race-condition

问题 Consider the following code: private static BackgroundWorker bg = new BackgroundWorker(); static void Main(string[] args) { bg.DoWork += bg_DoWork; bg.ProgressChanged += bg_ProgressChanged; bg.WorkerReportsProgress = true; bg.RunWorkerAsync(); Thread.Sleep(10000); } static void bg_ProgressChanged(object sender, ProgressChangedEventArgs e) { Console.WriteLine(e.ProgressPercentage); Thread.Sleep(100); Console.WriteLine(e.ProgressPercentage); } static void bg_DoWork(object sender,

问题 Consider the following code: private static BackgroundWorker bg = new BackgroundWorker(); static void Main(string[] args) { bg.DoWork += bg_DoWork; bg.ProgressChanged += bg_ProgressChanged; bg.WorkerReportsProgress = true; bg.RunWorkerAsync(); Thread.Sleep(10000); } static void bg_ProgressChanged(object sender, ProgressChangedEventArgs e) { Console.WriteLine(e.ProgressPercentage); Thread.Sleep(100); Console.WriteLine(e.ProgressPercentage); } static void bg_DoWork(object sender,

问题 My question refers specifically to why it was designed that way, due to the unnecessary performance implication. When thread T1 has this code: cv.acquire() cv.wait() cv.release() and thread T2 has this code: cv.acquire() cv.notify() # requires that lock be held cv.release() what happens is that T1 waits and releases the lock, then T2 acquires it, notifies cv which wakes up T1. Now, there is a race-condition between T2's release and T1's reacquiring after returning from wait() . If T1 tries to

问题 My question refers specifically to why it was designed that way, due to the unnecessary performance implication. When thread T1 has this code: cv.acquire() cv.wait() cv.release() and thread T2 has this code: cv.acquire() cv.notify() # requires that lock be held cv.release() what happens is that T1 waits and releases the lock, then T2 acquires it, notifies cv which wakes up T1. Now, there is a race-condition between T2's release and T1's reacquiring after returning from wait() . If T1 tries to

来源： https://stackoverflow.com/questions/54058007/how-to-programatically-prove-this-code-has-a-race-condition

来源： https://stackoverflow.com/questions/64108775/mutex-on-boostasiowrite-not-works

来源： https://stackoverflow.com/questions/64108775/mutex-on-boostasiowrite-not-works

来源： https://stackoverflow.com/questions/64108775/mutex-on-boostasiowrite-not-works

来源： https://stackoverflow.com/questions/56384817/why-wait-with-predicate-solves-the-lost-wakeup-for-condition-variable

来源： https://stackoverflow.com/questions/56384817/why-wait-with-predicate-solves-the-lost-wakeup-for-condition-variable