IndexOutOfBoundsException when adding to ArrayList at index

Question

I get exception Exception in thread \"main\" java.lang.IndexOutOfBoundsException: Index: 1, Size: 0 for the below code. But couldn\'t understand why.

Answer 1

ArrayList index starts from 0(Zero)

Your array list size is 0, and you are adding String element at 1st index. Without adding element at 0th index you can't add next index positions. Which is wrong.

So, Simply make it as

 s.add("Elephant");

Or you can

s.add(0,"Elephant");

Answer 2

You must add elements to ArrayList serially, starting from 0, 1 and so on.

If you need to add elements to specific position you can do the following -

String[] strings = new String[5];
strings[1] = "Elephant";

List<String> s = Arrays.asList(strings);
System.out.println(s); 

This will produce the sollowing output

[null, Elephant, null, null, null]

Answer 3

You can initialize the size (not the capacity) of an ArrayList in this way:

ArrayList<T> list = new ArrayList<T>(Arrays.asList(new T[size]));

in your case:

ArrayList<String> s = new ArrayList<String>(Arrays.asList(new String[10]));

this creates an ArrayList with 10 null elements, so you can add elements in random order within the size (index 0-9).

Answer 4

Don't add index as 1 directly in list If you want to add value in list add it like this s.add("Elephant"); By default list size is 0 If you will add any elements in list, size will increased automatically you cant add directly in list 1st index. //s.add(0, "Elephant");