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