c++ - "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实现这一点？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，提前感谢您的帮助。

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