打印数组中的值，可以使用：|3330840|for+(int+i+=+0;+i+<+values.length;+i%2B%2B)+{
++++System.out.println(values[i]);
}|3330841|如果您想要向后打印它，那么您应该从它的最后一个索引开始，并减少i，直到它大于或等于数组的起始索引为止：|3330842|for+(int+i+=+values.length+-+1;+i+>=+0;+i--)+{
++++System.out.println(values[i]);
}|3330843|每当您尝试访问超出有效索引范围的索引时，它将导致IndexOutofboundsException。因此，如果您喜欢在计算索引时调用任何算术，请注意这样一个事实，即您对它只有很少的有效值。|3330844|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Image that the array is like a chain of objects. You can always access any of those objects if you know its index. So in the case you have an array of length 9 that contains the first 9 positive integer you will have valid indexes through 0 to 8.

<pre><code> 0 1 2 3 4 5 6 7 8
[1][2][3][4][5][6][7][8][9]
</code></pre>

to print out the the values inside the array you could use: 

<pre><code>for (int i = 0; i &lt; values.length; i++) {
 System.out.println(values[i]);
}
</code></pre>

and if you would like to backward print it then you should start from it's last index and decrement the i until it greater or equal to the starting index of the array:

<pre><code>for (int i = values.length - 1; i &gt;= 0; i--) {
 System.out.println(values[i]);
}
</code></pre>

Whenever you try to access an index that is out of the bound of valid indexes it will result to IndexOutofboundsException. So if you like to invoke any arithmetics with calculating the indexes be aware of the fact that you have only few valid values for it.

I've been learning to code recently, but unfortunately iterating through arrays and understanding their Length property and not going out of bounds eludes me. 

I have the following exercise that I need to solve in Java - input a few numbers, parse them and put them into an array then get the three largest numbers in it and print them in descending order. I managed to do it, but I did need to look up the last part, as I was getting ArrayIndexOutOfBoundsException. I have hardcoded the values in the Array, in order to try and understand it, but I'd appreciate some help on the matter.

Here is my solution:

<pre><code> int[] numbers = {1, 2, 3, 4, 5, 6};

 Arrays.sort(numbers);

 int count = Math.min(3, numbers.length);
 for (int i = 0; i &lt; count; i++) {
 System.out.println(numbers[i + numbers.length - 1]);
</code></pre>

What I am trying to do here is print them in ascending order, as printing them in descending will be numbers[numbers.length - 1 - i]. On my example I am getting OutOfBounds. If I do <code>i + numbers.length - 3</code> it works.

Will someone be so kind as to explain why? I know it's a stupid question but I just can't seem to understand it.

Cannot understand Array bounds

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我最近一直在学习代码，但不幸的是，迭代数组和理解数组的长度属性，而不是越界。我需要在Java中解决以下问题--输入几个数字，解析它们并将它们放入一个数组中，然后得到其中的三个最大数字，然后按降序打印它们。我设法做到了，但我确实需要查找最后一部分，因为我得到了ArrayIndexOutOfBoundsException。...

问无法理解数组边界
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