On Wed, May 31, 2017 at 4:36 PM, Johannes Berg <[email protected]> wrote: > Hi, > >> > #include <linux/kernel.h> >> > #include <linux/mutex.h> >> > #include <linux/workqueue.h> >> > #include <linux/module.h> >> > #include <linux/delay.h> >> > >> > DEFINE_MUTEX(mtx); >> > static struct workqueue_struct *wq; >> > static struct work_struct w1, w2; >> > >> > static void w1_wk(struct work_struct *w) >> > { >> > mutex_lock(&mtx); >> > msleep(100); >> > mutex_unlock(&mtx); >> > } >> > >> > static void w2_wk(struct work_struct *w) >> > { >> > } >> > >> > /* >> > * if not defined, then lockdep should warn only, >> >> I guess when DEADLOCK not defined, there is no >> work is queued nor executed, therefore, no lock >> dependence is recorded, and there is no warn >> either. >> >> > * if defined, the system will really deadlock. >> > */ >> > >> > //#define DEADLOCK >> > >> > static int init(void) >> > { >> > wq = create_singlethread_workqueue("test"); >> > if (!wq) >> > return -ENOMEM; >> > INIT_WORK(&w1, w1_wk); >> > INIT_WORK(&w2, w2_wk); >> > >> >> /* add lock dependence, the lockdep should warn */ >> queue_work(wq, &w1); >> queue_work(wq, &w2); >> flush_work(&w1); >> >> > #ifdef DEADLOCK >> > queue_work(wq, &w1); >> > queue_work(wq, &w2); >> > #endif >> > mutex_lock(&mtx); >> > flush_work(&w2); >> > mutex_unlock(&mtx); >> > >> > #ifndef DEADLOCK >> > queue_work(wq, &w1); >> > queue_work(wq, &w2); >> > #endif > > This was "ifndef", so it does in fact run here, just like you > suggested. It doesn't warn though. > > I don't think the order of queue/flush would matter, in fact, if you > insert it like you did, with the flush outside the mutex, no issue > exists (until the later flush) >
the @w2 is not queued before flush_work(&w2), it is expected that @w2 is not associated with @wq, and the dependence mtx -> wq will not be recorded. And it is expected no warning. > Also, even if DEADLOCK *is* defined, lockdep doesn't report anything. Uhhh..... I have no idea about it yet.
