我有一些 C 代码，其中有两个链表(比如 A 和 B)，A 被插入到 B 的特定位置，而 A 仍然有元素.
I have some C code, where in there are two linked lists(say A and B) and A is inserted at a particular position into B and A still has elements.
如何使用 C++ STL 有效地模拟相同的行为?如果我尝试拼接，它会使第二个为空.
How do I simulate the same behavior effectively using the C++ STL? If I try splice, it makes the second one empty.
谢谢，古库尔.
您需要复制元素.考虑这样的事情:
You need to copy the elements. Consider something like this:
std::copy(a.begin(), a.end(), std::inserter(b, b_iterator));
如果您希望两个列表共享相同的节点，std::list 根本不支持这一点(STL 容器始终具有独占所有权).您可以通过在列表中存储指针或使用 boost::ptr_list 来避免重复元素，它在内部存储指针但提供了更好的 API.
If you want the same nodes shared by two lists, this is simply not supported by std::list (STL containers always have exclusive ownership). You can avoid duplicating the elements by storing pointers in the list, or by using boost::ptr_list, which internally stores pointers but offers a nicer API.
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