C++ 对生产者和多个消费者来说是 std::queue 线程安全的
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is std::queue thread safe with producer and multiple consumers
提问by unikat
how can I make a queue thread safe? I need to push / pop / front / back and clear. is there something similar in boost?
如何使队列线程安全？我需要推/弹出/前/后并清除。boost中有类似的东西吗？
I have one producer and one or more consumer.
我有一个生产者和一个或多个消费者。
采纳答案by mindo
You must protect access to std::queue. If you are using boost protect it using boost::mutex. Now if you have multiple readers and one writer thread look at boost::shared_lock(for readers) and boost::unique_lock(for writer).
您必须保护对std::queue. 如果您正在使用 boost 保护它使用boost::mutex. 现在，如果您有多个读者和一个作者线程，请查看boost::shared_lock（对于读者）和boost::unique_lock（对于作者）。
However if you will encounter writer thread starvation look at boost::shared_mutex.
但是，如果您会遇到作家线程饥饿，请查看boost::shared_mutex.
回答by juanchopanza
std::queueis not thread safe if one or more threads are writing. And its interface is not conducive to a thread safe implementation, because it has separate methods such as pop(), size()and empty()which would have to be synchronized externally.
std::queue如果一个或多个线程正在写入，则不是线程安全的。并且它的接口不利于线程安全的实现，因为它具有单独的方法，例如pop(),size()并且empty()必须在外部进行同步。
A common approach*is to implement a queue type with a simpler interface, and use locking mechanisms internally to provide synchronization.
一种常见的方法*是用更简单的接口实现一个队列类型，并在内部使用锁定机制来提供同步。
* A search for "concurrent queue C++" should yield many results. I implemented a very simple toy one here, where the limitation was to use only standard C++. See also Anthony Williams' book C++ concurrency in action, as well as his blog.
* 搜索“并发队列 C++”应该会产生很多结果。我在这里实现了一个非常简单的玩具，其中的限制是只能使用标准的 C++。另请参阅 Anthony Williams 的书C++ concurrency in action以及他的博客。
回答by Mike Minaev
in boost 1.53 there is a lockfee queue http://www.boost.org/doc/libs/1_53_0/doc/html/boost/lockfree/queue.html, no mutex or smth like this.
在 boost 1.53 中有一个 lockfee 队列http://www.boost.org/doc/libs/1_53_0/doc/html/boost/lockfree/queue.html，没有像这样的互斥锁或 smth。
回答by MSalters
You have to protect it, e.g. with a std::mutex, on everyoperation. Boost would be an alternative if you don't have C++11 yet.
您必须std::mutex在每次操作时保护它，例如使用, 。如果您还没有 C++11，Boost 将是一个替代方案。
