"data ready" 标志的同步机制？

Question

考虑以下 C++ 伪代码：

// somewhere in common code, properly scoped
boost::mutex data_ready_lock;
bool data_ready;

// Thread 1:
void SomeThreadFunc() {
  // ... push data onto a shared data structure that is properly locked
  data_ready_lock.lock();
  data_ready = true;
  data_ready_lock.unlock();
}

// Thread 2:  (actually a function called from the main() thread)
// Returns the number of bytes written to output_data
size_t RequestData(uint8_t* const output_data) {
  data_ready_lock.lock();
  if (data_ready) {
    // reset the flag, so I don't read out the same data twice
    data_ready = false;
    data_ready_lock.unlock();
    // copy over data, etc.
    return kDataSize;
  } else {
    data_ready_lock.unlock();
    return 0;
  }
}

有没有更好的方法来完成这个？我在考虑条件变量，但我需要能够重置标志以确保对 RequestData() 的背靠背调用不会产生相同的数据。

一如既往，在此先感谢您的帮助。

Answer 1

我不知道您的最终目标是什么，但也许使用实际的线程安全队列会简化您的代码。这是一个：

http://www.boost.org/doc/libs/1_53_0/doc/html/boost/lockfree/queue.html

Answer 2

如果标志是您唯一关心的问题，那么您可以尝试使用 atomic。

// somewhere in common code, properly scoped
boost::atomic< bool > data_ready(false); // can be std::atomic and std::memory_order_* below

// Thread 1:
void SomeThreadFunc() {
  // ... push data onto a shared data structure that is properly locked
  data_ready.store(true, boost::memory_order_release);
}

// Thread 2:  (actually a function called from the main() thread)
// Returns the number of bytes written to output_data
size_t RequestData(uint8_t* const output_data) {
  if (data_ready.exchange(false, boost::memory_order_acquire)) {
    // copy over data, etc.
    return kDataSize;
  } else {
    return 0;
  }
}

但是，在实际代码中，您将在 'push data' 和 'copy over data' 段代码之间竞争，除非它们单独同步。

"data ready" 标志的同步机制？

Synchronization Mechanism For "data ready" Flag?

c++

multithreading

boost

locking