真正的代码方式更复杂,但我认为我设法制作了一个mcve。C++线程安全和notify_all()
我试图做到以下几点:
- 有一些线程做的工作
- 把他们都进入暂停状态
- 唤醒他们第一次,等待它完成,再醒来第二个,等待它完成,唤醒了第三个。等等。
我使用的代码下面,它似乎工作
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notify_all()
调用(一个我标有(*))可能会导致以下情况:
- 所有线程都睡着了
- 所有线程都从唤醒主线程通过调用
notify_all()
- 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■找被唤醒,但他们有没有选中原子变量YET
- 主线程发出第二
notify_all()
,这都丢失了(因为线程都惊醒的是,他们并没有简单地检查了原子能还)
这会发生吗?我找不到notify_all()
的任何措辞,如果它的电话是缓冲或与实际检查条件变量的函数同步的顺序。
我没有仔细研究过你的代码,但是你描述的一般情况是可能的。条件变量不存储信号。只有那些在发送信号时等待条件变量的线程才能看到该信号。由信号唤醒的线程不再等待,即使它们还没有从'wait()'函数返回。 –