Iterating through a Collection, avoiding ConcurrentModificationException when removing objects in a loop

Question

We all know you can't do the following because of ConcurrentModificationException:

for (Object i : l) {
    if (condition(i)) {
        l.remove(i);
    }
}

But this apparently works sometimes, but not always. Here's some specific code:

public static void main(String[] args) {
    Collection<Integer> l = new ArrayList<>();

    for (int i = 0; i < 10; ++i) {
        l.add(4);
        l.add(5);
        l.add(6);
    }

    for (int i : l) {
        if (i == 5) {
            l.remove(i);
        }
    }

    System.out.println(l);
}

This, of course, results in:

Exception in thread "main" java.util.ConcurrentModificationException

Even though multiple threads aren't doing it. Anyway.

What's the best solution to this problem? How can I remove an item from the collection in a loop without throwing this exception?

I'm also using an arbitrary Collection here, not necessarily an ArrayList, so you can't rely on get.

Answer 1

Iterator.remove() is safe, you can use it like this:

List<String> list = new ArrayList<>();

// This is a clever way to create the iterator and call iterator.hasNext() like
// you would do in a while-loop. It would be the same as doing:
//     Iterator<String> iterator = list.iterator();
//     while (iterator.hasNext()) {
for (Iterator<String> iterator = list.iterator(); iterator.hasNext();) {
    String string = iterator.next();
    if (string.isEmpty()) {
        // Remove the current element from the iterator and the list.
        iterator.remove();
    }
}

Note that Iterator.remove() is the only safe way to modify a collection during iteration; the behavior is unspecified if the underlying collection is modified in any other way while the iteration is in progress.

Source: docs.oracle > The Collection Interface

---

And similarly, if you have a ListIterator and want to add items, you can use ListIterator#add, for the same reason you can use Iterator#remove — it's designed to allow it.

---

In your case you tried to remove from a list, but the same restriction applies if trying to put into a Map while iterating its content.

Answer 2

This works:

Iterator<Integer> iter = l.iterator();
while (iter.hasNext()) {
    if (iter.next() == 5) {
        iter.remove();
    }
}

I assumed that since a foreach loop is syntactic sugar for iterating, using an iterator wouldn't help... but it gives you this .remove() functionality.

Answer 3

With Java 8 you can use the new removeIf method. Applied to your example:

Collection<Integer> coll = new ArrayList<>();
//populate

coll.removeIf(i -> i == 5);

Answer 4

Since the question has been already answered i.e. the best way is to use the remove method of the iterator object, I would go into the specifics of the place where the error "java.util.ConcurrentModificationException" is thrown.

Every collection class has a private class which implements the Iterator interface and provides methods like next(), remove() and hasNext().

The code for next looks something like this...

public E next() {
    checkForComodification();
    try {
        E next = get(cursor);
        lastRet = cursor++;
        return next;
    } catch(IndexOutOfBoundsException e) {
        checkForComodification();
        throw new NoSuchElementException();
    }
}

Here the method checkForComodification is implemented as

final void checkForComodification() {
    if (modCount != expectedModCount)
        throw new ConcurrentModificationException();
}

So, as you can see, if you explicitly try to remove an element from the collection. It results in modCount getting different from expectedModCount, resulting in the exception ConcurrentModificationException.

Answer 5

You can either use the iterator directly like you mentioned, or else keep a second collection and add each item you want to remove to the new collection, then removeAll at the end. This allows you to keep using the type-safety of the for-each loop at the cost of increased memory use and cpu time (shouldn't be a huge problem unless you have really, really big lists or a really old computer)

public static void main(String[] args)
{
    Collection<Integer> l = new ArrayList<Integer>();
    Collection<Integer> itemsToRemove = new ArrayList<>();
    for (int i=0; i < 10; i++) {
        l.add(Integer.of(4));
        l.add(Integer.of(5));
        l.add(Integer.of(6));
    }
    for (Integer i : l)
    {
        if (i.intValue() == 5) {
            itemsToRemove.add(i);
        }
    }

    l.removeAll(itemsToRemove);
    System.out.println(l);
}

Answer 6

In such cases a common trick is (was?) to go backwards:

for(int i = l.size() - 1; i >= 0; i --) {
  if (l.get(i) == 5) {
    l.remove(i);
  }
}

That said, I'm more than happy that you have better ways in Java 8, e.g. removeIf or filter on streams.

Answer 7

Same answer as Claudius with a for loop:

for (Iterator<Object> it = objects.iterator(); it.hasNext();) {
    Object object = it.next();
    if (test) {
        it.remove();
    }
}

Answer 8

Make a copy of existing list and iterate over new copy.

for (String str : new ArrayList<String>(listOfStr))     
{
    listOfStr.remove(/* object reference or index */);
}

Answer 9

With Eclipse Collections, the method removeIf defined on MutableCollection will work:

MutableList<Integer> list = Lists.mutable.of(1, 2, 3, 4, 5);
list.removeIf(Predicates.lessThan(3));
Assert.assertEquals(Lists.mutable.of(3, 4, 5), list);

With Java 8 Lambda syntax this can be written as follows:

MutableList<Integer> list = Lists.mutable.of(1, 2, 3, 4, 5);
list.removeIf(Predicates.cast(integer -> integer < 3));
Assert.assertEquals(Lists.mutable.of(3, 4, 5), list);

The call to Predicates.cast() is necessary here because a default removeIf method was added on the java.util.Collection interface in Java 8.

Note: I am a committer for Eclipse Collections.

Answer 10

People are asserting one can't remove from a Collection being iterated by a foreach loop. I just wanted to point out that is technically incorrect and describe exactly (I know the OP's question is so advanced as to obviate knowing this) the code behind that assumption:

for (TouchableObj obj : untouchedSet) {  // <--- This is where ConcurrentModificationException strikes
    if (obj.isTouched()) {
        untouchedSet.remove(obj);
        touchedSt.add(obj);
        break;  // this is key to avoiding returning to the foreach
    }
}

It isn't that you can't remove from the iterated Colletion rather that you can't then continue iteration once you do. Hence the break in the code above.

Apologies if this answer is a somewhat specialist use-case and more suited to the original thread I arrived here from, that one is marked as a duplicate (despite this thread appearing more nuanced) of this and locked.

Answer 11

With a traditional for loop

ArrayList<String> myArray = new ArrayList<>();

for (int i = 0; i < myArray.size(); ) {
    String text = myArray.get(i);
    if (someCondition(text))
        myArray.remove(i);
    else
        i++;   
}

Answer 12

ConcurrentHashMap or ConcurrentLinkedQueue or ConcurrentSkipListMap may be another option, because they will never throw any ConcurrentModificationException, even if you remove or add item.

Answer 13

Another way is to use a copy of your arrayList just for iteration:

List<Object> l = ...
    
List<Object> iterationList = ImmutableList.copyOf(l);
    
for (Object curr : iterationList) {
    if (condition(curr)) {
        l.remove(curr);
    }
}

Answer 14

A ListIterator allows you to add or remove items in the list. Suppose you have a list of Car objects:

List<Car> cars = ArrayList<>();
// add cars here...

for (ListIterator<Car> carIterator = cars.listIterator();  carIterator.hasNext(); )
{
   if (<some-condition>)
   { 
      carIterator().remove()
   }
   else if (<some-other-condition>)
   { 
      carIterator().add(aNewCar);
   }
}

Answer 15

I have a suggestion for the problem above. No need of secondary list or any extra time. Please find an example which would do the same stuff but in a different way.

//"list" is ArrayList<Object>
//"state" is some boolean variable, which when set to true, Object will be removed from the list
int index = 0;
while(index < list.size()) {
    Object r = list.get(index);
    if( state ) {
        list.remove(index);
        index = 0;
        continue;
    }
    index += 1;
}

This would avoid the Concurrency Exception.

Answer 16

I know this question is too old to be about Java 8, but for those using Java 8 you can easily use removeIf():

Collection<Integer> l = new ArrayList<Integer>();

for (int i=0; i < 10; ++i) {
    l.add(new Integer(4));
    l.add(new Integer(5));
    l.add(new Integer(6));
}

l.removeIf(i -> i.intValue() == 5);

Answer 17

The Best way(Recommended) is use of java.util.Concurrent package . By using this package you can easily avoid this Exception . refer Modified Code

public static void main(String[] args) {
    Collection<Integer> l = new CopyOnWriteArrayList<Integer>();

    for (int i=0; i < 10; ++i) {
        l.add(new Integer(4));
        l.add(new Integer(5));
        l.add(new Integer(6));
    }

    for (Integer i : l) {
        if (i.intValue() == 5) {
            l.remove(i);
        }
    }

    System.out.println(l);
}

Answer 18

You can use a while loop.

Iterator<Map.Entry<String, String>> iterator = map.entrySet().iterator();
while(iterator.hasNext()){
    Map.Entry<String, String> entry = iterator.next();
    if(entry.getKey().equals("test")) {
        iterator.remove();
    } 
}

Answer 19

In case ArrayList:remove(int index)- if(index is last element's position) it avoids without System.arraycopy() and takes not time for this.

arraycopy time increases if(index decreases), by the way elements of list also decreases!

the best effective remove way is- removing its elements in descending order: while(list.size()>0)list.remove(list.size()-1);//takes O(1) while(list.size()>0)list.remove(0);//takes O(factorial(n))

//region prepare data
ArrayList<Integer> ints = new ArrayList<Integer>();
ArrayList<Integer> toRemove = new ArrayList<Integer>();
Random rdm = new Random();
long millis;
for (int i = 0; i < 100000; i++) {
    Integer integer = rdm.nextInt();
    ints.add(integer);
}
ArrayList<Integer> intsForIndex = new ArrayList<Integer>(ints);
ArrayList<Integer> intsDescIndex = new ArrayList<Integer>(ints);
ArrayList<Integer> intsIterator = new ArrayList<Integer>(ints);
//endregion

// region for index
millis = System.currentTimeMillis();
for (int i = 0; i < intsForIndex.size(); i++) 
   if (intsForIndex.get(i) % 2 == 0) intsForIndex.remove(i--);
System.out.println(System.currentTimeMillis() - millis);
// endregion

// region for index desc
millis = System.currentTimeMillis();
for (int i = intsDescIndex.size() - 1; i >= 0; i--) 
   if (intsDescIndex.get(i) % 2 == 0) intsDescIndex.remove(i);
System.out.println(System.currentTimeMillis() - millis);
//endregion

// region iterator
millis = System.currentTimeMillis();
for (Iterator<Integer> iterator = intsIterator.iterator(); iterator.hasNext(); )
    if (iterator.next() % 2 == 0) iterator.remove();
System.out.println(System.currentTimeMillis() - millis);
//endregion

• for index loop: 1090 msec
• for desc index: 519 msec---the best
• for iterator: 1043 msec

Answer 20

for (Integer i : l)
{
    if (i.intValue() == 5){
            itemsToRemove.add(i);
            break;
    }
}

The catch is the after removing the element from the list if you skip the internal iterator.next() call. it still works! Though I dont propose to write code like this it helps to understand the concept behind it :-)

Cheers!

Answer 21

Example of thread safe collection modification:

public class Example {
    private final List<String> queue = Collections.synchronizedList(new ArrayList<String>());

    public void removeFromQueue() {
        synchronized (queue) {
            Iterator<String> iterator = queue.iterator();
            String string = iterator.next();
            if (string.isEmpty()) {
                iterator.remove();
            }
        }
    }
}

Answer 22

I know this question assumes just a Collection, and not more specifically any List. But for those reading this question who are indeed working with a List reference, you can avoid ConcurrentModificationException with a while-loop (while modifying within it) instead if you want to avoid Iterator (either if you want to avoid it in general, or avoid it specifically to achieve a looping order different from start-to-end stopping at each element [which I believe is the only order Iterator itself can do]):

*Update: See comments below that clarify the analogous is also achievable with the traditional-for-loop.

final List<Integer> list = new ArrayList<>();
for(int i = 0; i < 10; ++i){
    list.add(i);
}

int i = 1;
while(i < list.size()){
    if(list.get(i) % 2 == 0){
        list.remove(i++);

    } else {
        i += 2;
    }
}

No ConcurrentModificationException from that code.

There we see looping not start at the beginning, and not stop at every element (which I believe Iterator itself can't do).

FWIW we also see get being called on list, which could not be done if its reference was just Collection (instead of the more specific List-type of Collection) - List interface includes get, but Collection interface does not. If not for that difference, then the list reference could instead be a Collection [and therefore technically this Answer would then be a direct Answer, instead of a tangential Answer].

FWIWW same code still works after modified to start at beginning at stop at every element (just like Iterator order):

final List<Integer> list = new ArrayList<>();
for(int i = 0; i < 10; ++i){
    list.add(i);
}

int i = 0;
while(i < list.size()){
    if(list.get(i) % 2 == 0){
        list.remove(i);

    } else {
        ++i;
    }
}

Answer 23

One solution could be to rotate the list and remove the first element to avoid the ConcurrentModificationException or IndexOutOfBoundsException

int n = list.size();
for(int j=0;j<n;j++){
    //you can also put a condition before remove
    list.remove(0);
    Collections.rotate(list, 1);
}
Collections.rotate(list, -1);

Answer 24

Try this one (removes all elements in the list that equal i):

for (Object i : l) {
    if (condition(i)) {
        l = (l.stream().filter((a) -> a != i)).collect(Collectors.toList());
    }
}

Answer 25

you can also use Recursion

Recursion in java is a process in which a method calls itself continuously. A method in java that calls itself is called recursive method.

Answer 26

Now, You can remove with the following code

l.removeIf(current -> current == 5);

Answer 27

Concurrent Modification Exception

1. Single thread

Iterator<String> iterator = list.iterator();
while (iterator.hasNext()) {
    String value = iter.next()
    if (value == "A") {
        //throws ConcurrentModificationException
        list.remove(it.next());
    }
}

Solution: iterator remove() method

Iterator<String> iterator = list.iterator();
while (iterator.hasNext()) {
    String value = iter.next()
    if (value == "A") {
        it.remove()
    }
}

2. Multi thread

• copy/convert and iterate over another one collection. For small collections
• synchronize^[About]
• thread safe collection^[About]

Answer 28

I ended up with this ConcurrentModificationException, while iterating the list using stream().map() method. However the for(:) did not throw the exception while iterating and modifying the the list.

Here is code snippet , if its of help to anyone: here I'm iterating on a ArrayList<BuildEntity> , and modifying it using the list.remove(obj)

 for(BuildEntity build : uniqueBuildEntities){
            if(build!=null){
                if(isBuildCrashedWithErrors(build)){
                    log.info("The following build crashed with errors ,  will not be persisted -> \n{}"
                            ,build.getBuildUrl());
                    uniqueBuildEntities.remove(build);
                    if (uniqueBuildEntities.isEmpty()) return  EMPTY_LIST;
                }
            }
        }
        if(uniqueBuildEntities.size()>0) {
            dbEntries.addAll(uniqueBuildEntities);
        }

Answer 29

this might not be the best way, but for most of the small cases this should acceptable:

"create a second empty-array and add only the ones you want to keep"

_{I don't remeber where I read this from... for justiness I will make this wiki in hope someone finds it or just to don't earn rep I don't deserve.}