Why this problem print 3 4?

Question

public static void main(String[] args){

try{

int[] array = new int [4];
array[8] = 40/0;
System.out.println(“1”);

}

catch(ArrayIndexOutOfBoundsException e){
System.out.println(“2”);
}

catch(ArithmeticException e){
System.out.println(“3”);
}

finally{
System.out.println(“4”);
}

}

Answer 1

You try to do 40/0 which isn't allowed, so it doesn't print 1. It fails because it's not allowed by the rules of arithmetic, so it prints 3 and not 2. Then it finally prints 4.

EDIT: If you're expecting an out of bounds error, the reason you're not getting one is because it fails before it can assign 40/0 to the array.

Answer 2

you are try 40/0 it causes ArithmeticException
but it is catched and prints 3 instead,
then code continue after catch in finally block and prints 4
after finally there are no other code so its ends