On Fri, Dec 16, 2016 at 03:19:43PM +0100, Nicolai Hähnle wrote:
> Hi Peter and Chris,
>
> (trying to combine the handoff discussion here)
>
> On 06.12.2016 17:55, Peter Zijlstra wrote:
> >On Thu, Dec 01, 2016 at 03:06:48PM +0100, Nicolai Hähnle wrote:
> >>@@ -693,8 +748,12 @@ __mutex_lock_common(struct mutex *lock, long state,
> >>unsigned int subclass,
> >> * mutex_unlock() handing the lock off to us, do a trylock
> >> * before testing the error conditions to make sure we pick up
> >> * the handoff.
> >>+ *
> >>+ * For w/w locks, we always need to do this even if we're not
> >>+ * currently the first waiter, because we may have been the
> >>+ * first waiter during the unlock.
> >> */
> >>- if (__mutex_trylock(lock, first))
> >>+ if (__mutex_trylock(lock, use_ww_ctx || first))
> >> goto acquired;
> >
> >So I'm somewhat uncomfortable with this. The point is that with the
> >.handoff logic it is very easy to accidentally allow:
> >
> > mutex_lock(&a);
> > mutex_lock(&a);
> >
> >And I'm not sure this doesn't make that happen for ww_mutexes. We get to
> >this __mutex_trylock() without first having blocked.
>
> Okay, took me a while, but I see the problem. If we have:
>
> ww_mutex_lock(&a, NULL);
> ww_mutex_lock(&a, ctx);
>
> then it's possible that another currently waiting task sets the HANDOFF flag
> between those calls and we'll allow the second ww_mutex_lock to go through.
Its worse, __mutex_trylock() doesn't check if MUTEX_FLAG_HANDOFF is set,
if .handoff == true && __owner_task() == current, we 'acquire'.
And since 'use_ww_ctx' is unconditionally true for ww_mutex_lock(), the
sequence:
ww_mutex_lock(&a, ...);
ww_mutex_lock(&a, ...);
will 'work'.
