Question

I was given the following code and asked to write the Solution class extending from TestList. I wrote a constructor for it (which just called super) and the printSecond2() method invoked in the last line of the code below. All other methods are inherited. Here’s the code:

public class Test3A {
    public static void main(String[] args) {
        TestList tl = new Solution();
        tl.loadList();
        ((Solution) (tl)).printSecond2();//prints every second element
    }
}

However, the damn thing was never printing anything out, so I went into the TestList class (which was provided) and put println statements after every single line of the loadList () method:

   public void loadList ()
   {
      if (input.hasNextInt ())//input is a Scanner object
      {
         int number = input.nextInt ();
         loadList ();
         add (number);
      }
   }

I discovered that I can continue to add whitespace, newlines and integers indefinitely, and that the add(number) method is only finally called when I input a character. So if I don’t do that, it just sort of hangs around waiting for more input instead of moving on.

I’m confused by this as the provided sample input/output is:

sample input

1 2 3 4 5

sample output

2 4

So there’s no character being inputted by the automatic marker.

I have tried overriding the method in Solution (we can’t touch the other classes) and:

1. ) changing if to while
2. ) adding an else block
3. ) adding an else if (!input.hasNextInt ())

None of these changed anything. I have no idea how the program is supposed to move on and get as far as calling printSecond2().

Any thoughts? I’d really like to pass my next prac test 😀

Answer 1

When user is supposed to enter a sequence of numbers either the number of items should be provided or the input should be terminated in some manner. 1 2 3 and 1 2 3 4 are both valid inputs so scanner can’t decide where to end on its own.
It can be assumed that the number sequence is terminated by EOF character Ctrl-Z on windows and Ctrl-D on unix as no other information is given.