Strange things with java...

Discussion in 'Java' started by Roberto Nicastro, May 11, 2007.

  1. Hi all,
    i have this piece of code

    Suppose that lstA is List of 3 Objects

    List<Object> lstA = ...something;
    List<Object> lstB = lstA;
    ....
    [block A] //this is a block of code
    ....

    Inside block A i remove one item from lstA, so lstA become a List of 2
    Objects. Inside block A i never touch lstB.
    Why at the end of block A also lstB is with 2 elements? How is this
    possible?

    Thanks in advance

    R
     
    Roberto Nicastro, May 11, 2007
    #1
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  2. Neither lstA nor lstB is a list. They are both reference variables that
    are initialized as pointers to the same List object.

    Although block A does not touch lstB, it does remove an element from the
    List it points to.

    Patricia
     
    Patricia Shanahan, May 11, 2007
    #2
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  3. Because lstA and lstB are pointers. All objects are. When you assign
    lstB = lstA, you make lstB point at the same list; you make lstB ==
    lstA. Perhaps what you wanted to do was make lstB a copy of lstA? Or
    perhaps you wanted to go further and make lstB a copy of lstA and all
    its members copies of the members in lstA? Have you thought about what
    happens if you change an object in either list?
     
    Richard Senior, May 11, 2007
    #3
  4. Roberto Nicastro

    Raghav Guest

    Actually here you have got one list in memory and 2 references to
    that. So when u modify the list, it is reflected in either of
    references pointing to that List.
    HTH.
     
    Raghav, May 11, 2007
    #4
  5. Roberto Nicastro

    Eric Sosman Guest

    Roberto Nicastro wrote On 05/11/07 13:51,:
    An analogy (I love analogies!) that may help you
    understand what others have pointed out: lstA is the
    number on the debit card in your wallet, and lstB is
    the same string of digits recorded in your password-
    protected on-line form-filler assistant. In [block A]
    you hand the card to a local merchant to pay for your
    new jPea, 150 silver splonders including VAT. Then
    you go to your computer, use the automatic form-filler
    to log in to your bank account, and check the balance.
    Are the 150 silver splonders still there to be spent?

    One bank account, two copies of the account number.
    One List instance, two references to it. Spend money
    with the debit card, and the balance change is visible
    when you use the copied number. Change the List via
    listA, and the change is visible through listB.
     
    Eric Sosman, May 11, 2007
    #5
  6. I want to create an exact copy of lstA. So, to create a real copy i
    have to create a new List object

    List lstA = ...something
    List lstB = new ArrayList();

    for (Iterator it; it = lstA.iterator(); it.hasNext(); ){
    lstB.add(it.next());
    }

    I think this will work for me.
    Thank you all

    R
     
    Roberto Nicastro, May 12, 2007
    #6
  7. Yes, but bear in mind that the two lists, while they are different
    lists, contain the same elements. If you change an object in lstA, it
    also changes in lstB.

    Have a look at this
    http://forum.java.sun.com/thread.jspa?threadID=487186&messageID=2281852
    and do some searches on "shallow copy" and "deep copy".
     
    Richard Senior, May 12, 2007
    #7
  8. There's an addAll method you could use instead of a loop of your own:

    List lstB = new ArrayList();
    lstB.addAll(lstA);

    Look it up in Javadoc.
     
    Jussi Piitulainen, May 12, 2007
    #8
  9. Or, even simpler:

    List lstB = new ArrayList(lstA);

    Patricia
     
    Patricia Shanahan, May 12, 2007
    #9
  10. Roberto Nicastro

    Chris Smith Guest

    While you've got several explanations of what's going on, you haven't
    gotten much by way of how to make it act as you'd like. To do that, you
    could change the second line there to:

    List<Object> lstB = new ArrayList(lstA); // or other List implementation

    That would cause lstB to point to a different List, which has its
    CONTENTS initialized from the list that lstA points to. Then you could
    modify the list that lstA points to, and the list that lstB points to
    would be unchanged.
     
    Chris Smith, May 12, 2007
    #10
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