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Java AbstractSequentialList | ListIterator() - GeeksforGeeks

Java AbstractSequentialList | ListIterator()

Last Updated : 11 Jul, 2025

AbstractSequentialList listIterator():

method in Java is used to get a listIterator over this list. It returns a list iterator over the elements in this list (in proper sequence).

Syntax:

public abstract ListIterator listIterator(int index)

Parameters:

This method takes a parameter

index

which is the index of first element to be returned from the list iterator (by a call to the next method)

Returns:

This method returns a list iterator over the elements in this list (in proper sequence).

Exceptions:

This method throws

IndexOutOfBoundsException

, if the index is out of range (index size()) Below is the code to illustrate ListIterator():

Program 1: Java


 // Java program to demonstrate
// add() method

import java.util.*;

public class GfG {

    public static void main(String[] args)
    {
        // Creating an instance of the AbstractSequentialList
        AbstractSequentialList<Integer> absl = new LinkedList<>();

        // adding elements to absl
        absl.add(5);
        absl.add(6);
        absl.add(7);
        absl.add(2, 8);
        absl.add(2, 7);
        absl.add(1, 9);
        absl.add(4, 10);

        // Getting ListIterator
        ListIterator<Integer> Itr = absl.listIterator(2);

        // Traversing elements
        while (Itr.hasNext()) {
            System.out.print(Itr.next() + " ");
        }
    }
}

Program 2:

To demonstrate IndexOutOfBoundException

Java


 // Java code to show IndexOutofBoundException

import java.util.*;

public class GfG {

    public static void main(String[] args)
    {

        // Creating an instance of the AbstractSequentialList
        AbstractSequentialList<Integer> absl = new LinkedList<>();

        // adding elements to absl
        absl.add(5);
        absl.add(6);
        absl.add(7);
        absl.add(2, 8);
        absl.add(2, 7);
        absl.add(1, 9);
        absl.add(4, 10);

        // Printing the list
        System.out.println(absl);

        try {
            // showing IndexOutOfBoundsException
            // Getting ListIterator
            ListIterator<Integer> Itr = absl.listIterator(15);

            // Traversing elements
            while (Itr.hasNext()) {
                System.out.print(Itr.next() + " ");
            }
        }
        catch (Exception e) {
            System.out.println("Exception: " + e);
        }
    }
}

Output:

[5, 9, 6, 7, 10, 8, 7]
Exception: java.lang.IndexOutOfBoundsException: Index: 15, Size: 7

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