C++ 线程安全和 notify_all()

Question

真正的代码要复杂得多，但我想我设法制作了一个 mcve。

我正在尝试执行以下操作:

1. 让一些线程正常工作
2. 让他们都进入暂停状态
3. 唤醒第一个，等待它完成，然后唤醒第二个，等待它完成，唤醒第三个......等等。

我使用的代码如下，它似乎有效

std::atomic_int which_thread_to_wake_up;
std::atomic_int threads_asleep;
threads_asleep.store(0);
std::atomic_bool ALL_THREADS_READY;
ALL_THREADS_READY.store(false);
int threads_num = .. // Number of threads
bool thread_has_finished = false;

std::mutex mtx;
std::condition_variable cv;

std::mutex mtx2;
std::condition_variable cv2;


auto threadFunction = [](int my_index) {

  // some heavy workload here..
  ....

  {
        std::unique_lock<std::mutex> lck(mtx);
        ++threads_asleep;
        cv.notify_all(); // Wake up any other thread that might be waiting
  } 

  std::unique_lock<std::mutex> lck(mtx);
  bool all_ready = ALL_THREADS_READY.load();
  size_t index = which_thread_to_wake_up.load();
  cv.wait(lck, [&]() {
      all_ready = ALL_THREADS_READY.load();
      index = which_thread_to_wake_up.load();
      return all_ready && my_index == index;
  });

  // This thread was awaken for work!
  .. do some more work that requires synchronization..

  std::unique_lock<std::mutex> lck2(mtx2);
  thread_has_finished = true;
  cv2.notify_one(); // Signal to the main thread that I'm done
};

// launch all the threads..
std::vector<std::thread> ALL_THREADS;
for (int i = 0; i < threads_num; ++i)
  ALL_THREADS.emplace_back(threadFunction, i);      


// Now the main thread needs to wait for ALL the threads to finish their first phase and go to sleep    

std::unique_lock<std::mutex> lck(mtx);
size_t how_many_threads_are_asleep = threads_asleep.load();
    while (how_many_threads_are_asleep < threads_num) {
      cv.wait(lck, [&]() {
        how_many_threads_are_asleep = threads_asleep.load();
        return how_many_threads_are_asleep == numThreads;
      });
    }

// At this point I'm sure ALL THREADS ARE ASLEEP!

// Wake them up one by one (there should only be ONE awake at any time before it finishes his computation)

for (int i = 0; i < threads_num; i++) 
{
  which_thread_to_wake_up.store(i);
  cv.notify_all(); // (*) Wake them all up to check if they're the chosen one
  std::unique_lock<std::mutex> lck2(mtx2);
  cv2.wait(lck, [&]() { return thread_has_finished; }); // Wait for the chosen one to finish
  thread_has_finished = false;
}

恐怕最后一次notify_all()调用(我用(*)标记的那个)可能会导致以下情况:

• 所有线程都在休眠
• 所有线程都通过调用notify_all()从主线程唤醒
• 具有正确索引的线程完成最后一次计算并释放锁
• 所有其他线程都已被唤醒，但它们尚未检查原子变量
• 主线程发出第二个 notify_all() 并且这会丢失(因为线程都被唤醒了，它们还没有简单地检查原子)

这会发生吗？如果它的调用以某种方式缓冲或与实际检查条件变量的函数同步的顺序，我找不到任何关于 notify_all() 的措辞。

Answer 1

根据 ( notify_all) 上的文档

notify_all只是继续一个线程的要求的一半。条件语句也必须为真。所以必须有一个交通警察来唤醒第一个，唤醒第二个，唤醒第三个。通知函数告诉线程检查该条件。

我的回答比代码更高级，但我希望对您有所帮助。