c++ - 如何确保在 notify_one 之前调用 wait_for

Question

条件变量的典型用法如下所示(参见下面的代码):http://en.cppreference.com/w/cpp/thread/condition_variable .

但是，似乎主线程可能会在工作线程调用 wait 之前调用 notify_one，这将导致死锁。我错了吗？如果不是，通常的解决方法是什么？

#include <iostream>
#include <string>
#include <thread>
#include <mutex>
#include <condition_variable>

std::mutex m;
std::condition_variable cv;
std::string data;
bool ready = false;
bool processed = false;

void worker_thread()
{
    // Wait until main() sends data
    std::unique_lock<std::mutex> lk(m);
    cv.wait(lk, []{return ready;});

    // after the wait, we own the lock.
    std::cout << "Worker thread is processing data\n";
    data += " after processing";

    // Send data back to main()
    processed = true;
    std::cout << "Worker thread signals data processing completed\n";

    // Manual unlocking is done before notifying, to avoid waking up
    // the waiting thread only to block again (see notify_one for details)
    lk.unlock();
    cv.notify_one();
}

int main()
{
    std::thread worker(worker_thread);

    data = "Example data";
    // send data to the worker thread
    {
        std::lock_guard<std::mutex> lk(m);
        ready = true;
        std::cout << "main() signals data ready for processing\n";
    }
    cv.notify_one();

    // wait for the worker
    {
        std::unique_lock<std::mutex> lk(m);
        cv.wait(lk, []{return processed;});
    }
    std::cout << "Back in main(), data = " << data << '\n';

    worker.join();
}

Answer 1

注意 definition of wait使用条件(您应该使用的唯一等待):

while (!pred()) {
    wait(lock);
}

如果 notify 已经触发，则意味着条件已经为真(在信号线程中的 notify_one 之前排序)。因此，当接收方获取互斥量并查看 pred() 时，它将为真并继续进行。