Re: For each and remove

From:
"Daniel Dyer" <"You don't need it">
Newsgroups:
comp.lang.java.help
Date:
Mon, 11 Jun 2007 19:57:50 +0100
Message-ID:
<op.ttrtmoib8kxvgr@jack.local>
On Mon, 11 Jun 2007 00:05:32 +0100, Roedy Green =

<see_website@mindprod.com.invalid> wrote:

On Fri, 08 Jun 2007 15:58:37 -0700, grasp06110 <grasp06110@yahoo.com>
wrote, quoted or indirectly quoted someone who said :

for(int i=list.size();i>=0;i--) {
 Object obj = list.get(i);
   if(needsToBeRemoved(i)) {
     list.remove(i);
   }
 }


Your technique would skip processing the element after any removed
element.


That's what I thought first time I read it, but (s)he is using the rathe=
r =

devious trick of iterating over the collection backwards to avoid that =

problem. The main problem with this approach, as I mentioned the other =
 =

day, is that performance sucks if the list is long and does not support =
 =

random access (i.e. LinkedList).

here is the new entry in the Java Cheat Sheet

// I T E R A T O R - R E M O V E:
// efficiently removing elements from a List
// (ArrayList, LinkedList etc .
// or Collection.
// You can't remove elements with a for:each.
// This works faster than a get/remove.
// This approach avoids the effects of the List
// renumbering as you go which can cause you to
// inadvertently skip elements or run off the end.
for ( Iterator<Item> iter = Items.iterator(); iter.hasNext(); )
   {
   Item item = iter.next();
   if ( item.isUnwanted() )
       {
       // remove from underlying Collection
       iter.remove();
       }
    }


For purely aesthetic reasons, I prefer the while loop variant (although =
 =

this does slightly widen the scope of the iterator reference).

Dan.

-- =

Daniel Dyer
http//www.uncommons.org

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