ArrayIndexOutOfBoundsException when iterating through all the elements of an array

Question

how to handle this exception "ArrayIndexOutOfBoundsException" my code : I create an array of 64 length then I intialized every index then I print the indexes to make sure I am fulling all indexes but it prints up to 63 then gives the exception !! any idea

    public static void main(String [] arg) {
    int [] a=new int [64];
    for(int i=1;i<=a.length;i++){
        a[i]=i;
        System.out.println(i);
    }

}

Answer 1

The array indexes in Java start from 0 and go to array.length - 1 . So change the loop to for(int i=0;i<a.length;i++)

Answer 2

索引从0开始，因此最后一个索引是63.更改for循环，如下所示：
for(int i=0;i<a.length;i++){

Answer 3

See the JLS-Arrays :

If an array has n components, we say n is the length of the array; the components of the array are referenced using integer indices from 0 to n - 1, inclusive.

So you have to iterate through [0,length()-1]

for(int i=0;i<a.length;i++) {
    a[i]=i+1;  //add +1, because you want the content to be 1..64
    System.out.println(a[i]);

}

Answer 4

Need Complete Explanation? Read this

The index of an Array always starts from 0 . Therefore as you are having 64 elements in your array then their indexes will be from 0 to 63 . If you want to access the 64th element then you will have to do it by a[63] .

Now if we look at your code, then you have written your condition to be for(int i=1;i<=a.length;i++) here a.length will return you the actual length of the array which is 64.

Two things are happening here:

1. As you start the index from 1 ie i=1 therefore you are skipping the very first element of your array which will be at the 0th index.
2. In the last it is trying to access the a[64] element which will come out to be the 65th element of the array. But your array contains only 64 elements. Thus you get ArrayIndexOutOfBoundsException .

The correct way to iterate an array with for loop would be:

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

The index starting from 0 and going to < array.length .

Answer 5

You've done your math wrong. Arrays begin counting at 0. Eg int[] d = new int[2] is an array with counts 0 and 1.

You must set your integer 'i' to a value of 0 rather than 1 for this to work correctly. Because you start at 1, your for loop counts past the limits of the array, and gives you an ArrayIndexOutOfBoundsException.

Answer 6

In Java arrays always start at index 0. So if you want the last index of an array to be 64, the array has to be of size 64+1 = 65.

//                       start   length
int[] myArray = new int [1    +  64    ];

Answer 7

You can correct your program this way :

int i = 0;     // Notice it starts from 0
while (i < a.length) {
    a[i]=i;
    System.out.println(i++);
}