保持std :: list迭代器通过插入有效

注意：这不是我应该使用list还是deque的问题。这是一个关于迭代器在 insert（）面前的有效性的问题。

Note: This is not a question whether I should "use list or deque". It's a question about the validity of iterators in the face of insert().

这可能是一个简单的问题而且我太过密集而无法看到正确的方法。我正在实现（无论好坏）网络流量缓冲区作为 std :: list< char> buf ，我将我当前的读取位置保持为迭代器 readpos 。

This may be a simple question and I'm just too dense to see the right way to do this. I'm implementing (for better or worse) a network traffic buffer as a std::list<char> buf, and I'm maintaining my current read position as an iterator readpos.

当我添加数据时，我会执行类似

When I add data, I do something like

buf.insert(buf.end(), newdata.begin(), newdata.end());

我现在的问题是，如何保留 readpos 迭代器有效吗？如果它指向旧的 buf 的中间，那么它应该没问题（通过std :: list的迭代器保证），但通常我可能已经阅读并处理了所有数据和我有 readpos == buf.end（）。插入后，我希望 readpos 总是指向下一个未读的字符，如果插入的话应该是第一个插入的字符。

My question is now, how do I keep the readpos iterator valid? If it points to the middle of the old buf, then it should be fine (by the iterator guarantees for std::list), but typically I may have read and processed all data and I have readpos == buf.end(). After the insertion, I want readpos always to point to the next unread character, which in case of the insertion should be the first inserted one.

有什么建议吗？ （没有将缓冲区更改为 std :: deque< char> ，这似乎更适合任务，如下所示。）

Any suggestions? (Short of changing the buffer to a std::deque<char>, which appears to be much better suited to the task, as suggested below.)

更新：从GCC4.4的快速测试中我发现deque和list的行为与 readpos = buf.end相反（ ）：在最后插入后，readpos在列表中被破坏，但指向deque中的下一个元素。 这是标准保证吗？

Update: From a quick test with GCC4.4 I observe that deque and list behave differently with respect to readpos = buf.end(): After inserting at the end, readpos is broken in a list, but points to the next element in a deque. Is this a standard guarantee?

（根据 cplusplus ，任何deque :: insert（）无效所有迭代器。这没有用。可能使用计数器比跟踪迭代器更好在双端队列中的位置？）

(According to cplusplus, any deque::insert() invalidated all iterators. That's no good. Maybe using a counter is better than an iterator to track a position in a deque?)