Multiple choice technology programming languages

Given: 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.printIn("x = " + ++x); } } }

  1. x = 42

  2. x = 43

  3. x = 44

  4. Compilation fails

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

The variable x is declared inside the try block, meaning its scope is limited to that block. It is inaccessible in the catch and finally blocks, causing a compilation error when referenced there. Additionally, the typo X++ (uppercase) and printIn (capital I) would also prevent compilation.