在notify_all()之后和wait()之前获取锁之间会发生什么? [英] what happen between after notify_all() and before wait() get the lock?

问题描述

我使用以下代码测试 std :: condition_variable :

class CondWait{
public:
    std::condition_variable cv;
    std::mutex mu;
    int i=0;
public:
    void mainTask(){
        std::unique_lock<std::mutex> lk(mu);
        cv.wait(lk);
        i++;
        std::cout<<"main task, "<<i<<std::endl;
    }
    void notifyTask(){

        std::unique_lock<std::mutex> lk(mu);
        i = 0;
        std::cout<<"notify task, "<<i<<std::endl;
        cv.notify_one();
        std::cout<<"notify task, sleep 5 sec"<<std::endl;
        std::this_thread::sleep_for(std::chrono::seconds(5));
    }
};
int main()
{
    CondWait condwait;
    std::thread t1(&CondWait::mainTask,&condwait);
    std::thread t2(&CondWait::notifyTask,&condwait);
    t1.join();
    t2.join();

    return 0;
}

有时，输出在下面，程序被阻止:

Sometimes, the output is below and the program is blocked:

notify task, 0
notify task, sleep 5 sec

有时，该程序会运行良好，即睡眠5秒钟后，它将输出主要任务1 ，完整的输出为:

Sometimes, the program will run well,i.e. after 5 seconds of sleep, it will output main task, 1, the complete output is:

notify task, 0
notify task, sleep 5 sec
main task, 1

在我看来，在 notifyTask 线程中，互斥锁仍在 notify_one 之后使用，所以 mainTask中的 wait 无法锁定互斥锁.但是我不知道接下来会发生什么，为什么这个例子会有模糊的表现.您能提供一些建议吗?非常感谢！

In my opinion, int the notifyTask thread, the mutex is still used after notify_one, so the wait in the mainTask cannot lock the mutex. But I don't know what will happen next, why the example will have an ambiguity performance. Could you please provide some advice? Thanks a lot!

推荐答案

由于同时启动两个线程，可能会在 cv.wait()在 mainTask 中，并且由于没有线程在等待，因此该功能没有任何作用.之后，执行 wait()并挂起，等待通知

Since you start both threads concurrently perhaps a situation occurs when cv.notify_one() is called before cv.wait() in mainTask and since no threads are waiting yet, the function does nothing. After that, wait() is executed and hangs waiting for a notification

这篇关于在notify_all()之后和wait()之前获取锁之间会发生什么?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！