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

Java Exception Handling (Try-catch)

  • amarvadla27 + 3 comments
    try{
            Scanner sc = new Scanner(System.in);
            int x = sc.nextInt();
            int y = sc.nextInt();
            if(y==0)
                throw  new ArithmeticException("/ by zero");
            else
                System.out.println(x/y);
        }
        catch(InputMismatchException e){
            System.out.println(e.getClass().getName());
        }
        
        catch(ArithmeticException e){
            System.out.println(e);
        }

    `

    • bmk9727 + 1 comment

      can please explain this System.out.println(e.getClass().getName());

      • quirkudoe213 + 1 comment

        you prolly already understand this

        • syndicate27 + 0 comments

          Thank you ^_^

    • syndicate27 + 0 comments

      I look this it looks a lot cleaner then writing out the Arithmetic exception in the catch block. Thank you

    • allforcatmohit + 0 comments

      This also works:

      import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

      public class Solution {

      public static void main(String[] args) {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
    Scanner sc = new Scanner(System.in);
    try{
        int x = sc.nextInt();
        int y = sc.nextInt();
        try{
            int result = x/y;
            System.out.println(result);
        } catch(Exception e){
            System.out.println("java.lang.ArithmeticException: / by zero");
        }
    } catch(Exception e){
        System.out.println("java.util.InputMismatchException");
    }
}

      }