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

Java Exception Handling (Try-catch)

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  • Schuetzl + 8 comments

    Suggestion: Make the fourth testcase less strict. I had my catch outting...
    java.util.InputMismatchException: For input string: "2147483648"
    When it expected...
    java.util.InputMismatchException
    I simply made another catch for that specific exception and output "java.util.InputMismatchException" directy to work around it but I feel like that shouldn't be necessary. Unless simply printing out the Exception directly is the wrong way to go about it?

    • abdulhannanali + 3 comments

      Same here bro same here

      // These were my catches
catch (InputMismatchException e){

	System.out.println("java.util.InputMismatchException");

}
catch (Exception e){
	System.out.println(e);
}
      • Tuxdude + 0 comments

        If you use the Scanner class, you only need to catch the InputMismatchException and ArithmeticException to pass all the test cases. There is no need for a catch-all Exception.


        Using a catch-all Exception clause is generally a bad idea when writing programs.

      • Farahmand + 4 comments

        Instead of

        System.out.println("java.util.InputMismatchException");

        we can use:

        System.out.println(e.getClass().getName());
        • taffarelbergamin + 0 comments

          Fine, but what's the point of showing an exception message in a different way than the other? If you use that, it will give you an error when in the division by zero test case.

          Why one exception should show the extra content and the other not? This is not even part of the exercise.

          What they could do if they really want a nice representation of the exceptions, could be we print something like "This is an input mismatch exception" and "This is an arithmetic exception".

          This would force people to know which exceptions could happen and don't use the generic Exception.

        • 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 + 0 comments

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

            • 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);
        }
        • ItzMv + 1 comment

          Hey can you tell me why the Test case 4th not showing:-

          java.util.InputMismatchException: For input string: "2147483648"

          After using e.getClass().getName()

          • allforcatmohit + 0 comments

            Here when you use e.getClass().getName(), .getClass() method will be returning the class of the exception and as we have used getName() method. we will get the name of the exception class.

            so here :java.util.InputMismatchException: For input string: "2147483648"

            class name is just "java.util.InputMismatchException"

        • Aakash1021 + 0 comments

          What is the use of e.getClass().getName()

      • JimB6800 + 0 comments

        Yes, it's a poor problem.

        I ended up doing...

        } catch (InputMismatchException e) {
  System.out.println(new InputMismatchException());
}

        as a way of stripping off the message.

    • ShafaetChallenge Author + 3 comments

      It was intentional.

      • Schuetzl + 0 comments

        I had assumed it was. It just seemed to me like an unnecessary hoop to have one jump through is all.

      • samplenhold + 0 comments

        If this was in school, I would have told the teacher to scew off. This is garabage. I could have had this challenge done in three lines of code. And having detailed exceptions for one thing and not another thing is a jerky thing to do. It goes against all coding conventions I was taught in school.

      • b0kater + 0 comments

        And the purpose is?

    • SSPIDER + 0 comments

      thanx dude its work

    • giaym + 0 comments

      wait you actually had to have try catches? I was just throwing exceptions around :P, no fun if you just print the stuff without crashing the program

    • manisha_9801 + 0 comments

      Yes I have also done the same thing

    • amannepid + 0 comments

      Another way without any concern of the exception class: 

      catch(Exception e){
            System.out.println(e.getClass().getName() + (e.getMessage() != null && e instanceof ArithmeticException  ? (": " + e.getMessage()) : ""));
        }
    • vinaynit11 + 1 comment

      can't it be possible that , we print the required output by using the object of exception we passed in the catch block .

      Ex :

      try {

      }catch(InputMismatchException i) { System.out.println(i); }

      above program print Exception with the causal , but the required output is without causal ?

      • manisha_9801 + 1 comment

        By this u handled InputMismatchException......but what about other types of exception How u will handle them

        • vinaynit11 + 0 comments

          for other exceptions , we will have to first put specific catch block for that specific exceptions still if there is some possibility that other exception might get raised put a generic catch block for handeling all exceptions .

    • Anu_cool_007 + 0 comments

      Still not fixed :(

  • AdHominem + 6 comments

    Again a great example of how bad the test cases are written. Any professional would have caught the exceptions together and logged / souted them in a uniform way like this:

    try (Scanner scanner = new Scanner(System.in)) {
	System.out.println(scanner.nextInt()/scanner.nextInt());
} catch (ArithmeticException | InputMismatchException ex) {
	System.out.println(ex);
}

    This is best practise for any team I worked with and definitely the cleanest way to solve this. But thanks to the nitpicking test cases, beginners learn to catch exceptions in an utterly bloated way:

    try (Scanner scanner = new Scanner(System.in)) {
	System.out.println(scanner.nextInt()/scanner.nextInt());
} catch (ArithmeticException aex) {
	System.out.println(aex);
} catch (InputMismatchException imex) {
	System.out.println(imex.getClass().getName());
}
    • wave_33 + 1 comment

      what is the difference between both ???

      • AdHominem + 0 comments

        They produce the same output, thats the problem. You can separate catch blocks IF you want to handle exceptions individually but if they are caught together, you can just use one catch block and catch all of them.

    • hackeranshu + 0 comments
      [deleted]
    • mssvai + 0 comments

      Thanks. I didn't know I can pass objects to try block.

      4th test case got failed tho, so I fixed it a lil bit.

      public static void main(String[] args)
{
    try(Scanner in = new Scanner(System.in))
    {
        System.out.println(in.nextInt()/in.nextInt());
    } catch(InputMismatchException ex)
    {
        System.out.println(ex.getClass().getName());
    }
    catch(ArithmeticException ae)
    {
        System.out.println(ae);
    }
}
    • 148W1A1237 + 0 comments

      Thank you brother. I have got it.

    • sb_pt + 0 comments

      Imagine people writing actual production code like this!

    • i140394 + 0 comments

      I learned a new thing :)

  • murtuza89 + 3 comments

    This worked for me:

    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) {
        Scanner sc=new Scanner(System.in);
        try{
            int x=sc.nextInt();
            int y=sc.nextInt();
            int z=x/y;
            System.out.println(z);
        }catch(InputMismatchException ime){
            System.out.println(ime.getClass().getName());
        }catch(ArithmeticException ae){
            System.out.println(ae.getClass().getName()+": "+ae.getMessage());
        }catch(Exception i){
            System.out.println(i.getMessage());

        }
    }
}
    • kanmanim + 1 comment

      this does not work

      • murtuza89 + 1 comment

        Worked for me :( Its java program by the way. Hope you selected Java in language option on top right side of the editor. Which test case did it failed for you? Did it gave any compilation error? Can you elaborate?

        • satish_paluvai + 2 comments

          perfect solution in java for passing all the test cases :

          public static void main(String[] args) { try{ Scanner s = new Scanner(System.in); System.out.println(s.nextInt()/s.nextInt()); }catch(ArithmeticException e) { System.out.println(e);

          }catch(InputMismatchException e)
    {
    System.out.println(e.getClass().getName());
}
    catch(Exception e)
{
    System.out.println(e.getClass());
}
}
          • pavima1965 + 0 comments

            i used this stub..but,,,while i type by my own way..it doesnn't work :(// how did i get all testcases passed...when i do copy paste method... :(

          • pavima1965 + 2 comments

            Scanner s= new Scanner(System.in); int x=s.nextInt(); int y=s.nextInt();

                try{
        int z=x/y;
        System.out.println(z);
    }
    catch(InputMismatchException i){
        System.out.println(i.getClass().getName());
    }catch(Exception ae){
        System.out.println(ae+": / by zero");
    }   

    ?? why it is not workig for me..since the arithmetic exception already catched..nd hence am just going to print the next following statements?!
            • viktor_kirjanov + 1 comment
              int x=s.nextInt(); 
int y=s.nextInt();

should be within try{}
              • damini123 + 1 comment

                yeah it worked.. but can you tell me why is it so??

                • asheslaw + 0 comments

                  InputMismatchException is an Exception thrown by Scanner class while taking inputs. The statements in which x and y are taking values from object of Scanner class should be within try block to catch that exception.

            • Priyanka_das + 0 comments

              You don't need append ": / by zero"

    • simran1707 + 0 comments

      Thanks a lot.It worked for me.But It will also work if u don't use the last catch.

    • Kajalgoel + 0 comments

      why do we need to print - System.out.println(ae.getClass().getName()+": "+ae.getMessage()); and not directly the exception object?

  • 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

  • arhankundu99 + 0 comments

    Here is my code...

    import java.util.*;
public class solution
{
    public static void main(String[] args)
    {
        Scanner scan=new Scanner(System.in);
        try
        {
            int a=scan.nextInt();
            int b=scan.nextInt();
            System.out.println(a/b);
        }
        catch(InputMismatchException e)
        {
            System.out.println("java.util.InputMismatchException");
        }
        catch(Exception e)
        {
            System.out.println(e);
        }
    }
}

    Instead of printing java.util.InputMismatchException I got java.util.InputMismatchException: For input string: "2147483648" ,So I made another catch to directly print the exception

  • kailiu + 0 comments

    I think the most important part is how to retrieve the exception

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
		try {
			int a = scan.nextInt();
			int b = scan.nextInt();
			int c = a/b;
			System.out.println(c);
		
            } catch (Exception e) {
                String s = e.getClass().getCanonicalName();
                if (s=="java.lang.ArithmeticException") {
                    s+= ": / by zero";                
                }
                System.out.println(s);
            }
        }

  • RodneyShag + 0 comments

    Java solution - passes 100% of test cases

    From my HackerRank solutions.

    import java.util.Scanner;
import java.util.InputMismatchException;

public class Solution {
    public static void main(String[] args) {
    	Scanner scan = new Scanner(System.in);
        try {
            int x = scan.nextInt();
            int y = scan.nextInt();
            System.out.println(x/y);
        } catch (InputMismatchException e) {
            System.out.println(e.getClass().getName());
        } catch (ArithmeticException e) {
            System.out.println(e.getClass().getName() + ": / by zero");
        }
    }
}

    Let me know if you have any questions.

  • yagiz + 0 comments

    Writing a test case that expects an exception thrown is a better approach than one that expects the exception message. I'm not very familiar with HackerRank so this may not be possible, but it's not a good practice to force the developer to catch an exception that was going to be thrown anyway for the sake of testing.

  • tao_zhang + 0 comments

    Didn't mentioned need to handle different exceptions differently in the question, but apparently that's how they wanted. 

        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 scanner = new Scanner(System.in);
        try{
            int x = scanner.nextInt();
            int y = scanner.nextInt();
            System.out.println(x/y);
        }catch(Exception e){
            if (e instanceof InputMismatchException){
                System.out.println(e.getClass().getName());
            } else {
                System.out.println(e);
            }
            
        }

    }

  • sironcel + 1 comment

    I think this is not a right exception handling. I have to write all exceptions. Because I don't know test cases inputs

    • skm14Asked to answer + 0 comments

      Yeah, i'm also not fully satisfied with the solution. There should be a better explanation to it.

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