Question

This is a sample implementation of indexOf() in LinkedList on Oracle’s website. I’m a little confused on how the if loop works here:

public int indexOf(E e) {
    for (ListIterator<E> it = listIterator(); it.hasNext(); )
        if (e == null ? it.next() == null : e.equals(it.next()))
            return it.previousIndex();
    // Element not found
    return -1;
}

So the ListIterator object is created at the head of the list. for loop goes on until the iterator reaches the list’s end, and the if loop checks if the target object is found. The part I did not understand though, is that why does the if loop check it.next() == null when e == null? Could someone help walk me through how it’s done when input e is null?

Answer 1

The loop checks if it.next() == null only if e == null this is done to avoid NullPointerException when evaluating e.equals(it.next()).

if e != null, then the regular e.equals() method is invoked.

null is a valid “element” that can be inserted to a LinkedList, so this must be taken into consideration.

The position of the last element is not inserted. Note that unlike the text book data structure where the last element in a Linked List is null, in here – when you get to the last element – it.hasNext() will be evaluated to false, without letting you see this “junk” element.