In Java, what will the following program print when no exception is thrown in the try block? public class X { public static void main(String [] args) { try { badMethod(); System.out.print("A"); } catch (Exception ex) { System.out.print("B"); } finally { System.out.print("C"); } System.out.print("D"); } public static void badMethod() {} }

Introduction / Context:
This question tests control flow through try/catch/finally when no exception occurs. The finally block always executes, regardless of whether an exception is thrown or caught.

Given Data / Assumptions:

• badMethod() does nothing and throws no exception.
• try prints "A" after calling badMethod().
• catch (Exception) prints "B" only if an Exception is thrown.
• finally always prints "C". After the try-catch-finally, "D" is printed.

Concept / Approach:
No exception is thrown, so the catch block is skipped. The finally block executes and then control continues after the try/catch/finally.

Step-by-Step Solution:

Verification / Alternative check:
Add a throw in badMethod() and see how the flow changes (catch would run, but finally still runs).

Why Other Options Are Wrong:
"AC" misses the final "D"; "ABCD" would require an exception that is caught (printing "B"), which does not happen.

Common Pitfalls:
Thinking finally is conditional; forgetting the last print after the try/catch/finally.

Final Answer:
ACD