Re: Java Arrays.sort throws exception
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