java Java中的并发修改异常
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Concurrent Modification Exception in Java
提问by muneikh
I am getting the ConcurrentModificationException while executing this code. I am unable to figure out why it is happening?
我在执行此代码时收到 ConcurrentModificationException。我无法弄清楚为什么会这样？
private void verifyBookingIfAvailable(ArrayList<Integer> list, int id) {
Iterator<Integer> iterator = list.iterator();
while (iterator.hasNext()) {
int value = iterator.next();
if (value == id) {
int index = list.indexOf(id);
if (index != -1) {
list.remove(index);
}
}
}
}
Thanks in advance.
提前致谢。
回答by Rohit Jain
You are removing the element in the list using the listreference itself, which can throw ConcurrentModificationException. Note that, this might work sometimes, but not always, and is not guaranteed to work perfectly.
您正在使用list引用本身删除列表中的元素，这可能会抛出ConcurrentModificationException. 请注意，这有时可能会起作用，但并非总是如此，并且不能保证完美运行。
Also, even though you use Iteratorto iterate your list, you still shouldn't use list.remove, you should only use iterator.remove()to remove the elements, else it won't make any difference, whether you use iterators or enhanced for-loop.
此外，即使您用于Iterator迭代列表，您仍然不应该使用list.remove，您应该只用于iterator.remove()删除元素，否则无论您使用迭代器还是增强的 for 循环，它都不会产生任何区别。
So, use iterator.remove()to remove elements.
因此，用于iterator.remove()删除元素。
if (index != -1) {
iterator.remove(value);
}
See this post: - java-efficient-equivalent-to-removing-while-iterating-a-collectionfor more detailed explanation.
请参阅这篇文章： - java-efficient-equivalent-to-removing-while-iterating-a-collection以获得更详细的解释。
回答by Egor
Simply because you're trying to remove elements from the ArrayListwhile iterating over them. To overcome this issue, use the java.util.concurrent.CopyOnWriteArrayList. Hope this helps.
仅仅是因为您试图ArrayList在迭代它们时从中删除元素。要解决此问题，请使用java.util.concurrent.CopyOnWriteArrayList。希望这可以帮助。
回答by Denys Séguret
What happens is that the ArrayList iterator isn't designed to enable modification while you're iterating on it.
发生的情况是 ArrayList 迭代器并非旨在在您对其进行迭代时启用修改。
So, to avoid more serious bugs coming from incoherent data, it has a modification count which is updated when you remove an item and checked when you iterate :
所以，为了避免来自不连贯数据的更严重的错误，它有一个修改计数，当你删除一个项目时会更新，并在你迭代时检查：
From ArrayList.java :
从 ArrayList.java ：
411 public E remove(int index) {
412 rangeCheck(index);
413
414 modCount++;
415 E oldValue = elementData(index);
416
417 int numMoved = size - index - 1;
418 if (numMoved > 0)
419 System.arraycopy(elementData, index+1, elementData, index,
420 numMoved);
421 elementData[--size] = null; // Let gc do its work
422
423 return oldValue;
424 }
...
779
780 final void checkForComodification() {
781 if (modCount != expectedModCount)
782 throw new ConcurrentModificationException();
783 }
As specified in the javadoc :
如 javadoc 中所述：
The returned list iterator is fail-fast.
返回的列表迭代器是快速失败的。
To avoid this problem, use the iterator to remove the current element, not the list directly. The removemethod of the iterator ensures the iterator is kept coherent.
为了避免这个问题，使用迭代器移除当前元素，而不是直接移除列表。remove迭代器的方法确保迭代器保持一致。
回答by someone
Try this
试试这个
private void verifyBookingIfAvailable(ArrayList<Integer> list, int id) {
List<Integer> tempList =new ArrayList<Integer>();
tempList.addAll(list);
for(Integer value :tempList) {
if (value == 1) {
int index = tempList.indexOf(1);
if (index != -1) {
list.remove(index);
}
}
}
}
while iteration you are removing objects
在迭代时您正在删除对象
