Re: Java Arrays.sort throws exception

From:
Roedy Green <see_website@mindprod.com.invalid>
Newsgroups:
comp.lang.java.programmer
Date:
Wed, 14 May 2008 20:55:31 GMT
Message-ID:
<31km24pnvf83he76fvffrgop1vom8c3apo@4ax.com>
On Wed, 14 May 2008 11:32:07 -0700 (PDT), captain
<madison223@gmail.com> wrote, quoted or indirectly quoted someone who
said :

 List adminItems = new ArrayList();

 loop:
    adminItems.add(administeredItem);

 Collections.sort(adminItems);

------------
Collections.class (Sun Code):

  public static void sort(List list) {
    Object a[] = list.toArray();
    Arrays.sort(a); // THROWS EXCEPTION
    ListIterator i = list.listIterator();
    for (int j=0; j<a.length; j++) {
       i.next();
       i.set(a[j]);
    }
   }


You did not tell us the type of administeredItem. I assume for
purposes of argument AdministeredItem.

You want something like this:

 List<AdministeredItem> adminItems
= new ArrayList<AdministeredItem>( 100 );

  loop:
     adminItems.add(administeredItem);

  Collections.sort(adminItems);

Your class AdministeredItem class will need to implement
Comparable<AdministeredItem>

See http://mindprod.com/jgloss/comparable.html
for how.

If administeredItem is a String then you need

 List<String> adminItems
= new ArrayList<String>( 100 );

The piece of code of most interest is the compareTo method of
AdministeredItem. You did not reveal it yet.
--

Roedy Green Canadian Mind Products
The Java Glossary
http://mindprod.com

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