C++ 是否存在循环列表的标准实现？
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Does a standard implementation of a Circular List exist for C++?
提问by Runcible
I want to use a circular list.
我想使用循环列表。
Short of implementing my own (like this person did) what are my options?
如果没有实现我自己的（就像这个人所做的那样），我的选择是什么？
Specifically what I want to do is iterate over a list of objects. When my iterator reaches the end of the list, it should automatically return to the beginning. (Yes, I realize this could be dangerous.)
具体来说，我想做的是迭代对象列表。当我的迭代器到达列表的末尾时，它应该自动返回到开头。（是的，我意识到这可能很危险。）
See Vladimir's definition of a circular_iterator: "A circular_iterator will never be equal with CircularList::end(), thus you can always dereference this iterator."
参见 Vladimir 对 a 的定义circular_iterator：“Circular_iterator 永远不会与 CircularList::end() 相等，因此您始终可以取消引用此迭代器。”
回答by Naaff
There's no standard circular list.
没有标准的循环列表。
However, there is a circular bufferin Boost, which might be helpful.
但是，Boost 中有一个循环缓冲区，这可能会有所帮助。
If you don't need anything fancy, you might consider just using a vectorand accessing the elements with an index. You can just modyour index with the size of the vector to achieve much the same thing as a circular list.
如果您不需要任何花哨的东西，您可以考虑只使用 avector并使用索引访问元素。您可以仅mod使用向量大小的索引来实现与循环列表大致相同的功能。
回答by Captain Segfault
If you want something looking like an iterator you can roll your own, looking something like
如果你想要一个看起来像迭代器的东西，你可以自己滚动，看起来像
template <class baseIter>
class circularIterator {
private:
baseIter cur;
baseIter begin;
baseIter end;
public:
circularIterator(baseIter b, baseIter e, baseIter c=b)
:cur(i), begin(b), end(e) {}
baseIter & operator ++(void) {++cur; if(cur == end) {cur = begin;}}
};
(Other iterator operations left as exercise to reader).
（其他迭代器操作留给读者练习）。
回答by Mahmoud Khaled
list<int>::iterator circularNext(list<int> &l, list<int>::iterator &it)
{
return std::next(it) == l.end() ? l.begin() : std::next(it);
}
回答by fuzzyTew
In addition to @captain-segfault and @mahmoud-khaled's iterator-focused answers, you can also use std::listas a circular list by altering what you do to retrieve elements from it. Use spliceto move one end of the list to the other end as you process it.
除了@captain-segfault 和@mahmoud-khaled 以迭代器为中心的答案之外，您还可以std::list通过更改从列表中检索元素的操作来将其用作循环列表。在处理列表时，使用splice将列表的一端移动到另一端。
template <typename T>
T & circularFront(std::list<T> & l)
{
l.splice(l.end(), l, l.begin());
return l.back();
}
template <typename T>
T & circularBack(std::list<T> & l)
{
l.splice(l.begin(), l, l.rbegin());
return l.front();
}
