blocks|key|1289017|text|这是JVM的一个常见问题，如果您想在迭代集合时从集合中删除一个项，则需要使用迭代器|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1289018|例：|1289019|val+myCollection+=+mutableListOf(1,2,3,4)
val+iterator+=+myCollection.iterator()
while(iterator.hasNext()){
++++val+item+=+iterator.next()
++++if(item+==+3){
++++++++iterator.remove()
++++}
}|code-block|syntax|javascript|1289020|这将避免ConcurrentModificationExceptions|1289021|我希望这回答了你的问题，祝你今天愉快。|1289022|编辑：您可以找到另一个解释，即使是这里代码，问题也是一样的|style|BOLD|1289023|编辑n°2，leonardkraemer的作者，向您展示了一种更友好的方法。|1289024|entityMap|0|LINK|mutability|MUTABLE|url|https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/-mutable-iterator/index.html|1|https://stackoverflow.com/questions/223918/iterating-through-a-collection-avoiding-concurrentmodificationexception-when-re^0|12|3|0|0|0|0|0|0|0|2|H|2|1|0|0|5|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]|$1|D|3|E|5|6|7|18|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|19|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|1B|8|@]|9|@]|C|$]]|$1|O|3|P|5|6|7|1C|8|@$A|1D|B|1E|Q|R]]|9|@$A|1F|B|1G|1|1H]]|C|$]]|$1|S|3|T|5|6|7|1I|8|@$A|1J|B|1K|Q|R]]|9|@]|C|$]]|$1|U|3|-4|5|6|7|1L|8|@]|9|@]|C|$]]]|V|$W|$5|X|Y|Z|C|$10|11]]|12|$5|X|Y|Z|C|$10|13]]]]

That's a common problem with the JVM, if you want to remove an item from a collection while iterating through it, you need to use the <a href="https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/-mutable-iterator/index.html" rel="noreferrer">Iterators</a>

exemple:

<pre><code>val myCollection = mutableListOf(1,2,3,4)
val iterator = myCollection.iterator()
while(iterator.hasNext()){
 val item = iterator.next()
 if(item == 3){
 iterator.remove()
 }
}
</code></pre>

this will avoid ConcurrentModificationExceptions

I hope this answered your question, have a good day

Edit: you can find another explanation <a href="https://stackoverflow.com/questions/223918/iterating-through-a-collection-avoiding-concurrentmodificationexception-when-re">here</a>, even if it is Java code the problem is the same 
Edit n°2 the anwser of leonardkraemer show you a more kotlin-friendly way to do so

blocks|key|178956|text|在Kotlin，您可以使用removeIf{+predicate+}。这是使用Iterator的速记。全文声明：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|178957|listAccounts.removeIf{+it+==+account.id+}|code-block|syntax|javascript|178958|有关解释，请参见遍历集合，在循环中删除对象时避免使用ConcurrentModificationException|178959|Update：Kotlin-stdlib引入了removeAll+{+predicate+}，正如Aguragorn在他的回答中所指出的那样，它同样可以用于运行时环境中不存在的removeIf+(即Java1.6或Android级别24)。|BOLD|178960|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/util/Collection.html|1|https://stackoverflow.com/questions/223918/iterating-through-a-collection-avoiding-concurrentmodificationexception-when-re|2|https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/remove-all.html|3|https://stackoverflow.com/a/54210042/4265739^0|D|L|13|8|D|L|0|0|0|8|1D|1|0|0|6|N|N|2G|8|N|N|2|1P|2|3|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@$9|18|A|19|1|1A]]|E|$]]|$1|F|3|G|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1C|8|@]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|M|3|N|5|6|7|1G|8|@$9|1H|A|1I|B|O]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]]|D|@$9|1N|A|1O|1|1P]|$9|1Q|A|1R|1|1S]]|E|$]]|$1|P|3|-4|5|6|7|1T|8|@]|D|@]|E|$]]]|Q|$R|$5|S|T|U|E|$V|W]]|X|$5|S|T|U|E|$V|Y]]|Z|$5|S|T|U|E|$V|10]]|11|$5|S|T|U|E|$V|12]]]]

In Kotlin you can use <a href="https://docs.oracle.com/javase/8/docs/api/java/util/Collection.html" rel="noreferrer"><code>removeIf{ predicate }</code></a>. Which is a shorthand to using the <code>Iterator</code>. Full statement:

<pre><code>listAccounts.removeIf{ it == account.id }
</code></pre>

for the explanation see <a href="https://stackoverflow.com/questions/223918/iterating-through-a-collection-avoiding-concurrentmodificationexception-when-re">Iterating through a Collection, avoiding ConcurrentModificationException when removing objects in a loop</a>

Update: Kotlin-stdlib introduced <a href="https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/remove-all.html" rel="noreferrer"><code>removeAll { predicate }</code></a> which, as Aguragorn pointed out in his <a href="https://stackoverflow.com/a/54210042/4265739">answer</a>, does the same and can be used if <code>removeIf</code> is not present on your runtime environment (i.e. Java 1.6 or Android pre API level 24).

blocks|key|1289089|text|with(listAccounts.iterator())+{
++++forEach+{
++++++++if+(it.id+==+someObj.id)+{
++++++++++++//+do+some+stuff+with+it
++++++++++++oldSubscription+=+it
++++++++++++remove()
++++++++}
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1289090|与SeekDaSky相同的解，但更多的Kotlin'y|unstyled|1289091|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>with(listAccounts.iterator()) {
 forEach {
 if (it.id == someObj.id) {
 // do some stuff with it
 oldSubscription = it
 remove()
 }
 }
}
</code></pre>

Same solution as SeekDaSky but more Kotlin'y

blocks|key|178996|text|kotlin引入的实际上是removeAll+{+predicate+}。所以您的代码应该如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|178997|listAccounts.removeAll+{+it.id+==+idParamether+}|code-block|syntax|javascript|178998|请参阅：https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/remove-all.html|178999|注意:在用kotlin编写代码时，我更喜欢坚持使用kotlin+API，这样可以避免诸如“调用需要API级别24”之类的问题。|179000|entityMap|0|LINK|mutability|MUTABLE|url|https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/remove-all.html^0|D|N|0|0|4|27|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@$9|11|A|12|1|13]]|E|$]]|$1|M|3|N|5|6|7|14|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

It is actually <code>removeAll { predicate }</code> that kotlin stdlib introduced.
So your code should look like this:

<pre><code>listAccounts.removeAll { it.id == idParamether }
</code></pre>

see: <a href="https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/remove-all.html" rel="noreferrer">https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/remove-all.html</a>

Note: when coding in kotlin I prefer sticking to kotlin APIs, it avoids problems like "Call requires API level 24"

blocks|key|1416470|text|尝试使用ConcurrentLinkedQueue而不是list来避免此异常。正如在ConcurrentLinkedQueue.Java中提到的，它命令元素FIFO+(先进先出).So，它将避免在迭代列表时修改列表时出现的任何问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1416471|例如：|1416472|val+list+=+ConcurrentLinkedQueue<String>()
list.add("toto")
list.add("tata")
list.add("titi")
list.forEachIndexed+{+index,+it+->
++++if+(index+%25+2+==+0)+{
++++++++list.remove("tata")
++++++++System.out.println(it)
++++}
}|code-block|syntax|javascript|1416473|结果是：|1416474|托托
提提|blockquote|1416475|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|R|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|S|8|@]|9|@]|A|$]]|$1|K|3|L|5|M|7|T|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|U|8|@]|9|@]|A|$]]]|O|$]]

Try to use ConcurrentLinkedQueue instead of list to avoid this exception. 
As mentioned in ConcurrentLinkedQueue.Java, it orders elements FIFO (first-in-first-out).So it will avoid any problem with modifying a list while iterating it. 
For exemple :

<pre class="lang-cs prettyprint-override"><code>val list = ConcurrentLinkedQueue&lt;String&gt;()
list.add("toto")
list.add("tata")
list.add("titi")
list.forEachIndexed { index, it -&gt;
 if (index % 2 == 0) {
 list.remove("tata")
 System.out.println(it)
 }
}
</code></pre>

the out put is : 

<blockquote>
 toto 
 titi 
</blockquote>

blocks|key|1416672|text|我也遇到了这个问题，我简单地通过克隆起始列表来解决这个问题，所以我会检查这个问题，并在原来的列表中添加或删除元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1416673|这段代码给了我一个例外：|1416674|+for(account+in+listAccounts){
++++....
++++listAccounts.add(anotherAccount)
++++....

}|code-block|syntax|javascript|1416675|所以用这个代替吧：|1416676|val+listAccountCloned+=+listAccounts.toMutableList()
+for(account+in+listAccountCloned){
++++....
++++listAccounts.add(anotherAccount)
++++....

}|1416677|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I also had this problem and I solved it simply by cloning the starting list, so I go through that and add or remove elements in the original one.
This code gave me the exception:
<pre><code> for(account in listAccounts){
 ....
 listAccounts.add(anotherAccount)
 ....

}
</code></pre>
So just replace it with this:
<pre><code>val listAccountCloned = listAccounts.toMutableList()
 for(account in listAccountCloned){
 ....
 listAccounts.add(anotherAccount)
 ....

}

</code></pre>

blocks|key|1289173|text|在迭代列表之前，可以先复制一份列表。这是一个小列表的有效解决方案。例如，在迭代侦听器之前，这就是他们通常在Android上所做的事情。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1289174|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You can make a copy of your list before iterating through it. It's a valid solution for small lists. That's what they are typically doing on Android before iterating through listeners for instance.

blocks|key|1416567|text|我想补充一下赖安的回答。如果您想在迭代期间添加到列表中，而不仅仅是删除，则需要在集合中调用.listIterator()而不是.iterator()。然后您也将得到add方法，这要归功于这个接口。完整代码：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1416568|with(listAccounts.listIterator())+{
++++forEach+{
++++++++if+(it.id+==+someObj.id)+{
++++++++++++//+do+some+stuff+with+it
++++++++++++oldSubscription+=+it
++++++++++++remove()
++++++++++++add(obj)
++++++++}
++++}
}|code-block|syntax|javascript|1416569|注意:我知道OP只是想删除，但是标题比较笼统，如果您在这种情况下搜索添加，也会发现这个问题。|1416570|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/50032000/how-to-avoid-concurrentmodificationexception-kotlin/53435698#53435698|1|https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/-mutable-list-iterator/^0|19|F|1R|B|2A|3|6|5|0|2L|4|1|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]|$9|16|A|17|1|18]]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

I'd like to supplement <a href="https://stackoverflow.com/questions/50032000/how-to-avoid-concurrentmodificationexception-kotlin/53435698#53435698">Ryan's answer</a>. If you'd like to add to your list during iteration, not just remove, you'd need to call <code>.listIterator()</code> on your collection instead of <code>.iterator()</code>. Then you'll have the <code>add</code> method too thanks to <a href="https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/-mutable-list-iterator/" rel="nofollow noreferrer">this interface</a>.
Complete code:
<pre><code>with(listAccounts.listIterator()) {
 forEach {
 if (it.id == someObj.id) {
 // do some stuff with it
 oldSubscription = it
 remove()
 add(obj)
 }
 }
}
</code></pre>
Note: I know the OP just wanted to remove, but the title is more general and this is also the question you find if you search for adding in this situation too.

blocks|key|1289268|text|解决此问题的最简单方法是在删除后添加“断开”。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1289269|for+(i+in+accounts)+{
++++++++++++++++++++if+(nick+==+i.nick)+{
++++++++++++++++++++++++print("Enter+PIN:+")
++++++++++++++++++++++++var+pin+=+scan.nextInt()
++++++++++++++++++++++++if+(pin+==+i.pin)+{
++++++++++++++++++++++++++++accounts.remove(i)
++++++++++++++++++++++++++++println("Account+has+been+deleted+successfully!")
++++++++++++++++++++++++++++break
++++++++++++++++++++++++}
++++++++++++++++++++}
++++++++++++++++}|code-block|syntax|javascript|1289270|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

the simplest way to solve this issue would be just adding &quot;break&quot; after the remove
<pre><code>for (i in accounts) {
 if (nick == i.nick) {
 print(&quot;Enter PIN: &quot;)
 var pin = scan.nextInt()
 if (pin == i.pin) {
 accounts.remove(i)
 println(&quot;Account has been deleted successfully!&quot;)
 break
 }
 }
 }
</code></pre>

blocks|key|1289303|text|作为kotlin扩展的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1289304|inline+fun+<T>+List<T>.forEachIterable(block:+(T)+->+Unit)+{
+++with(iterator())+{
++++++while+(hasNext())+{
+++++++++block(next())
++++++}
++}
}|code-block|syntax|javascript|1289305|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

solution as kotlin extension:
<pre><code>inline fun &lt;T&gt; List&lt;T&gt;.forEachIterable(block: (T) -&gt; Unit) {
 with(iterator()) {
 while (hasNext()) {
 block(next())
 }
 }
}
</code></pre>

blocks|key|179277|text|当另一个线程也修改集合时，迭代器并不总是有用的。我尝试过许多方法，但后来我意识到手动遍历集合更安全(向后移除)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|179278|for+(i+in+listAccounts.size-1+downTo+0)+{
++++myList.getOrNull(i)?.apply+{
+++++++if+(id+==+idParameter)
++++++++++listAccounts.remove(this)
++++}
}|code-block|syntax|javascript|179279|entityMap^0|D|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Iterators are not always helpful when another thread also modifies the collection. I had tried many ways but then I realized traversing manually the collection is much safer (backward for remove):
<pre><code>for (i in listAccounts.size-1 downTo 0) {
 myList.getOrNull(i)?.apply {
 if (id == idParameter)
 listAccounts.remove(this)
 }
}
</code></pre>

I have a list of accounts, and when i make the long click, I want to remove the item from the arraylist. I'm trying to remove it from a alertdialog, but i'm getting the ConcurrentModificationException. This is where is crashing:

<pre class="lang-kotlin prettyprint-override"><code>listAccounts.forEachIndexed { index, account -&gt;
 if (idParamether == account.id) {
 listAccounts.remove(account)
 }
}
</code></pre>

how to avoid ConcurrentModificationException kotlin

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个帐户列表，当我进行长时间单击时，我想从列列表中删除该项目。我试图从一个警报对话框中删除它，但是我得到了ConcurrentModificationException。这就是崩溃的地方：listAccounts.forEachIndexed { index, account ->    if (idParamether == account.id) {        listAccounts

问如何避免ConcurrentModificationException kotlin
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