Multiple choice technology programming languages

class Scoop { static int thrower() throws Exception { return 42; } public static void main(String [] args) { try { int x = thrower(); } catch (Exception e) { x++; } finally { System.out.println("x = " + ++x); } } } What is the result?

  1. x = 42

  2. x = 43

  3. x = 44

  4. Compilation fails.

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

To solve this question, the user needs to understand the concepts of exceptions and the try-catch-finally block.

The thrower() method is defined to throw an Exception, but it returns the integer 42. In the main() method, we try to call thrower(), but since it throws an exception, we catch the exception in the catch block. However, the variable x is not defined outside of the try block, so attempting to increment it in the catch block will result in a compilation error.

Therefore, the correct answer is:

The Answer is: D