Why is Java giving my ArrayList an IndexOutOfBoundsException?

Question

Why is Java giving me an error message? Im using JavaSE-12 and don't understand why I have an IndexOutOfBoundsException. Do I have to specify the index of the ArrayList or what am I missing?

    import java.util.ArrayList;


public class DSArrayList {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        ArrayList<String> namensListe = new ArrayList<String>();

        namensListe.add("Hans");
        namensListe.add("Peter");

        for(int i = 0; i <= namensListe.size(); i++) {
            System.out.println("Name: " + namensListe.get(i));
        }

    }

}

Here `i <= namensListe.size()` with `<=` You are iterating beyond the list. Change this to `i < namensListe.size()` — Nikhil, Jun 02 '20 at 10:30
`<`, not `<=`. With `<=`, `i` would have the value `2` in the last iteration. — Johannes Kuhn, Jun 02 '20 at 10:30
Also, you don't use `i` inside the loop, so it might be better to use an enhanced for loop: `for (String name : namensListe) { System.out.println("Name: " + name); }` — Johannes Kuhn, Jun 02 '20 at 10:33
whaddya mean @JohannesKuhn? He uses the i: `namensListe.get(i)` — see sharper, Jun 02 '20 at 10:59

score 0 · Answer 1 · answered Jun 02 '20 at 10:30

for(int i = 0; i < namensListe.size(); i++)

(note the < operator)

To explain, arrays and like structures are indexed from zero, not one. Obviously we count, however, from one, not zero. So the length of an array is always one greater than the index of the last item it contains.

Why is Java giving my ArrayList an IndexOutOfBoundsException?

1 Answers1