Predict the output of following Java programclass Test extends Exception {}class Main { public static void main(String args[]) { try { throw new Test(); } catch (Test t) { System.out.println("Got the Test Exception"); } finally { System.out.println("Inside finally block "); } }}OptionsGot the Test ExceptionGot the Test Exception Inside finally blockInside finally blockCompiler Error
Question
Predict the output of following Java programclass Test extends Exception {}class Main { public static void main(String args[]) { try { throw new Test(); } catch (Test t) { System.out.println("Got the Test Exception"); } finally { System.out.println("Inside finally block "); } }}OptionsGot the Test ExceptionGot the Test Exception Inside finally blockInside finally blockCompiler Error
Solution
The output of the Java program will be:
Got the Test Exception Inside finally block
Here's the step by step explanation:
- The main method of the Main class is executed.
- Inside the main method, a try block is started. Inside this try block, a new Test exception is thrown.
- Since the Test exception is thrown, the control is passed to the catch block that can handle this type of exception. In this case, the catch block that handles the Test exception is executed. Inside this block, "Got the Test Exception" is printed to the console.
- After the try and catch blocks are executed, the finally block is executed regardless of whether an exception was thrown or not. Inside this block, "Inside finally block" is printed to the console.
- Therefore, the output of the program is:
Got the Test Exception Inside finally block
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