Java：数组索引越界异常
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Java: Array Index Out of Bounds Exception
提问by Arly
System.out.print("Enter an integer: ");
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
int lArray = x - 2;
int[] newArray = new int[lArray];
System.out.println("Let's display all possible integers...");
for (int i = 0; i <= newArray.length; i++) {
newArray[i] = i + 2;
System.out.print(newArray[i] + " ");
}
I've just started Java recently, but I sure that if I coded similarly in another language, I would face the same problem. This is an excerpt from an application where it lists all the prime numbers up until the user's input.
我最近才开始使用 Java，但我确信如果我用另一种语言进行类似的编码，我会面临同样的问题。这是一个应用程序的摘录，其中列出了用户输入之前的所有素数。
The reason why x-2 is used as the definition of lArray is because the length of the array will be all the integers from 2 until the number {2, 3, 4, 5... x}.
之所以使用 x-2 作为 lArray 的定义是因为数组的长度将是从 2 到数字 {2, 3, 4, 5 ... x} 的所有整数。
I noticed that for the line
我注意到这一行
for (int i = 0; i <= newArray.length; i++) {
if I change i <= newArrayto i < newArray, the code works without error. However, the user's input, x, is left out which is a problem if x is prime.
如果我更改i <= newArray为i < newArray，则代码可以正常工作。然而，用户的输入 x 被遗漏了，如果 x 是素数，这是一个问题。
回答by codaddict
You should use <and not <=in:
您应该使用<而不是<=在：
for (int i = 0; i <= newArray.length; i++)
^^
If fooany array, valid index of fooare [0,foo.length-1]
如果有foo任何数组，有效索引foo是[0,foo.length-1]
Using foo.lengthas an index will cause ArrayIndexOutofBoundsException.
使用foo.length作为索引会导致ArrayIndexOutofBoundsException。
And also lArraywhich contains number of natural numbers <=xbut excluding only onenumber 1, its value should be x-1and not x-2.
并且lArray其中包含多个自然数<=x但只排除一个数字1，它的值应该是x-1而不是x-2。
回答by Ani
Change the array length to (x - 1)instead, and go with the <condition, which you've already found is necessary to avoid the out-of-bounds exception.
将数组长度改为(x - 1)改为，并使用<您已经发现的条件，以避免越界异常。
The reason you need an array that is 1 element larger than what you're currently using is because there are (n - 1)candidates that must be considered between 2 and n , not (n - 2).
您需要一个比您当前使用的元素大 1 个元素的数组的原因是(n - 1)，必须考虑 2 和 n 之间的候选对象，而不是(n - 2)。
For example, there are two candidates less than or equal to three (2 and 3), both of which, coincidentally, happen to be prime.
例如，有两个小于或等于 3 的候选者（2 和 3），巧合的是，它们都是素数。
回答by Pablo Santa Cruz
You need to use:
您需要使用：
int lArray = x - 1;
And change your condition to use <instead of <=.
并更改您的条件以使用<而不是<=.
In Java as in C/C++, arrays are ZERO based. So your array of Nvalues will go from index 0to N-1.
在 Java 和 C/C++ 中，数组都是基于零的。所以你的N值数组将从索引0到N-1。
Taking your example: {2, 3, 4, 5... x}.
以你为例：{2, 3, 4, 5... x}。
You will need N-1values to store all positive numbers but 1in an integer array. So, if Nequals to 4, your array will be:
您将需要N-1值将除1 之外的所有正数存储在整数数组中。所以，如果N等于4，你的数组将是：
newArray[0] = 2;
newArray[1] = 3;
newArray[2] = 4;
Hence, array lenght must be 3(N-1).
因此，数组长度必须为3( N-1)。
回答by a1ex07
for (int i = 0; i <= newArray.length; i++) //should be <, not <=
for (int i = 0; i < newArray.length; i++)
