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.
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.
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.
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);
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.
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);
}
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.
Same answer as Claudius with a for loop:
for (Iterator<Object> it = objects.iterator(); it.hasNext();) {
Object object = it.next();
if (test) {
it.remove();
}
}
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.
Make a copy of existing list and iterate over new copy.
for (String str : new ArrayList<String>(listOfStr))
{
listOfStr.remove(/* object reference or index */);
}
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.
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++;
}
ConcurrentHashMap or ConcurrentLinkedQueue or ConcurrentSkipListMap may be another option, because they will never throw any ConcurrentModificationException, even if you remove or add item.
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);
}
}
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);
}
}
Now, You can remove with the following code
l.removeIf(current -> current == 5);
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);
Java Concurrent Modification Exception
Single thread
Iterator<String> iterator = list.iterator();
while (iterator.hasNext()) {
String value = iter.next()
if (value == "A") {
list.remove(it.next()); //throws ConcurrentModificationException
}
}
Solution: iterator remove() method
Iterator<String> iterator = list.iterator();
while (iterator.hasNext()) {
String value = iter.next()
if (value == "A") {
it.remove()
}
}
Multi thread
copy/convert and iterate over another one collection. For small collections
synchronize[About]
thread safe collection[About]
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.
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!
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();
}
}
}
}
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;
}
}
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);
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());
}
}
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();
}
}
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);
}
If using HashMap, in newer versions of Java (8+) you can select each of 3 options:
public class UserProfileEntity {
private String Code;
private String mobileNumber;
private LocalDateTime inputDT;
// getters and setters here
}
HashMap<String, UserProfileEntity> upMap = new HashMap<>();
// remove by value
upMap.values().removeIf(value -> !value.getCode().contains("0005"));
// remove by key
upMap.keySet().removeIf(key -> key.contentEquals("testUser"));
// remove by entry / key + value
upMap.entrySet().removeIf(entry -> (entry.getKey().endsWith("admin") || entry.getValue().getInputDT().isBefore(LocalDateTime.now().minusMinutes(3)));
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);
}
Iterators are not always helpful when another thread also modifies the collection. I had tried many ways but then realized traversing the collection manually is much safer (backward for removal):
for (i in myList.size-1 downTo 0) {
myList.getOrNull(i)?.also {
if (it == 5)
myList.remove(it)
}
}
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
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.
This is the main file:
public static void main(String[] args) {
BTFunction bt=new BTFunction();
bt.insert(5);
bt.insert(15);
bt.insert(10);
bt.insert(7);
}
This is the class BTFunction:
public class BTFunction {
BTNode root=null;
void insert(int data){
root=BTinsertion(data,root);
}
BTNode BTinsertion(int data,BTNode n){
if(n==null){
n=new BTNode(data);
}
else{
if(n.right==null)
n.right=BTinsertion(data,n.right);
else
n.left=BTinsertion(data,n.left);
}
return n;
}
I understand the first 3 insertions(i.e. root, right and left) but when a new value(i.e. 7) is inserted, how does the function works.
According to me when insert(7) is implemented, it should just search for root.next and root.right which both are not null now. So, it should not do anything.
Can you explain the recursive process, specially when more values are added.
Here's what happens when bt.insert(7); is called:
n is not null, so we move to the else block. n.right is not null, so we move to the next else block. Here we call BTinsertion with n.left as the root.
Running BTinsertion a level down, data is still 7, and n is our node with value 10.
n is not null, but n.right is null, so the 7 node will be added to the right of node 10.
return n is called, so our node with value 10 is now a node with a root value and a value for n.right (7). Hope that makes sense.
Here is what I've done so far:
struct rep_list {
struct node *head;
struct node *tail;
}
typedef rep_list *list;
int length(const list lst) {
if (lst->head == NULL) {
return 0;
}
else {
lst->head = lst->head->next;
return 1 + length(lst);
}
}
This works, but the head of the list the function accepts as a parameter gets changed. I don't know how to fix that.
I'm not allowed to change the function definition so it should always accept a list variable.
Any ideas?
EDIT: I tried to do what Tyler S suggested in the comments but I encountered another problem. If I create a node* variable at the beginning, it should point to lst->head. But then every recursive call to the function changes the value back to lst->head and I cannot move forward.
You don't need a local node: just don't change the list head. Instead, pass the next pointer as the recursion head.
int length(const list lst) {
if (lst->head == NULL) {
return 0;
}
else {
return 1 + length(lst->head-next);
}
}
I see. Okay; this gets a bit clunky because of the chosen representation. You need a temporary variable to contain the remaining list. This iscludes changing the head.
int length(const list lst) {
if (lst->head == NULL) {
return 0;
}
else {
new_lst = new(list)
new_lst->head = lst->head->next;
var result = 1 + length(new_lst);
free(new_lst)
return result
}
}
At each recursion step, you create a new list object, point it to the 2nd element of the current list, and continue. Does this do the job for you?
Although this solution is clunky and I hate it, its the only way I can see to accomplish what you're asking without modifying the method signature. We create a temporary node * as member data of the class and modify it when we start.
struct rep_list {
struct node *head;
struct node *tail;
}
node *temp = NULL;
bool didSetHead = false;
typedef rep_list *list;
int length(const list lst) {
if ((didSetHead) && (lst->head != temp)) {
temp = lst->head;
didSetHead = false;
}
if (temp == NULL) {
didSetHead = true;
return 0;
}
else {
temp = temp->next;
return 1 + length(temp);
}
}
Please note, I haven't tested this code and you may have to play with a bit, but the idea will work.
Following is the code that I found online to find the minimum depth of a binary search tree:
public class Solution {
public int minDepth(TreeNode root) {
if(root == null){
return 0;
}
LinkedList<TreeNode> nodes = new LinkedList<TreeNode>();
LinkedList<Integer> counts = new LinkedList<Integer>();
nodes.add(root);
counts.add(1);
while(!nodes.isEmpty()){
TreeNode curr = nodes.remove();
int count = counts.remove();
if(curr.left != null){
nodes.add(curr.left);
counts.add(count+1);
}
if(curr.right != null){
nodes.add(curr.right);
counts.add(count+1);
}
if(curr.left == null && curr.right == null){
return count;
}
}
return 0;
}
}
What I do not understand is the extra return statement at the end- return 0. Why is this needed?
It's for the case where the root isn't null, but it's the only node in the tree (the root is at depth 0). That return statement is needed because if the tree is empty, then something must be returned. It returns 0, because the depth is 0.
Similar to ghostofrasputin, the return statement is there because if the while condition is not met, then there is still a value to return.
Now the more important question, why do we need the last return if the program will never reach that return statement? (which I believe is this case for this)
Even though you are able to tell that the return statement wont be used, the compiler is not sophisticated enough to determine that, so it requires a return statement just in case the while loop is exited.
It's similar to the following code
public boolean getTrueOrFalse() {
if(Math.random() < 1) {
return true;
}
return false;
}
Though we know that this will always return true because Math.random() is always less than 1, the compiler isn't able to figure that out and thus the return statement is required if the if-statement is not met.
I wrote countLeaf method in my binary tree class to count every leaves from root.
However, it gave me stack overflow error, but I couldn't figure what I did wrong.
this is the countLeaf class from my binaryTree
public int countLeaf(Node node){
if(root == null){return 0;} // this part work when I create null Tree
else if(root.left == null && root.right == null){
return 1; //this work when I create tree without left and right
}
else {
System.out.print(root.data); // check infinite loop
return countLeaf(root.left) + countLeaf(root.right);
}
}
And this is my main
public static void main (String[] args){
BinaryTree a = new BinaryTree("A?",
new BinaryTree("B?",
new BinaryTree("D"),
new BinaryTree("E")),
new BinaryTree("C?",
new BinaryTree("E"),
new BinaryTree("F")));
System.out.print(a);
int n = a.countLeaf(a.root);
}
and when I run it, it gave me
A?A?A?A?A?A?A?A?A?A?A?A?A? ... and stackoverflow error
why it keep repeating original root instead of follow left or right??
Replace public int countLeaf(Node node) with public int countLeaf(Node root).
I believe it will help.
Anyways, variable node is never used.