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

Java Exception Handling (Try-catch)

  • kencluff + 3 comments

    I did that, but then testcase 2 failed because it wanted "java.lang.ArithmeticException: / by zero" and you just get "java.lang.ArithmeticException" with this trick.

    • sergeief + 0 comments

      you can get "/ by zero" from ecxeption details message

    • 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");
    }
}

        }

    • SamratV + 0 comments

      May be because you are using older version of java. In java 8 this code passes all the testcases successfully.

      try
        {
            int a=obj.nextInt();
            int b=obj.nextInt();
            System.out.print(a/b);
        }
        catch(InputMismatchException e)
        {
            System.out.print(e.getClass().getName());
        }
        catch(Exception e)
        {
            System.out.print(e);
        }