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是否允许std :: map在只读操作之后重新平衡（如Splay树）

[英]Is std::map allowed to re-balance after read-only operations (like a Splay tree)

原文 2011-08-29 03:56:22 8 2 c++/ c++11/ splay-tree

Some binary tree data structures (such as Splay trees) will re-balance on reads to move recently accessed items toward the root, such that the subsequent look-up time may be reduced. 一些二叉树数据结构（例如Splay树）将在读取上重新平衡以将最近访问的项目移动到根，从而可以减少随后的查找时间。

Are the standard containers ( std::map , std::set ) allowed to do this? 标准容器（ std::map ， std::set ）是否允许这样做？

At least one concern is thread safety. 至少有一个问题是线程安全。 Previously, I'd thought that as long as you were only doing read-only operations on standard containers, it was safe to do this from multiple threads without introducing mutexes/locks etc. Maybe I need to re-think this? 以前，我认为只要你只对标准容器进行只读操作，就可以安全地从多个线程执行此操作而不引入互斥锁/锁等等。也许我需要重新考虑一下？

I know that typically red-black trees are used for the standard tree containers, and that these data structures aren't usually modified on reads. 我知道通常使用红黑树作为标准树容器，并且这些数据结构通常不会在读取时修改。 But would a hypothetical implementation that did modify be conforming? 但是，修改过的假设实现是否符合要求？

My c++-standards-foo needs improvement, but I'm not sure whether the current standard addresses thread-safety for containers. 我的c ++ - standards-foo需要改进，但我不确定当前的标准是否解决了容器的线程安全问题。 Is this different in c++0x ? 这在c++0x是不同的吗？

2 个解决方案

From the c++0x draft: 从c++0x草案：

§23.2.2/1: §23.2.2/ 1：

For purposes of avoiding data races (17.6.5.9), implementations shall consider the following functions to be const: begin, end, rbegin, rend, front, back, data, find, lower_bound, upper_bound, equal_range, at and, except in associative or unordered associative containers, operator[]. 为避免数据争用（17.6.5.9），实现应将以下函数视为const：begin，end，rbegin，rend，front，back，data，find，lower_bound，upper_bound，equal_range，at和，除了关联或无序的关联容器，operator []。

Note that c++03 does not say anything about multi-threading, but as you say, most implementations use RB-trees, which will not rebalance on a read operation. 请注意， c++03没有提及任何有关多线程的内容，但正如您所说，大多数实现使用RB树，它不会在读取操作上重新平衡。

Read functions on maps etc. are required to have a const function defined. 地图等上的读取函数需要定义const函数。 Hence you get the guarantee that the object hasn't changed. 因此，您可以保证对象没有更改。

This is true both for C++11 ( 23.4.4.1 ) as well as C++03 ( 23.3.1 ). 对于C ++ 11（23.4.4.1）和C ++ 03（23.3.1）都是如此。

23.2.2 of the new C++11 standard may be of special interest here: 新的C ++ 11标准的23.2.2在这里可能特别感兴趣：

For purposes of avoiding data races (17.6.5.9), implementations shall consider the following functions to be const: begin, end, rbegin, rend, front, back, data, find, lower_bound, upper_bound, equal_range, at and, except in associative or unordered associative containers, operator[]. 为避免数据争用（17.6.5.9），实现应将以下函数视为const：begin，end，rbegin，rend，front，back，data，find，lower_bound，upper_bound，equal_range，at和，除了关联或无序的关联容器，operator []。

Notwithstanding (17.6.5.9), implementations are required to avoid data races when the contents of the contained object in different elements in the same sequence, excepting vector<bool> , are modified concurrently. 尽管如此（17.6.5.9），除了vector<bool> ，同时修改同一序列中不同元素中包含对象的内容时，还需要实现以避免数据争用。