为什么 ArrayIndexOutOfBoundsException 用于 2D 数组而 NPE 用于 1D 数组

Question

Please help me to understand why below code throws ArrayIndexOutOfBoundsException

Integer[][] arr1 = { { 1, 2, 3 }, { null }, { 7, 8, 9 } };
System.out.println("value = " + arr1[1][1].intValue());

Similar code when executed for single dimensional array throws NullPointerException.

Integer[] arr2 = { new Integer(1) , null , new Integer(2) };
System.out.println("value = " + arr2[1].intValue());

In my understanding I should get NPE for both 1D and 2D array.

Answer 1

It's because the array at index 1 ( arr1[1] ) is a valid array with 1 null in it. If you checked arr1[1][0] it would return null without an exception. You're getting an ArrayIndexOutOfBoundsException because it's a valid array but ends at index 0 .

If you switched the code to:

Integer[][] arr1 = { { 1, 2, 3 }, null, { 7, 8, 9 } };

in which index arr1[1] is actually null (instead of an array with null in it), then you would get a NullPointerException .

Answer 2

If you want to produce a NullPointerException in the first case, you have to replace:

Integer[][] arr1 = { { 1, 2, 3 }, { null }, { 7, 8, 9 } };
System.out.println("value = " + arr1[1][1].intValue());

with:

Integer[][] arr1 = { { 1, 2, 3 }, { null, null }, { 7, 8, 9 } };
System.out.println("value = " + arr1[1][1].intValue());

to make the individual array1 element arr1[1][1] null.

Otherwise arr1[1][1] is not instantiated and you get ArrayIndexOutOfBoundsException .