如何在迭代時從“ ArrayList”中刪除元素時如何避免“ ConcurrentModificationException”？ [重複]

本文翻譯自：How to avoid “ConcurrentModificationException” while removing elements from `ArrayList` while iterating it? [duplicate]

This question already has an answer here: 這個問題已經在這裏有了答案：

• Iterating through a Collection, avoiding ConcurrentModificationException when removing objects in a loop 24 answers 遍歷Collection，在循環中刪除對象時避免ConcurrentModificationException 24答案

I'm trying to remove some elements from an ArrayList while iterating it like this: 我正在嘗試從ArrayList刪除一些元素，同時進行如下迭代：

for (String str : myArrayList) {
    if (someCondition) {
        myArrayList.remove(str);
    }
}

Of course, I get a ConcurrentModificationException when trying to remove items from the list at the same time when iterating myArrayList . 當然，當嘗試在迭代myArrayList的同時從列表中刪除項目時，會收到ConcurrentModificationException 。 Is there some simple solution to solve this problem? 有解決此問題的簡單方法嗎？

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#1樓

參考：https://stackoom.com/question/1FPLD/如何在迭代時從-ArrayList-中刪除元素時如何避免-ConcurrentModificationException-重複

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#2樓

If you want to modify your List during traversal, then you need to use the Iterator . 如果要在遍歷期間修改列表，則需要使用Iterator 。 And then you can use iterator.remove() to remove the elements during traversal. 然後您可以使用iterator.remove()在遍歷期間刪除元素。

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#3樓

Use an Iterator and call remove() : 使用Iterator並調用remove() ：

Iterator<String> iter = myArrayList.iterator();

while (iter.hasNext()) {
    String str = iter.next();

    if (someCondition)
        iter.remove();
}

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#4樓

You have to use the iterator's remove() method, which means no enhanced for loop: 您必須使用迭代器的remove（）方法，這意味着沒有增強的for循環：

for (final Iterator iterator = myArrayList.iterator(); iterator.hasNext(); ) {
    iterator.next();
    if (someCondition) {
        iterator.remove();
    }
}

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#5樓

As an alternative to everyone else's answers I've always done something like this: 作爲其他所有人的答案的替代方案，我總是這樣做：

List<String> toRemove = new ArrayList<String>();
for (String str : myArrayList) {
    if (someCondition) {
        toRemove.add(str);
    }
}
myArrayList.removeAll(toRemove);

This will avoid you having to deal with the iterator directly, but requires another list. 這樣可以避免您必須直接處理迭代器，但需要另一個列表。 I've always preferred this route for whatever reason. 無論出於什麼原因，我總是喜歡這條路線。

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#6樓

List myArrayList  = Collections.synchronizedList(new ArrayList());

//add your elements  
 myArrayList.add();
 myArrayList.add();
 myArrayList.add();

synchronized(myArrayList) {
    Iterator i = myArrayList.iterator(); 
     while (i.hasNext()){
         Object  object = i.next();
     }
 }