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  1. Practice
  2. Java
  3. Exception Handling
  4. Java Exception Handling (Try-catch)
  5. Discussions

Java Exception Handling (Try-catch)

  • Vlad_Tkachenko + 2 comments

    I found only such way: 

    try (Scanner scanner = new Scanner(System.in)) {
	   System.out.println(scanner.nextInt()/scanner.nextInt());
        } catch (InputMismatchException ex) {
	   System.out.println("java.util.InputMismatchException");
        } catch (ArithmeticException ex){
            System.out.println(ex);
        }
    • shru19981 + 0 comments

      but a point in here is taht what if i give the Eception class instead of the specific class it has no chang except that Exceptionn class is the superclass.it should give the same ans but what is the diff.When i tried your way i got the answer.

    • yeikel + 1 comment

      This solution is actually better than mine because you´re using try with resources. Good one

      • shru19981 + 0 comments

        thank you