【菜鸟记坑篇】Iterator遍历时，二次遍历或遍历完成后Iterator转List报空指针异常/或转换无值的情况

继上一篇文章后又报新问题，废话不多说上代码！

       Iterator it = null;
        //说明我们有要除去这些有的
        if (oldPO.size() > 0) {

          for (it = userListVO.iterator(); it.hasNext(); ) {
                OauthUserVO userVO = (OauthUserVO) it.next();
                for (int i = 0; i < oldPO.size(); i++) {
                    if (oldPO.get(i).getPhone().equals(userVO.getPhone())) {
                        returnFalsList.add(userVO);
                        it.remove();
                        continue;
                    }
                }
            }
        }
        userListVO = IteratorUtils.toList(it);//这里转换后并没有值空的只有数组[]
        //以下时各种转list的方式
         // userListVO = (List<OauthUserVO>) CollectionUtils.collect(myIterator, TransformerUtils.nopTransformer());
       // ArrayList<OauthUserVO> elements = Lists.newArrayList(myIterator);

查找了好多相关资料终于捋清楚了。迭代器类似与指针一样的东西。他只是模拟了指针的一些功能，贴源码；

	    public Iterator<E> iterator() {
	        return new Itr();
	    }

// ---------------------------------------------------我是分割线-------------------------------------------------------------------

    private class Itr implements Iterator<E> {
    /**
     * Index of element to be returned by subsequent call to next.
     */
    int cursor = 0;

    /**
     * Index of element returned by most recent call to next or
     * previous.  Reset to -1 if this element is deleted by a call
     * to remove.
     */
    int lastRet = -1;

    /**
     * The modCount value that the iterator believes that the backing
     * List should have.  If this expectation is violated, the iterator
     * has detected concurrent modification.
     */
    int expectedModCount = modCount;

    public boolean hasNext() {
        return cursor != size();
    }

    public E next() {
        checkForComodification();
        try {
            int i = cursor;
            E next = get(i);
            lastRet = i;
            cursor = i + 1;
            return next;
        } catch (IndexOutOfBoundsException e) {
            checkForComodification();
            throw new NoSuchElementException();
        }
    }

    public void remove() {
        if (lastRet < 0)
            throw new IllegalStateException();
        checkForComodification();

        try {
            AbstractList.this.remove(lastRet);
            if (lastRet < cursor)
                cursor--;
            lastRet = -1;
            expectedModCount = modCount;
        } catch (IndexOutOfBoundsException e) {
            throw new ConcurrentModificationException();
        }
    }

    final void checkForComodification() {
        if (modCount != expectedModCount)
            throw new ConcurrentModificationException();
    }
} 

通过源码可以看出

		int cursor;     //当前索引          
		int lastRet = -1;   //前一位索引 
		int expectedModCount = modCount; //Iterator 修改次数 = collection修改次数 
		hasNext() //返回 cursor != size()
		next() //获取cursor指向的对象,并lastRet=cursor++
		remove() //移除lastRet指向的对象,并cursor-- 
		checkForComodification() //判断集合的修改次数是否合法

当Collection调用iterator方法后，根据当前Collection对象,返回一个新的Iterator,
hasNext():
返回 cursor!=size()结果;
checkForComodification():
判断 modCount!=expectedModCount ,为true则抛出ConcurrentModificationException();
next()：
调用Collection.get(cursor)，返回cursor值指向的索引的元素，并lastRet=cursor,cursor++(这里就是第二次遍历无法进行的原因，
当Iterator遍历完成，cursor == Collection.size(),调用hasNext()时返回false)；
上述解释是摘抄另外一篇博客
我理解的上述意思就是当我们在使用next（）方法时指针会向下指一位当你遍历完成之后指针指向最后一位，这时你在调用hasnext（）方法时就会返回false ，而我们最初创建时迭代器指向的时上一位元素指向一个空值

终于找到了原因。好了原因找到了。只要想办法把指向重置回去就可以了。奈何java没有自带自能自己写重置。对于我来说这个有点难，如果有大神可以写出来希望能发我一份嘿嘿！！

不过话说回来我这里的问题已经解决因为我完全没必要再转一次

本来我一位删除的是Iterator里的数据没想到他删除的也是list里面的数据。终于结束了！
感谢下面博主
https://blog.csdn.net/weixin_40163047/article/details/79303751

标签：遍历,Iterator,int,菜鸟,lastRet,next,cursor,checkForComodification
来源： https://blog.csdn.net/AlanZhoujun/article/details/97933161