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.

Solution 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.

Solution 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.

Solution 3

The NullPointerException is thrown out in the static block, where you are trying to assign a value 1 to the first element of array (x[0] = 1). Be aware, the int[] array named x is still not intilized.

public class Test {

static int x[];

static {
    x[0] = 1;// Here is the place where NullPointException is thrown.
}

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

There are 2 ways for you to fix it.

1 Use static int x[] = new int[5]; instead of static int x[] ;

2

Change

static {
    x[0] = 1;
}

To

static {
    x= new int[5];
    x[0] = 1;
}

Remember: Initialize the array before you use it.

Solution 4

You didn't initialize your x array. There is a difference between declaration and initialization of variables. When you write int x[]; you just declare a variable which, as an instance field, is initialized with a default value of null. To actually create an array you must write int x[] = new int[10]; or the size you need.

The reason for getting a NullPointerException instead of ArrayIndexOutOfBounds is that the latter is thrown when you do have an array and try to address a position out of its bounds, but in your case you don't have an array at all and try to put something into a non-exsting array. That's why an NPE

Solution 5

static int x[];

static {
    x[0] = 1;
}

Results in NullPointerException, because your x array in not initialised (is null)

ArrayIndexOutOfBoundException would happen if you accessed an index that is out of bounds:

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

static {
    x[20] = 1; //<-----accessing index 20
}

View more solutions