Hi Manuel, How would a processing line be broken between the CAS and the release of the seat?
This will happen *only if* the S.O. preempt the thread on that point and never get back. This is what I mean with the "my entire system is broken" *Only a S.O. scheduling failure can produce that.* Otherwise is impossible that a thread do not terminate that processing line. Can you think another possibility? Em terça-feira, 14 de janeiro de 2020 06:55:14 UTC-2, Manuel Pöter escreveu: > > The lock-free property guarantees that at any time at least one thread is > making progress in a finite number of steps. Or to put this more generally: > a stalled thread must not cause all other threads to stall indefinitely. > The arguments about lock-freedom (or the lack thereof) are usually based > on the (somewhat artificial) assumption that any thread can fail at any > time - i.e., simply stop executing. If such a failed thread causes the > whole system to grind to a halt, then it is not lock-free. > > You wrote yourself that "If a thread dies at that point, my entire system > is broken...", so it is definitely not lock-free. > That being said, lock-freedom is a nice property, but by no means > indispensable for scalability. Several of Dmitry's algorithms are not > lock-free (like the bounded MPMC queue), which does not mean that they do > not scale. > > Alistarh et al. showed that lock-free algorithms are practically > wait-free. I suppose the same could be shown for several concurrent > algorithms that are not strictly lock-free. So it is not so much important > whether an algorithm is lock-free or not, but whether it works well in > practice for the use case it is designed for. > > > On Tuesday, 14 January 2020 03:16:23 UTC+1, bittnkr wrote: >> >> > >> Well, if producer is preempted, consumers are blocked from progressing. >> >> Sorry, but this isn't true. >> >> The consumers are preempted only in case of a empty queue. Which isn't a >> lock. >> >> Meaning that there is nothing to do. If you don't have active producers, >> three is nothing to consume. >> >> How can you see a lock there? >> >> Please >> >> Em seg, 13 de jan de 2020 04:35, Dmitry Vyukov <dvy...@gmail.com> >> escreveu: >> >>> +lock-free group again, please don't drop it >>> >>> On Mon, Jan 13, 2020 at 8:19 AM bittnkr <bit...@gmail.com> wrote: >>> > >>> > If a thread dies at that point, my entire system is broken... >>> >>> Which means it's not lock-free. >>> >>> > But It can preempts without any problem at near zero cost. >>> >>> Well, if producer is preempted, consumers are blocked from >>> progressing. This is 100% equivalent to a mutex. If a thread is >>> preempted while holding a mutex, it also does not result in any >>> correctness problems. >>> >>> >>> > Em seg, 13 de jan de 2020 03:55, Dmitry Vyukov <dvy...@gmail.com> >>> escreveu: >>> >> >>> >> On Sun, Jan 12, 2020 at 7:33 PM bittnkr <bit...@gmail.com> wrote: >>> >> > >>> >> > >>> >> > > This blocks consumes from progressing (consuming next produced >>> items) >>> >> > and is effectively a mutex. >>> >> > >>> >> > Suppose the thread A got a local copy of tail in t then is >>> preempted, >>> >> > >>> >> > another thread will get the same tail an get the seat normally. >>> >> > >>> >> > When the previous thread retakes the line, the CAS will fail, >>> because the seat was taken. >>> >> > >>> >> > Restarting the while without any kind of blocking. >>> >> > >>> >> > Where do you see a mutex here? >>> >> >>> >> I mean preemption between succeeding CAS and writing element /NULL to >>> the array. >>> >> >>> >> If a thread is terminated at that point, the whole queue is broken (no >>> >> termination-safety). >>> >> >>> >> >>> >> > Em dom, 12 de jan de 2020 04:49, Dmitry Vyukov <dvy...@gmail.com> >>> escreveu: >>> >> >> >>> >> >> On Sat, Jan 11, 2020 at 9:26 AM bittnkr <bit...@gmail.com> wrote: >>> >> >> > >>> >> >> > Good observations. Thank you. >>> >> >> > >>> >> >> > If the thread preempts on those points, the seat position will >>> be held on local variables h and t. >>> >> >> >>> >> >> This blocks consumes from progressing (consuming next produced >>> items) >>> >> >> and is effectively a mutex. This makes algorithm >>> non-termination-safe >>> >> >> and non-lock-free. >>> >> >> >>> >> >> >>> >> >> > After the thread line is restored, the CAS can fail, and the >>> loop will just restart in normal flow, without any locking. >>> >> >> > >>> >> >> > I updated the docs, I think is clearer now. >>> >> >> > >>> >> >> > Em sáb., 11 de jan. de 2020 às 05:38, Dmitry Vyukov < >>> dvy...@gmail.com> escreveu: >>> >> >> >> >>> >> >> >> On Sat, Jan 11, 2020 at 4:09 AM bittnkr <bit...@gmail.com> >>> wrote: >>> >> >> >> > >>> >> >> >> > Hello Dmitry and fellows from the group. >>> >> >> >> > >>> >> >> >> > If you look carefully, you will see that there is no >>> blocking, just simple CAS, even the buffer is a common buffer. >>> >> >> >> >>> >> >> >> But consider you look inside of a spin lock, or you take a >>> >> >> >> spin-lock-based algorithm and inline all spin lock code into the >>> >> >> >> algorithm. What you will see is exactly the same -- no >>> blocking, just >>> >> >> >> a CAS. However, it does not really change anything, it's still >>> >> >> >> blocking mutex-based algorithm. >>> >> >> >> One can also combine spin lock state with some data, e.g. a >>> particular >>> >> >> >> value of a data member means "locked" and blocks progress of >>> other >>> >> >> >> threads. It makes spin lock even less visible, but again does >>> not >>> >> >> >> change anything on conceptual level. >>> >> >> >> >>> >> >> >> If I am reading the algorithm correctly, if a thread is >>> preempted >>> >> >> >> between these 2 operations: >>> >> >> >> >>> >> >> >> } while ( (data[t & mask]) || (CompareExchange(tail, t+1, t) >>> != t) ) >>> >> >> >> data[t & mask] = item // now is safe to update the buffer >>> >> >> >> >>> >> >> >> It effectively locks a mutex on the queue and blocks progress >>> of all consumers. >>> >> >> >> >>> >> >> >> There is also a similar blocks on consumer side here: >>> >> >> >> >>> >> >> >> } while ( !(data[h & mask]) || CompareExchange(head, h+1, h) >>> != h ) >>> >> >> >> data[h] = 0 // release the seat >>> >> >> >> >>> >> >> >> >>> >> >> >> > The queue only sleeps if it is empty out of it is full. But >>> this is not locking. >>> >> >> >> > >>> >> >> >> > All protection is done by the two atomic variables head an >>> tail. >>> >> >> >> > >>> >> >> >> > The cost is constant near a single CAS. For any number of >>> threads. >>> >> >> >> > >>> >> >> >> > Take a look on this benchmarks. They speaks for themselves. >>> >> >> >> > >>> >> >> >> > >>> https://github.com/bittnkr/uniq/blob/master/README.md#benchmarks >>> >> >> >> > >>> >> >> >> > Besides I don't have a formal proof, we have a test with zero >>> checksum. >>> >> >> >> > >>> >> >> >> > This is the results I have for a producer/consumer test >>> pushing random numbers is the queue. >>> >> >> >> > >>> >> >> >> > Creating 4 producers & 4 consumers >>> >> >> >> > to flow 10.000.000 items trough the queue. >>> >> >> >> > >>> >> >> >> > Produced: 10.743.668.245.000.000 >>> >> >> >> > Consumed: 5.554.289.678.184.004 >>> >> >> >> > Produced: 10.743.668.245.000.000 >>> >> >> >> > Consumed: 15.217.833.969.059.643 >>> >> >> >> > Produced: 10.743.668.245.000.000 >>> >> >> >> > Consumed: 7.380.542.769.600.801 >>> >> >> >> > Produced: 10.743.668.245.000.000 >>> >> >> >> > Consumed: 14.822.006.563.155.552 >>> >> >> >> > >>> >> >> >> > Checksum: 0 (it must be zero) >>> >> >> >> > >>> >> >> >> > The producers increase total and the consumers decrease. The >>> result for 10M random numbers is zero. >>> >> >> >> > >>> >> >> >> > Thanks for the advices, I'll investigate about this tools. >>> >> >> >> > >>> >> >> >> > Em qui, 9 de jan de 2020 04:58, Dmitry Vyukov < >>> dvy...@gmail.com> escreveu: >>> >> >> >> >> >>> >> >> >> >> On Wed, Jan 8, 2020 at 9:29 PM bittnkr <bit...@gmail.com> >>> wrote: >>> >> >> >> >> > >>> >> >> >> >> > Dear Dmitry, >>> >> >> >> >> > >>> >> >> >> >> > I found a nice solution for the problem called 3 thread >>> consensus, considered impossible on the book The art of the multiprocessor >>> programming. I think that is a breakthrough. >>> >> >> >> >> > >>> >> >> >> >> > Debating on S.O. with someone about if the solution is >>> solid or no, If it is possible to occur data races, he referred relacy. >>> >> >> >> >> > >>> >> >> >> >> > So I'm writing you to ask your opinion about the solution. >>> >> >> >> >> > >>> >> >> >> >> > Can you take a little look on it? >>> >> >> >> >> >>> >> >> >> >> +lock-free mailing list >>> >> >> >> >> >>> >> >> >> >> Hi bittnkr, >>> >> >> >> >> >>> >> >> >> >> At first glance the algorithm at >>> https://github.com/bittnkr/uniq looks >>> >> >> >> >> blocking and non-linearizable to me. >>> >> >> >> >> Very similar in nature to: >>> >> >> >> >> >>> http://www.1024cores.net/home/lock-free-algorithms/queues/bounded-mpmc-queue >>> >> >> >> >> >>> >> >> >> >> -- >>> >> >> >> >> Dmitry Vyukov >>> >> >> >> >> >>> >> >> >> >> All about lockfree/waitfree algorithms, multicore, >>> scalability, >>> >> >> >> >> parallel computing and related topics: >>> >> >> >> >> http://www.1024cores.net >>> >> >> >> >>> >> >> >> >>> >> >> >> >>> >> >> >> -- >>> >> >> >> Dmitry Vyukov >>> >> >> >> >>> >> >> >> All about lockfree/waitfree algorithms, multicore, scalability, >>> >> >> >> parallel computing and related topics: >>> >> >> >> http://www.1024cores.net >>> >> >> >>> >> >> >>> >> >> >>> >> >> -- >>> >> >> Dmitry Vyukov >>> >> >> >>> >> >> All about lockfree/waitfree algorithms, multicore, scalability, >>> >> >> parallel computing and related topics: >>> >> >> http://www.1024cores.net >>> >> >>> >> >>> >> >>> >> -- >>> >> Dmitry Vyukov >>> >> >>> >> All about lockfree/waitfree algorithms, multicore, scalability, >>> >> parallel computing and related topics: >>> >> http://www.1024cores.net >>> >>> >>> >>> -- >>> Dmitry Vyukov >>> >>> All about lockfree/waitfree algorithms, multicore, scalability, >>> parallel computing and related topics: >>> http://www.1024cores.net >>> >> -- --- You received this message because you are subscribed to the Google Groups "Scalable Synchronization Algorithms" group. 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