Java switch with default, break, and multiple passes — what is the full output for z from 0 to 3? public class Switch2 { final static short x = 2; public static int y = 0; public static void main(String [] args) { for (int z = 0; z < 4; z++) { switch (z) { case x: System.out.print("0 "); default: System.out.print("def "); case x-1: System.out.print("1 "); break; case x-2: System.out.print("2 "); } } } } Select the correct sequence.

Introduction / Context:This problem tests how default behaves when reached (with or without a match) and how break truncates fall-through. You must trace four iterations of z and respect the textual order of cases.

Given Data / Assumptions:

• x = 2 → cases are 2, default, 1 (with break), and 0.
• Loop z iterates 0, 1, 2, 3.
• Only the case for 1 has a break; others fall through.

Concept / Approach:On a match, execution starts at that case and continues forward. If no case matches, execution starts at default and continues forward. The break after printing "1 " stops further fall-through for that iteration.

Step-by-Step Solution:

Verification / Alternative check:Removing the break after case (x-1) would allow fall-through into case (x-2), changing the sequence.

Why Other Options Are Wrong:

• They omit required parts of fall-through or misplace default participation.

Common Pitfalls:Believing default executes only when no case matches; it also falls through to later cases unless a break stops it.

Final Answer:2 1 0 def 1 def 1