public void fun(){ List <Integer>lis = new ArrayList<Integer>(); lis.add(1); lis.add(2); for(Integer st:lis){ lis.remove(1); System.out.println(lis.size()); } } public static void main(String[] args) { test t = new test(); t.fun(); }
解决方法
[1,2]
然后在列表中启动一个迭代器:
[1,2] ^
您的for循环然后删除位置1的元素,它是数字2:
[1] ^
迭代器在下一个隐含的hasNext()调用中返回false,循环终止.
如果您向列表中添加更多元素,则会得到ConcurrentModificationException.那么隐含的next()将会抛出.
作为一个注释,从Javadoc的ArrayList从JCF:
Note that the fail-fast behavior of an iterator cannot be guaranteed as it is,generally speaking,impossible to make any hard guarantees in the presence of unsynchronized concurrent modification. fail-fast iterators throw
ConcurrentModificationException
on a best-effort basis. Therefore,it would be wrong to write a program that depended on this exception for its correctness: the fail-fast behavior of iterators should be used only to detect bugs.
这可能实际上是Oracle ArrayList迭代器实现中的一个错误; hasNext()不检查修改:
public boolean hasNext() { return cursor != size; }
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