Array access

Learn access with arr[i] and length.

Array access

Array elements are accessed by index. Indexes start at 0: arr[0] is the first, arr[1] the second element.

int[] arr = {10, 20, 30};
System.out.println(arr[0]);   // 10
System.out.println(arr.length); // 3

The index is between 0 and length - 1. Going beyond throws ArrayIndexOutOfBoundsException. arr.length is a field (no parens); ArrayList.size() is a method.

Index bounds

Array indexes start at 0 and go up to length - 1. Going beyond throws:

int[] arr = new int[3];  // indexes 0, 1, 2
arr[0] = 10;  // valid
arr[2] = 30;  // valid (last)
arr[3] = 40;  // ERROR!

So loops use i < arr.length, not i <= arr.length. The array size is fixed; arr.length never changes.

Exercise Print the first element and size

Write a Main class with a main method. Assign the literal {10, 20, 30} to int[] arr. Print the first element with System.out.println(arr[0]) and the size with System.out.println(arr.length).

Show example solution

public class Main {
    public static void main(String[] args) {
        int[] arr = {10, 20, 30};
        System.out.println(arr[0]);
        System.out.println(arr.length);
    }
}

Test yourself

Access the first element

int first = arr[____];

Show answer

int first = arr[0];

int[] arr = new int[3]; arr[3] = 5; — what happens?
- Changes the last element
- Changes the first element
- Throws an error
- Valid assignment
Show answer

Throws an error — arr[3] is out of bounds; it throws.
Match the access concepts

arr[0]

First element
arr[i]

i-th element
arr.length

Size
length-1

Last index

Next lesson: Filling arrays

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