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int[] = null; help

adeeb alexander
Ranch Hand

Joined: May 29, 2008
Posts: 268
Hi all.
As normal how we declare a string array when we dont know the size i.e how much data should be inserted then i give as String[] str = null. After that i can perform all the operations. When i try to do the same with int that is int[] arr = null. I dont get compilation error, when i try to access its elements i am getting the java.nullpointer exception.



Any help would be greatly appreciated.
Thanks
alexander
Greg Charles
Sheriff

Joined: Oct 01, 2001
Posts: 2861
    
  11

I'm not sure I understand. If you set something to null, then try to access it, of course you will get a null pointer exception. In your first example, with the String array, the same thing would happen.
Mamata Chaudhari
Greenhorn

Joined: Apr 26, 2009
Posts: 23
hi..
If you want dynamic size then you can use ArrayList<String>

but it is not possible to define string array without size.
adeeb alexander
Ranch Hand

Joined: May 29, 2008
Posts: 268
Thanks for your replies guyz. See the code below. I declared a String[] temp and used it like this below. It works for me.


Rob Spoor
Sheriff

Joined: Oct 27, 2005
Posts: 19755
    
  20

That's because your giving it a value first, on line 5.


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adeeb alexander
Ranch Hand

Joined: May 29, 2008
Posts: 268
yes. In the same way i have given values for int[] also. For example consider.
int [] exp= null;
for(int i = 0;i<5;i++);
{
exp[i]=i;
}

I have to do somewhat this thing in my program also.

when i use System.out.println(exp[0]);

i get error during execution. i need the reason please. As you said i have given values to exp isnt it?


Thanks
Rob Spoor
Sheriff

Joined: Oct 27, 2005
Posts: 19755
    
  20

You don't assign anything to exp itself. Nowhere are you calling "exp = X" where X is something other than null. That means that the array itself remains null, and there are no elements to access.

With the String[] you are doing just that and that's why that example is working and this one isn't.

The easiest solution is adding "exp = new int[5];" just before the loop.
adeeb alexander
Ranch Hand

Joined: May 29, 2008
Posts: 268
Hi
Look at the code below. I am using this to solve given input separated by ",". I think i have used some extra code. i.e i guess there is some other short way to do this. If any one know please share with me.


Thanks and Regards
alexander
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