Reputation: 4020
I am working on an application which has some legacy code. Here, there is a linkedlist and the code iterates that linklist using an iterator in a while loop.
LinkedList ll = grammarSection.getSectionsAsLinkList();
Iterator iter = ll.iterator();
int i=0;
while (iter.hasNext()) {
1. GrammarSection agrammarSection = (GrammarSection) iter.next();
2. grammarLineWithMatches = m_grammarLineMatcher.getMatch(agrammarSection, p_line);
3. if (grammarLineWithMatches != null) { //condition a
4. if (getPeek(ll)!=agrammarSection)
5. ll.addFirst(ll.remove(i)); //changing the linkedlist Line5
return grammarLineWithMatches;
}
i++;
}
In the while loop, if condition a is true, then the linkedlist is modified as in line5. However, in this case, the next method on line1 throws a ConcurrentModificationException. How to add and delete the linkedlist without getting any ConcurrentModificationException
Upvotes: 4
Views: 2068
Reputation: 14558
Iterators are fail-fast iterators so it may throw ConcurrentModificationException in your case.
The simplest solution would be;
instead of deleting items from list while iterating, add the ones that are going to be deleted to a new list and after the loop, you can remove all those using the removeAll method of list.
or vice versa, keep the ones you need to retain and assign the new list to the old one
Upvotes: 1
Reputation: 2348
The short answer is: you can't.
The JDK's List
implementations are designed to be modified by the iterator, to preserve order of iteration (there's no way to tell whether an arbitrary list change will do this, so the iterator assumes the worst).
The solution, in your case, is to create a new LinkedList
. As you iterate through, either add the iterated elements to the end or beginning of the new list. Then throw away the old.
Upvotes: 1
Reputation: 597422
You can't change the collection you are currently iterating. You can:
list.size()
. But with LinkedList
this is not efficient.If it was only about removal, you can use iter.remove()
, but you also have addFirst(..)
Upvotes: 3