java.lang.ExceptionInInitializerError Caused by: java.lang.NullPointerException

Question

This is from OCJP example. I have written a following code

public class Test {

  static int x[];

  static {
     x[0] = 1;
  }

  public static void main(String... args) {
  }        
}       

Output: java.lang.ExceptionInInitializerError

Caused by: java.lang.NullPointerException at x[0] = 1;

Why it is throwing NullPointerException and not ArrayIndexOutOfBoundException.

Answer 1

Why it is throwing NullPointerException and not ArrayIndexOutOfBoundException.

Because you did not initialize the array.

Initialize array

   static int x[] = new int[10];

Reason for NullPointerException:

Thrown when an application attempts to use null in a case where an object is required. These include:

• Calling the instance method of a null object.
• Accessing or modifying the field of a null object.
• Taking the length of null as if it were an array.
• Accessing or modifying the slots of null as if it were an array.
• Throwing null as if it were a Throwable value.

You hit by the bolder point, since the array is null.

Answer 2

It's throwing NullPointerException because x is null.

x[] is declared, but not initialized.
Before initialization, objects have null value, and primitives have default values (e.g. 0, false etc)

So you should initialize as shown below:

static int x[] = new int[20]; //at the time of declaration of x
or
static int x[];
x = new int[20]; //after declaring x[] and before using x[] in your code

ArrayIndexOutOfBoundException will occur if array is initialized and accessed with an illegal index.

e.g :
x contains 20 elements, so index numbers 0 to 19 are valid, if we access with any index < 0 or
index > 19, ArrayIndexOutOfBoundException will be thrown.