What will be the output of the following code?class Test extends Exception { }public class Main { public static void main(String args[]) { try { throw new Test(); } catch(Test t) { System.out.print("Catch the Test Exception,"); } finally { System.out.print("Inside finally block "); } }}Options: Pick one correct answer from belowCatch the Test Exception,Inside finally blockRuntime errorCatch the Test Exception, Inside the finally block
Question
What will be the output of the following code?class Test extends Exception { }public class Main { public static void main(String args[]) { try { throw new Test(); } catch(Test t) { System.out.print("Catch the Test Exception,"); } finally { System.out.print("Inside finally block "); } }}Options: Pick one correct answer from belowCatch the Test Exception,Inside finally blockRuntime errorCatch the Test Exception, Inside the finally block
Solution
The correct answer is "Catch the Test Exception, Inside finally block".
Here's the step by step explanation:
- The main method throws a new instance of the Test exception.
- The catch block catches this exception and prints "Catch the Test Exception,".
- Regardless of whether an exception was thrown or caught, the finally block always executes. It prints "Inside finally block ".
- Therefore, the output of the program is "Catch the Test Exception, Inside finally block ".
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