Java ArrayIndexOutOfBoundsException: Code Analysis and Output

Tom wrote the following code:

try {
    int []array = new int[] {1, 2, 3, 4, 5};
    int result = array[5];
    System.out.printin('result value : ' + result++);
} catch(Exception e) {
    e.printStackTrace();
} catch(ArrayIndexOutOfBoundsException e) {
    e.printStackTrace();
} finally {
    System.out.printin('finally block');
}

Select the correct output for the following code:

A) ArrayIndexOutOfBoundsException finally block B) Compilation error cannot run C) result value : 5 finally block D) result value : 6 finally block

Correct Answer: B) Compilation error cannot run.

This is because the catch block for ArrayIndexOutOfBoundsException will never be reached since it is more specific than the catch block for Exception. In Java, exception handling follows a hierarchical order. Since ArrayIndexOutOfBoundsException is a subclass of Exception, the catch block for the more specific exception (ArrayIndexOutOfBoundsException) should be placed before the catch block for the more general exception (Exception). Therefore, the code results in a compilation error due to this ordering issue.