15.12.2 Examples: Array Access Evaluation Order

In an array access, the expression to the left of the brackets appears to be fully evaluated before any part of the expression within the brackets is evaluated. For example, in the (admittedly monstrous) expression a[(a=b)[3]], the expression a is fully evaluated before the expression (a=b)[3]; this means that the original value of a is fetched and remembered while the expression (a=b)[3] is evaluated. This array referenced by the original value of a is then subscripted by a value that is element 3 of another array (possibly the same array) that was referenced by b and is now also referenced by a.

Thus, the example:

class Test {
	public static void main(String[] args) {
		int[] a = { 11, 12, 13, 14 };
		int[] b = { 0, 1, 2, 3 };
		System.out.println(a[(a=b)[3]]);
	}
}

prints:

14

because the monstrous expression's value is equivalent to a[b[3]] or a[3] or 14.

If evaluation of the expression to the left of the brackets completes abruptly, no part of the expression within the brackets will appear to have been evaluated. Thus, the example:

class Test {
	public static void main(String[] args) {
		int index = 1;
		try {
			skedaddle()[index=2]++;
		} catch (Exception e) {
			System.out.println(e + ", index=" + index);
		}
	}
	static int[] skedaddle() throws Exception {
		throw new Exception("Ciao");
	}
}

prints:

java.lang.Exception: Ciao, index=1

because the embedded assignment of 2 to index never occurs.

If the array reference expression produces null instead of a reference to an array, then a NullPointerException is thrown at run time, but only after all parts of the array reference expression have been evaluated and only if these evaluations completed normally. Thus, the example:

class Test {

	public static void main(String[] args) {
		int index = 1;
		try {
			nada()[index=2]++;
		} catch (Exception e) {
			System.out.println(e + ", index=" + index);
		}
	}
	static int[] nada() { return null; }
}

prints:

java.lang.NullPointerException, index=2

because the embedded assignment of 2 to index occurs before the check for a null pointer. As a related example, the program:

class Test {

	public static void main(String[] args) {
		int[] a = null;
		try {
			int i = a[vamoose()];
			System.out.println(i);
		} catch (Exception e) {
			System.out.println(e);
		}
	}

	static int vamoose() throws Exception {
		throw new Exception("Twenty-three skidoo!");
	}
}

always prints:

java.lang.Exception: Twenty-three skidoo!

A NullPointerException never occurs, because the index expression must be completely evaluated before any part of the indexing operation occurs, and that includes the check as to whether the value of the left-hand operand is null.