Re: single-threaded wq lockdep is broken

Johannes Berg
2017-05-31T08:36:41Z

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)

johannes