Re: [PATCH RFC v7 00/23] DEPT(Dependency Tracker)

Boqun Feng
2023-01-20T03:27:19Z

On Thu, Jan 19, 2023 at 07:07:59PM -0800, Boqun Feng wrote:
> On Thu, Jan 19, 2023 at 06:23:49PM -0800, Boqun Feng wrote:
> > On Fri, Jan 20, 2023 at 10:51:45AM +0900, Byungchul Park wrote:
> > > Boqun wrote:
> > > > On Thu, Jan 19, 2023 at 01:33:58PM +0000, Matthew Wilcox wrote:
> > > > > On Thu, Jan 19, 2023 at 03:23:08PM +0900, Byungchul Park wrote:
> > > > > > Boqun wrote:
> > > > > > > *Looks like the DEPT dependency graph doesn't handle the
> > > > > > > fair/unfair readers as lockdep current does. Which bring the
> > > > > > > next question.
> > > > > >
> > > > > > No. DEPT works better for unfair read. It works based on wait/event. So
> > > > > > read_lock() is considered a potential wait waiting on write_unlock()
> > > > > > while write_lock() is considered a potential wait waiting on either
> > > > > > write_unlock() or read_unlock(). DEPT is working perfect for it.
> > > > > >
> > > > > > For fair read (maybe you meant queued read lock), I think the case
> > > > > > should be handled in the same way as normal lock. I might get it wrong.
> > > > > > Please let me know if I miss something.
> > > > >
> > > > > From the lockdep/DEPT point of view, the question is whether:
> > > > >
> > > > > read_lock(A)
> > > > > read_lock(A)
> > > > >
> > > > > can deadlock if a writer comes in between the two acquisitions and
> > > > > sleeps waiting on A to be released. A fair lock will block new
> > > > > readers when a writer is waiting, while an unfair lock will allow
> > > > > new readers even while a writer is waiting.
> > > > >
> > > >
> > > > To be more accurate, a fair reader will wait if there is a writer
> > > > waiting for other reader (fair or not) to unlock, and an unfair reader
> > > > won't.
> > >
> > > What a kind guys, both of you! Thanks.
> > >
> > > I asked to check if there are other subtle things than this. Fortunately,
> > > I already understand what you guys shared.
> > >
> > > > In kernel there are read/write locks that can have both fair and unfair
> > > > readers (e.g. queued rwlock). Regarding deadlocks,
> > > >
> > > > T0 T1 T2
> > > > -- -- --
> > > > fair_read_lock(A);
> > > > write_lock(B);
> > > > write_lock(A);
> > > > write_lock(B);
> > > > unfair_read_lock(A);
> > >
> > > With the DEPT's point of view (let me re-write the scenario):
> > >
> > > T0 T1 T2
> > > -- -- --
> > > fair_read_lock(A);
> > > write_lock(B);
> > > write_lock(A);
> > > write_lock(B);
> > > unfair_read_lock(A);
> > > write_unlock(B);
> > > read_unlock(A);
> > > read_unlock(A);
> > > write_unlock(B);
> > > write_unlock(A);
> > >
> > > T0: read_unlock(A) cannot happen if write_lock(B) is stuck by a B owner
> > > not doing either write_unlock(B) or read_unlock(B). In other words:
> > >
> > > 1. read_unlock(A) happening depends on write_unlock(B) happening.
> > > 2. read_unlock(A) happening depends on read_unlock(B) happening.
> > >
> > > T1: write_unlock(B) cannot happen if unfair_read_lock(A) is stuck by a A
> > > owner not doing write_unlock(A). In other words:
> > >
> > > 3. write_unlock(B) happening depends on write_unlock(A) happening.
> > >
> > > 1, 2 and 3 give the following dependencies:
> > >
> > > 1. read_unlock(A) -> write_unlock(B)
> > > 2. read_unlock(A) -> read_unlock(B)
> > > 3. write_unlock(B) -> write_unlock(A)
> > >
> > > There's no circular dependency so it's safe. DEPT doesn't report this.
> > >
> > > > the above is not a deadlock, since T1's unfair reader can "steal" the
> > > > lock. However the following is a deadlock:
> > > >
> > > > T0 T1 T2
> > > > -- -- --
> > > > unfair_read_lock(A);
> > > > write_lock(B);
> > > > write_lock(A);
> > > > write_lock(B);
> > > > fair_read_lock(A);
> > > >
> > > > , since T'1 fair reader will wait.
> > >
> > > With the DEPT's point of view (let me re-write the scenario):
> > >
> > > T0 T1 T2
> > > -- -- --
> > > unfair_read_lock(A);
> > > write_lock(B);
> > > write_lock(A);
> > > write_lock(B);
> > > fair_read_lock(A);
> > > write_unlock(B);
> > > read_unlock(A);
> > > read_unlock(A);
> > > write_unlock(B);
> > > write_unlock(A);
> > >
> > > T0: read_unlock(A) cannot happen if write_lock(B) is stuck by a B owner
> > > not doing either write_unlock(B) or read_unlock(B). In other words:
> > >
> > > 1. read_unlock(A) happening depends on write_unlock(B) happening.
> > > 2. read_unlock(A) happening depends on read_unlock(B) happening.
> > >
> > > T1: write_unlock(B) cannot happen if fair_read_lock(A) is stuck by a A
> > > owner not doing either write_unlock(A) or read_unlock(A). In other
> > > words:
> > >
> > > 3. write_unlock(B) happening depends on write_unlock(A) happening.
> > > 4. write_unlock(B) happening depends on read_unlock(A) happening.
> > >
> > > 1, 2, 3 and 4 give the following dependencies:
> > >
> > > 1. read_unlock(A) -> write_unlock(B)
> > > 2. read_unlock(A) -> read_unlock(B)
> > > 3. write_unlock(B) -> write_unlock(A)
> > > 4. write_unlock(B) -> read_unlock(A)
> > >
> > > With 1 and 4, there's a circular dependency so DEPT definitely report
> > > this as a problem.
> > >
> > > REMIND: DEPT focuses on waits and events.
> >
> > Do you have the test cases showing DEPT can detect this?
> >
>
> Just tried the following on your latest GitHub branch, I commented all
> but one deadlock case. Lockdep CAN detect it but DEPT CANNOT detect it.
> Feel free to double check.
>

In case anyone else want to try, let me explain a little bit how to
verify the behavior of the detectors. With the change, the only test
that runs is

dotest(queued_read_lock_hardirq_RE_Er, FAILURE, LOCKTYPE_RWLOCK);

"FAILURE" indicates selftests think lockdep should report a deadlock,
therefore for lockdep if all goes well, you will see:

[...] hardirq read-lock/lock-read: ok |

If you expect lockdep to print a full splat in the test (lockdep is
silent by default), you can add "debug_locks_verbose=2" in the kernel
command line, "2" mean RWLOCK testsuite.

Regards,
Boqun

> Regards,
> Boqun