Question

This problem came up in a practice test: create a new string array, initialize it to null, then initializing the first element and printing it. Why does this result in a null pointer exception? Why doesn’t it print “one”? Is it something to do with string immutability?

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");
        }
    }

Thanks!

Answer 1

Because you made arr referring to null, so it threw a NullPointerException.

EDIT:

Let me explain it via figures:

After this line:

String arr[] = new String[10];

10 places will be reserved in the heap for the array arr:

and after this line:

arr = null;

You are removing the reference to the array and make it refer to null:

So when you call this line:

arr[0] = "one";

A NullPointerException will be thrown.