我得到ArrayIndexOutofBoundsException - 反转数组

Question

I get the output for the program mentioned below. In addition Ii also encounter an exception as:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 7
    at ReverseOrder.main(ReverseOrder.java:15)

Why does this happen?

public class ReverseOrder {
    public static void main(String[] args)
    {
        int anArray[]={1,2,3,4,5,6};
        int anotherArray[]=new int[6];
        for(int i=5,j=0;i>=0;i--,j++)
        {
            anotherArray[j]=anArray[i];
        }
        //System.out.println(anotherArray.length);
        //System.out.println(anotherArray[2]);
        for(int j=0;j<=anotherArray.length;j++)
        {
            System.out.println(anotherArray[j]);
        }
    }
}

Answer 1

The problem is here:

 for(int j=0;j<=anotherArray.length;j++)
    {
        System.out.println(anotherArray[j]);
    }

you are accessing a position out of the array. This happen because method length gives you the number of elements in the array, and since the first position of an array is 0 and not 1 you should end the loop on anotherArray.length - 1 and not anotherArray.length .

There are two possible solutions to this where you modify your loop to:

a) for(int j=0;j<=anotherArray.length - 1;j++) or

b) for(int j=0;j<anotherArray.length;j++)

The latter (b) is preferable, since it has less arithmetic operations on it.

Answer 2

Change

for(int j=0;j<=anotherArray.length;j++)

to

for(int j=0;j<anotherArray.length;j++)

Since if it's <= you'll be going out of bounds.

In Java (and most languages), arrays are zero-based. If you have an array of size N then its indexes will be from 0 to N - 1 (total size of N ).

Answer 3

change

<=anotherArray.length

to

< anotherArray.length

For example, if array is

int arr[] = new int[2];
arr.length; // it will be 2, which is [0] and [1] so you can't go upto <=length,
// that way you will access [2] which is invalid and so the exception

Answer 4

for(int j=0;j<anotherArray.length;j++) 

instead of

for(int j=0;j<=anotherArray.length;j++) 

Because arrays are zero-based in Java.

Answer 5

You will get ArrayIndexOutOfBoundsException when you try access the element that's out of Array limit.

for(int j=0;j<anotherArray.length;j++) {
    System.out.println(anotherArray[j]);
}

Answer 6

Why do you use this way to reverse your array in first place. Any way

for(int j=0;j<=anotherArray.length;j++) should change to 

for(int j=0;j<anotherArray.length;j++)

Consider this too, It is easy.

    int anArray[]={1,2,3,4,5,6};
    int temp=0;
    for (int i=0;i<anArray.length/2;i++){
       temp=anArray[i];
       anArray[i]=anArray[anArray.length-(1+i)];
        anArray[anArray.length-(1+i)]=temp;
    }

Now your array is reversed.