Array Index Out Of Bounds Exception in Java (original) (raw)

Last Updated : 03 Dec, 2024

In Java, ArrayIndexOutOfBoundsException is a Runtime Exception thrown only at runtime. The Java Compiler does not check for this error during the compilation of a program. It occurs when we try to access the element out of the index we are allowed to, i.e. index >= size of the array.

Java supports the creation and manipulation of arrays as a data structure. The index of an array is an integer value that has a value in the interval [0, n-1], where n is the size of the array. If a request for a negative or an index greater than or equal to the size of the array is made, then Java throws an **ArrayIndexOutOfBounds Exception. This is unlike C/C++, where no index of the bound check is done.

**Example 1: Here, we are trying to access the index which is greater than or equal to the array length.

Java `

// Java program to show ArrayIndexOutOfBoundsException // when the index is greater than or equal to array length public class GFG {

public static void main(String[] args)
{

    // taking array of integers
    int a[] = { 1, 2, 3, 4, 5 };

    for (int i = 0; i <= a.length; i++)
        System.out.println(a[i]);
}

}

**Runtime Error Throws an Exception:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index 5 out of bounds for length 5
at GFG.main(GFG.java:11)

Here if you carefully see, the array is of size 5. Therefore while accessing its element using for loop, the maximum index value can be 4, but in our program, it is going till 5 and thus the exception.

**Example 2: Here, we are **trying to access the index of array which is negative.

Java `

// Java program to show ArrayIndexOutOfBoundsException // when we access the negative index of array public class GFG {

public static void main(String[] args)
{

    // taking array of integers
    int a[] = { 1, 2, 3 };

    // accessing the negative index of array
    System.out.println(a[-2]);
}

}

**Run-Time Exception:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index -2 out of bounds for length 3
at GFG.main(GFG.java:12)

Handling ArrayIndexOutOfBoundsException in Java

To handle ArrayIndexOutOfBoundsException, make sure that index of array is within the valid range. You can also use the **enhanced for-loop to automatically handle this exception.

**Example 1: Here, we are checking whether the index is valid or not by taking array length with in the index range i.e. [0, n-1]

Java `

// Java program to handle ArrayIndexOutOfBoundsException // by taking array index within valid range public class GFG {

public static void main(String[] args)
{

    // taking array of integers
    int a[] = { 1, 2, 3, 4, 5 };

    // here, we have remove equal to sign
    for (int i = 0; i < a.length; i++)
        System.out.println(a[i]);
}

}

**Example 2: Here, we are using enhanced for loop that automatically handles the accessing of array's index

Java `

// Java program to handle ArrayIndexOutOfBoundsException // by using enhanced for loop public class GFG {

public static void main(String[] args)
{

    // taking array of integers
    int a[] = { 1, 2, 3, 4, 5 };

    // using enhanced for loop
    for (int e : a) {
        System.out.println(e);
    }
}

}

**Example 3: Consider enclosing your code inside a try-catch statement and manipulate the exception accordingly. As mentioned, Java won’t let you access an invalid index and will definitely throw an ArrayIndexOutOfBoundsException. However, we should be careful inside the block of the catch statement because if we don’t handle the exception appropriately, we may conceal it and thus, create a bug in your application.

Java `

// Java program to handle ArrayIndexOutOfBoundsException // by using using try-catch block public class GFG {

public static void main(String[] args)
{

    // taking array of integers
    int a[] = { 1, 2, 3, 4, 5 };

    // using try catch block
    try {
        for (int i = 0; i <= a.length; i++)
            System.out.print(a[i] + " ");
    }
    catch (Exception e) {
        System.out.println("\nException Caught");
    }
}

}

Output

1 2 3 4 5 Exception Caught

Here in the above example, you can see that till index 4 (value 5), the loop printed all the values, but as soon as we tried to access the a[5], the program threw an exception which is caught by the catch block, and it printed the "Exception Caught" statement.