What will be the output of the following Java program?
class exception_handling
{
public static void main(String args[])
{
try
{
int a = 1;
int b = 10 / a;
try
{
if (a == 1)
a = a / a – a;
if (a == 2)
{
int c[] = {1};
c[8] = 9;
}
finally
{
System.out.print(“A”);
}
}
catch (NullPointerException e)
{
System.out.println(“B”);
}
}
}
a) A
b) B
c) AB
d) BA

Answer: a
Explanation: The inner try block does not have a catch which can tackle ArrayIndexOutOfBoundException hence finally is executed which prints ‘A’ the outer try block does have catch for NullPointerException exception but no such exception occurs in it hence its catch is never executed and only ‘A’ is printed.
Output:
A