In Java, analyze switch fall-through within a for loop and predict the exact output sequence.\n\nfor (int i = 0; i < 3; i++)\n{\n switch (i)\n {\n case 0: break;\n case 1: System.out.print("one ");\n case 2: System.out.print("two ");\n case 3: System.out.print("three ");\n }\n}\nSystem.out.println("done");

Introduction / Context:
This problem checks understanding of Java switch fall-through and its interaction with a for loop. Without break statements, control falls through to subsequent case labels, executing their statements even if the matched constant differs.

Given Data / Assumptions:

• i takes values 0, 1, 2.
• Only case 0 contains a break; others do not.
• Print statements occur in case 1, case 2, and case 3.

Concept / Approach:
For each loop iteration, identify the matched case and then follow fall-through execution to the end of the switch (unless a break is encountered). Then concatenate the outputs across iterations.

Step-by-Step Solution:
i = 0 ⇒ case 0 hits break immediately ⇒ prints nothing.i = 1 ⇒ enters case 1 ⇒ prints "one ", then falls through to case 2 and prints "two ", then falls through to case 3 and prints "three ".i = 2 ⇒ enters case 2 ⇒ prints "two ", falls through to case 3 and prints "three ".Finally, println("done") appends "done" with a newline.

Verification / Alternative check:
Add explicit break after each case to see the difference, or reorder cases to observe fall-through behavior.

Why Other Options Are Wrong:
They omit some fall-through prints or add impossible ones given the loop bounds.

Common Pitfalls:
Assuming a switch stops after one matching case without a break; forgetting that fall-through continues to all subsequent cases.

Final Answer:
one two three two three done