please tell me what will be the output of this and why?

[gunjan khanuja]

class Breaktest {
   public static void main(String args[]) {
        try {
            String arr[] = new String[10];
            arr = null;
            arr[0] = "one";
            System.out.print(arr[0]);
        }  catch(NullPointerException nex) { 
            System.out.print("null pointer exception"); 
        }catch(Exception ex) {
            System.out.print("exception");
        }
        
    }
}

[Added code tags - see UseCodeTags for details]

[Praveen Kumar M K]

Hi Gunjan,

Did you run this program? If yes, what was the output that you got?

[Matthew Brown]

Hi Gunjan. Welcome to the Ranch!

Praveen's right. The best way to find out what the output is is to run it yourself. Then, if you can't work out why that is the output, come back and we'll help you out.

[gunjan khanuja]

yes i tried and i got null
pointer exception ..but why did i get this output ..could you please explain..?

[Matthew Brown]

Well, on line 6 you're trying to access an element of a null array. What would you expect to happen?

[gunjan khanuja]

no but then i am also giving zeroth element a value in another line...

[Bear Bibeault]

You can't add anything to a null array.

[gunjan khanuja]

Thanks i get it...

[Praveen Kumar M K]

The array variable arr is pointing to null in line 5 and in the next line you are trying to "dereference" the null pointer. It is this line which is causing the exception.

In case you have not yet gone through what a null pointer is and what a NullPointerException is , do search this forum, there are plenty of posts on the same.

Thanks,
Praveen.