I can't get this piece of code to run. It is throwing NumberFormatException

Question

Getting NumberFormatException. What am I doing wrong here ? I'm trying to read an array of integers separated by spaces. Is there any other better way to do the same ? I did a lot of research and most people said that split() is better than StringTokenizer and BufferedReader.

public static void main (String args[]) 
    {
    @SuppressWarnings("resource")
    Scanner io = new Scanner(System.in);
    int a=io.nextInt();
    String input;

    for(int i=0; i<a; i++)
     {   
         input = io.nextLine();
         int x= input.length();
         String[] splitArr = input.split("\\s+");
         int p[]= new int[x];
         int q=0;
         for (String par : splitArr) {
           p[q++] = Integer.parseInt(par);}

            System.out.println(p);
         }
    }

    }

Updated Code, working now:

public static void main (String args[]) 
    {
    @SuppressWarnings("resource")
    Scanner io = new Scanner(System.in);
    int a=io.nextInt();
    io.nextLine();
    String input;


    for(int i=0; i<a; i++)
     {   
         int x= io.nextInt();    //size of array
         io.nextLine();
         input = io.nextLine();
         String[] splitArr = input.split("\\s+");
         int p[]= new int[x];
         int q=0;
         for (String par : splitArr) {
           p[q++] = Integer.parseInt(par);}
         for(int m=0; m<x; m++)
            System.out.print(p[m]+" ");
         }
    }

    }

@Lemonov How to fix that ? The array of integers user enters will be a string so how to change that ? — Harshit Trehan, Jun 29 '17 at 05:25
I don't see how we can really help you since we can't see your input string. Try doing `System.out.println(par)` just before you try to parse it, so that you can see just what string you're getting the error on. Or look at it in a debugger. — ajb, Jun 29 '17 at 05:27
io.nextInt(); will give you next integers, call it as many times as you need :) or try catching this exception if you still think that one of paramters can still be other than number — Lemonov, Jun 29 '17 at 05:27
What is the exact text that you entered when you got this error? — Erwin Bolwidt, Jun 29 '17 at 05:28
I removed the outer for loop and after that i entered this- 1 3 5 2 6. The output was- [I@1f96302 — Harshit Trehan, Jun 29 '17 at 05:31

Aakash Verma · Accepted Answer · 2017-06-29T05:43:35.473

1

Do a io.nextLine(); (without storing it anywhere) after int a=io.nextInt(); (just before any string input that you'd want to take).

Something like

    Scanner io = new Scanner(System.in);
    int a=io.nextInt();
    io.nextLine();
    String input;

Loop through your array to get the values properly; not System.out.prinltn(p);

edited Jun 29 '17 at 05:43

answered Jun 29 '17 at 05:31

Aakash Verma

2,459
1
18
45

Yeah this helped! I'm not getting the exception now but the parsed array is throwing some other value. Input - 1 4 23 Output- [I@1f96302 – Harshit Trehan Jun 29 '17 at 05:35
Can you tell me what are you getting on printing `input`? – Aakash Verma Jun 29 '17 at 05:37
Suppressing warnings != error handling – Lemonov Jun 29 '17 at 05:37
I am not suppressing errors :D It was just what I copied from his code as my solution is not related to that; might as well delete that. – Aakash Verma Jun 29 '17 at 05:38
Fair point ;) I focused to much on this – Lemonov Jun 29 '17 at 05:39
"System.out.println(p);" I think this was the problem. Now I have made another loop to print the array and it is giving proper output. – Harshit Trehan Jun 29 '17 at 05:40
Wut? What is that? :D I want the value stored in `input`..just do a sysout of it after `input = io.nextLine();` – Aakash Verma Jun 29 '17 at 05:42
Okay, so that was the problem! I thought you were printing the array properly. – Aakash Verma Jun 29 '17 at 05:44
@AakashVerma It's working now. Thanks :) Check out the updated code in the question and let me know if there is still something wrong. But it is working fine now. – Harshit Trehan Jun 29 '17 at 05:46
@HarshitTrehan You're welcome! You can mark my answer as 'accepted' if it did. It'd really help me a lot. – Aakash Verma Jun 29 '17 at 07:20
@AakashVerma I've done that already but I just registered here today so my reputation is less than 15 that's why my votes aren't public :) – Harshit Trehan Jun 29 '17 at 08:06
No, I mean to say accepting the answer by clicking on the tick mark present below the down arrow.. – Aakash Verma Jun 29 '17 at 08:08
Oh, sorry. Done :) – Harshit Trehan Jun 29 '17 at 08:10
Thanks man @HarshitTrehan – Aakash Verma Jun 29 '17 at 08:49

score -1 · Answer 2 · answered Jun 29 '17 at 05:33

-1

Try doing something like this

    while(io.hasNext()){ 
        try{
            somevar = io.nextInt()
        }
        catch (NumberFormatException e){
            //just pass through this non int value or do some error handlin
        }
    }

answered Jun 29 '17 at 05:33

Lemonov

446
4
15

I can't get this piece of code to run. It is throwing NumberFormatException

2 Answers2