我已经阅读了一篇关于从
this link中删除集合中的元素的文章
根据我的理解迭代器删除方法防止并发修改异常然后删除Collection.But的方法当我尝试运行下面的codde我无法得到concurrentmoficationexception
List dayList= new ArrayList();
dayList.add("Sunday");
dayList.add("Monday");
dayList.add("Tuesday");
dayList.add("Wednesday");
dayList.remove("Tuesday");
Iterator itr=dayList.iterator();
while(itr.hasNext())
{
Object testList=itr.next();
if(testList.equals("Monday"))
{
dayList.remove(testList);
}
}
System.out.println(dayList);
}
>根据javadoc,当我们尝试在iteartion期间进行任何修改时抛出ConcurrentModicationException.我正在使用集合remove方法,但仍然没有异常.但是如果我注释行dayList.remove(“Tuesday”);,则抛出异常.
任何人都可以解释这段代码中幕后发生的事情吗？
解决方法
如果我评论行dayList.remove(“星期二”);,抛出异常….
其实这不是问题.问题是仅针对中间值发生异常.
‘for each’循环的工作原理如下,
1.It gets the iterator.
2.Checks for hasNext().
public boolean hasNext()
{
return cursor != size(); // cursor is zero initially.
}
3.If true,gets the next element using next().
public E next()
{
checkForComodification();
try {
E next = get(cursor);
lastRet = cursor++;
return next;
} catch (IndexOutOfBoundsException e) {
checkForComodification();
throw new NoSuchElementException();
}
}
final void checkForComodification()
{
// Initially modCount = expectedModCount (our case 5)
if (modCount != expectedModCount)
throw new ConcurrentModificationException();
}
重复步骤2和3,直到hasNext()返回false.
如果我们从列表中删除一个元素,它的大小会减少并且modCount会增加.
如果我们在迭代时删除一个元素,则modCount！= expectedModCount得到满足并抛出ConcurrentModificationException.
但删除倒数第二个对象很奇怪.让我们看看它在你的情况下是如何工作的.
原来,
cursor = 0 size = 5 --> hasNext() succeeds and next() also succeeds without exception. cursor = 1 size = 5 --> hasNext() succeeds and next() also succeeds without exception. cursor = 2 size = 5 --> hasNext() succeeds and next() also succeeds without exception. cursor = 3 size = 5 --> hasNext() succeeds and next() also succeeds without exception.
在您删除“d”的情况下,大小会减少到4.
cursor = 4 size = 4 --> hasNext() does not succeed and next() is skipped.
在其他情况下,将抛出ConcurrentModificationException
as modCount！= expectedModCount.
在这种情况下,不会进行此检查.
如果在迭代时尝试打印元素,则只打印四个条目.跳过最后一个元素.
您的问题类似于this问题.
