Array access
- Java course
- Beginner
- Lesson 54/116
- 4 min
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); // 3The 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.
Practice 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).
Sample 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);
}
}Expected output
10 3
Questions and answers
Review the lesson questions together with their answers. To test yourself, solve the lesson interactively.
1 Access the first element
int first = arr[____];int first = arr[0];Explanation: arr[0] gives the first element.
2 int[] arr = new int[3]; arr[3] = 5; — what happens?
- Changes the last element
- Changes the first element
- Throws an error Correct answer
- Valid assignment
Explanation: arr[3] is out of bounds; it throws.
3 Match the access concepts
| Term | Meaning |
|---|---|
arr[0] |
First element |
arr[i] |
i-th element |
arr.length |
Size |
length-1 |
Last index |