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

Java Exception Handling (Try-catch)

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340 Discussions

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  • arthur_avelar
     + 0 comments
    public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        try {
            int x = sc.nextInt(), y = sc.nextInt(); 
            System.out.println(x / y); 
        } catch(InputMismatchException exc) {
            System.out.println(exc.getClass().getName());
        } catch(ArithmeticException exc) {
            System.out.println(exc.getClass().getName() + ": " + exc.getMessage());
        }
        finally {
            sc.close();
        }
        
    }
}

  • anshtomar184
     + 0 comments

    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) {
     try (Scanner scanner = new Scanner(System.in)) {
        int x = scanner.nextInt();
        int y = scanner.nextInt();

        System.out.println(x/y);
    } catch (Exception e) {
        System.out.println(getText(e));
    }
}

private static String getText(Exception e) {
    return e.toString().contains(": For input") ? e.toString().substring(0, e.toString().indexOf(":")) : e.toString();
}

    }

  • singhsarvagya6
     + 0 comments

    Scanner in = new Scanner(System.in);

        try{
        int x = in.nextInt();
        int y = in.nextInt();
        if(x>=0 && y>=0)
        System.out.println(x/y);
    }catch(InputMismatchException ee){
        System.out.println("java.util.InputMismatchException");
    }
        catch(Exception e){
        System.out.println(e.toString());
    }


}

    }

  • shoaibshakilsha1
     + 0 comments

    Initially in catch block i simply wrote sysout(e); to print exception message.

    But one of the test case failed for arithmetic exception so i used ternery operator like below instaed of writing catch block for all exceptions.

    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 = 0;
        int y = 0;


        x = sc.nextInt();
        y = sc.nextInt();

       System.out.print(x/y);

    }catch(Exception e){
        String emsg = (e.getClass().getName().equalsIgnoreCase("java.lang.ArithmeticException")) ? "java.lang.ArithmeticException: / by zero" : e.getClass().getName();
        System.out.print(emsg);
    }
    finally{
        sc.close();
    }



}

    }

  • yashdeora98294
     + 0 comments

    Here is java Exception Handlin (Try-catch) solution - https://programmingoneonone.com/hackerrank-java-exception-handling-try-catch-solution.html