ArrayOutOfBounds Exception

Question

public void delete(int index)
{
    if (index < 0 || index > _____)
    {
        throw new ArrayIndexOutOfBoundsException();
    }

    {
        Node<X> current = head;
        for (int i = 0; i < index; i++)
        {
            current = current.getLink();
        }

        Node<X> newNode = new Node<X>();
        newNode.setLink(current.getLink());
        current.setLink(newNode);
    }
}

So I can't seem to figure out the logic in writing an outofbounds exception for index when using node objects. It's not like i can do head.length, so would i use a .getLink loop first to find the Null? It's not like I can do index != null either.

Check if `getLink` returns null inside the loop and throw the exception if it does? — andars, Mar 19 '14 at 04:35
Why are you creating a newNode when you are deleting? It looks more like an insert operation. — anonymous, Mar 19 '14 at 04:36
the newnode becomes the links after the one i deleted, then putting that back to the head link — Jakkie Chan, Mar 19 '14 at 04:37

score 2 · Accepted Answer · answered Mar 19 '14 at 04:35

2

if (index < 0 || index >= sizeOfList)
{
    throw new ArrayIndexOutOfBoundsException();
}

If you have a function that computes the length of the collection.

answered Mar 19 '14 at 04:35

Rey Libutan

5,226
9
42
73

Does that function involve a loop to iterate over each node? – andars Mar 19 '14 at 04:38
yes, thats how i get it to return the size with a counter – Jakkie Chan Mar 19 '14 at 04:40
1

This works, but if you don't want to iterate twice you could throw the exception inside your existing loop. – andars Mar 19 '14 at 04:42

score 1 · Answer 2 · answered Apr 09 '14 at 22:54

1

You need to check for the edge cases, while writing a code. That's the best practice.

Since, you missed the edge cases, of size 0 and list full, in the mentioned example, you were getting ArrayIndexOutOfBoundsException.

answered Apr 09 '14 at 22:54

user3517318

11
2

ArrayOutOfBounds Exception

2 Answers2