Java switch fall-through with constant short case labels — what sequence is printed? public class Switch2 { final static short x = 2; public static int y = 0; public static void main(String [] args) { for (int z = 0; z < 3; z++) { switch (z) { case x: System.out.print("0 "); case x-1: System.out.print("1 "); case x-2: System.out.print("2 "); } } } } Choose the exact output.

Introduction / Context:This problem exercises switch fall-through behavior across a loop. The case labels are compile-time constants derived from a final short x, allowing x, x-1, and x-2 to be legal case expressions.

Given Data / Assumptions:

• x = 2, so cases are 2, 1, and 0 in that textual order.
• No break statements appear; fall-through applies.
• Loop variable z runs 0, 1, 2.

Concept / Approach:For each z, execution begins at the matching case label (if any) and then falls through to the end of the switch, executing all subsequent print statements in order.

Step-by-Step Solution:

Verification / Alternative check:Add break after each case to see "2 1 0" instead; current omission is deliberate to test fall-through.

Why Other Options Are Wrong:

• They place numbers in the wrong order or include/omit fall-through wrongly.

Common Pitfalls:Assuming the order of case labels in source must be numeric order—execution starts at the match and proceeds downward regardless of numeric value.

Final Answer:2 1 2 0 1 2