Java String split by comma

Question

I have this code down below that gives me this output

1,2,3,4,3,4,5,4,3,5,3,4,5,5,4,64,
[Ljava.lang.String;@3e25a5
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f
[I@19821f

The input.txt file contains 1,2,3,4,3,4,5,4,3,5,3,4,5,5,4,64,

The code is this. It is clear there is a simple error in splitting but I find it hard to find what.

public static void main(String[] args) {
    // TODO Auto-generated method stub



    try{
        FileInputStream fstream = new FileInputStream("input.txt");
        DataInputStream  dat = new DataInputStream (fstream);
        BufferedReader in = new BufferedReader(new InputStreamReader(dat));
        String[] str ;

        int arr[] = new int [100];


        String line;

        while ((line = in.readLine()) != null)
            {

                System.out.println(line);
                str = line.split(",");
                System.out.println(str);

                for(int i = 0 ;i<str.length ; i++)
                {
                    arr[i]= Integer.parseInt(str[i]);
                    System.out.println(arr);
                }
            }

        fstream.close();

    }

    catch(IOException e)
    {
        System.out.print(e);
    }


}

You're going to get a NumberFormatException because you have a trailing comma and the split() function is going to give you an empty string at the end of your array. — Qwerky
– Qwerky, Commented May 10, 2011 at 16:38

insumity · Accepted Answer · 2011-05-10 16:15:32Z

8

Change this

   System.out.println(arr);

to

   System.out.println(arr[i]);

before you were printing an array arr, not the array's elements.

answered May 10, 2011 at 16:15

insumity

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Comments

Kaypro II · Accepted Answer · 2011-05-10 16:19:51Z

3

It looks like you're repeatedly printing the array, which will cause Java to call the array's toString() method, that will give you output like you've provided.

To fix it, try changing your loop to this:

for(int i = 0 ;i<str.length ; i++)
{
    arr[i]= Integer.parseInt(str[i]);
    System.out.println(arr[i]];
}

To give you a clearer idea what's going on, this program will do something similar:

public class Test {
    public static void main(String[] args) {
        System.out.println(new int[0]);
    }
}

It will output:

[I@164f1d0d

answered May 10, 2011 at 16:19

Kaypro II

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Comments

Edwin Buck · Accepted Answer · 2013-08-20 03:37:12Z

By printing out the array in

 System.out.println(arr);

you effectively print out an Object, which happens to be the Array object. That has a toString() method, and it tends to print out the Object's Class and the hashcode() reference.

Now, since your array is of a primitive type, it looks like this

[I@19821f

where each item means something

[ => "an array of"
I => "primitive integer"
@19821f => "which has a hashcode of 0x19821f"

Note that this is a default convention, meaning that while it is what you are seeing, you can't count on this reading technique with the same sense of the word guarantee.

Like many have mentioned before, you need to print out the element of the array (which won't use this pseudo-object toString() representation, because the element is an integer.

sudmong · Accepted Answer · 2011-05-10 16:17:44Z

0

remove the unnecessary sysouts to get the exact output. and in last sysout print the element not complete array.

answered May 10, 2011 at 16:17

sudmong

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Comments

Java Devil · Accepted Answer · 2013-08-20 03:27:29Z

-2

str = line.split(",");

If the string contains comma, then you should not use split()

edited Aug 20, 2013 at 3:27

Java Devil

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answered Aug 20, 2013 at 3:05

kksuak

1

1 Comment

andyg0808 Over a year ago

-1: This makes no sense, and seems completely wrong. If there's a comma in the string, that's no reason not to use split...

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