7 ms·

Thanks for the link. If you look at the implementation no thread is ever stopped from completing. If a producing thread was killed mid operation by another mal

by mjpt777 11y ago

Thanks for the link. If you look at the implementation no thread is ever stopped from completing. If a producing thread was killed mid operation by another malicious thread then the stream is broken. Other threads are never blocked. Other writers can add via a finite number of steps and the stream will back pressure when flow control kicks in. The consumer is never blocked - it simply never sees more messages. The algorithm meets the definition. Unless I missed it that paper does not not state that the algorithm must cope with a malicious thread to be wait-free. Try applying your thinking to this other respected MPSC wait-free queue. http://www.1024cores.net/home/lock-free-algorithms/queues/intrusive-mpsc-node-based-queue http://www.1024cores.net/home/lock-free-algorithms/queues/in...

saurik 11y ago
What if, instead of being killed, the thread runs arbitrarily slowly (at glacial speeds) between the two critical steps? Do other threads have to wait during that time, (Note: I haven't slept in days, not did I read the algorithm as my brain isn't working well enough to understand it, but I did seem to get the gist of this back/forth, and this seemed like an important detail.)
- mjpt777 11y ago
  Please read the algorithm in code. No threads ever wait. :-)
danbruc 11y ago
Two threads want to send a single block message. Writer one advances the tail pointer by 100 bytes and then dies. Writer two then advances the tail pointer by another 100 bytes, copies its message into the buffer and writes its header. The reader will see the tail pointer 200 bytes past the last processed message but no valid header and this looks exactly like a writer currently in the process of placing a 200 byte message - including the header - into the buffer. Looks to me like thread one blocked thread two indefinitely because the message of thread two is never added to the buffer. Yes, the bytes are in the buffer, but only physically. The reader can not see them, i.e. the write operation did not logically complete.
- mjpt777 11y ago
  If you look at the code the reader does not read the tail. It skips forward reading the length fields. It would never see the writer two message and it never blocks. It just thinks there are no new messages.
  - danbruc 11y ago
    Writer two is blocked, not the reader. Writer two wants to add a message to the buffer but never succeeds because writer one fails to make the message of writer two visible to the reader.
    - mjpt777 11y ago
      Writer two is not blocked, it is not waiting, it is not being obstructed. It is wait-free. It returns and can do whatever work it wishes. However the data structure would be corrupt if writer one never completed. This is corruption and not the definition of wait-free.
      - danbruc 11y ago
        No, it is not wait-free. The operation of writer two does not even succeed. Some bytes get written to the buffer but the message is never really added to the log because to call a write a success it should at least become visible to the reader. Writer two returns from the write operation without noticing that the operation failed. To make sure that the write succeeded writer two would have to wait and verify that its message actually becomes visible to the reader. You can even go a step further and make writer two block. Just assume that the next thing it does is waiting for a response from the receiver of its message. Now because writer one failed the message never reaches the receiver, the receiver never sends a response and the writer waits indefinitely for the response, i.e. is blocked. Now the failed writer one is really and the only reason writer two is blocked, even though only indirectly. And you can not really call it a corruption of the data structure because it is the exact same stated that occurs during normal operation. If writer one would resume execution after two weeks suddenly the corruption would be gone.
wflfof 11y ago
Regarding applying the wait/lock/obstruction freedom definition to Vyukov's MPSC implementation you reference. It's clear that the MPSC design is not wait-free, nor is it lock-free. Dmitry identifies this fact in his initial postings about that design: https://groups.google.com/forum/#!topic/lock-free/Vd9xuHrLggE https://groups.google.com/forum/#!topic/lock-free/Vd9xuHrLgg... """ Push function is blocking wrt consumer. I.e. if producer blocked in (*), then consumer is blocked too. """ Between the XCHG and the update of next. If the producing thread completes the XCHG but fails to update next, all threads will fail to make progress. "Progress" in this case being the producers transferring data to the consumer. Put another way, in this algorithm a single misbehaving thread can prevent all threads from making progress. Regarding the idea of "maliciousness", specifically if a participant thread is "killed mid-operation". That idea is the very definition of wait-freedom: that all participants complete the algorithm in a bounded number of their own steps, irrespective of the activities of other threads. The definitions of wait/lock/obstruction freedom are well specified. I suggest the first half of _The Art of Multiprocessor Programming_ (the revised edition!) by Herlihy and Shavit for a deep dive.
- mjpt777 11y ago
  I've read the _The Art of Multiprocessor Programming_ and am happy to read it again. Maybe I have misinterpreted. So if "killed mid-operation" must be supported then I don't see how many algorithms can be said to make progress. Take for example the Lamport SPSC queue[1], if the producer gets killed mid operation between steps 4 and 5 of the push operation. Then the data is in the queue but the consumer is blocked from ever seeing it with this line of reasoning. The Lamport SPSC queue is considered wait-free by the concurrency community I know. I base my reasoning on this. What if the producer takes a pause for a long time between steps 4 and 5 before continuing? However if to be wait-free an algorithm must allow all other threads to continue using the data structure, not just continue making progress in other ways by being non-blocking and completing in a finite number of steps, then I stand corrected. If wait-free must include coping with any thread being killed mid operation is there a term for being lock-free but also having all threads not block and complete in a finite number of steps for their interaction with the algorithm? [1] - http://arxiv.org/pdf/1012.1824.pdf http://arxiv.org/pdf/1012.1824.pdf

by mjpt777 11y ago

saurik 11y ago

What if, instead of being killed, the thread runs arbitrarily slowly (at glacial speeds) between the two critical steps? Do other threads have to wait during that time, (Note: I haven't slept in days, not did I read the algorithm as my brain isn't working well enough to understand it, but I did seem to get the gist of this back/forth, and this seemed like an important detail.)

mjpt777 11y ago
Please read the algorithm in code. No threads ever wait. :-)

danbruc 11y ago

Two threads want to send a single block message. Writer one advances the tail pointer by 100 bytes and then dies. Writer two then advances the tail pointer by another 100 bytes, copies its message into the buffer and writes its header. The reader will see the tail pointer 200 bytes past the last processed message but no valid header and this looks exactly like a writer currently in the process of placing a 200 byte message - including the header - into the buffer. Looks to me like thread one blocked thread two indefinitely because the message of thread two is never added to the buffer. Yes, the bytes are in the buffer, but only physically. The reader can not see them, i.e. the write operation did not logically complete.

mjpt777 11y ago
If you look at the code the reader does not read the tail. It skips forward reading the length fields. It would never see the writer two message and it never blocks. It just thinks there are no new messages.
- danbruc 11y ago
  Writer two is blocked, not the reader. Writer two wants to add a message to the buffer but never succeeds because writer one fails to make the message of writer two visible to the reader.
  - mjpt777 11y ago
    Writer two is not blocked, it is not waiting, it is not being obstructed. It is wait-free. It returns and can do whatever work it wishes. However the data structure would be corrupt if writer one never completed. This is corruption and not the definition of wait-free.
    - danbruc 11y ago
      No, it is not wait-free. The operation of writer two does not even succeed. Some bytes get written to the buffer but the message is never really added to the log because to call a write a success it should at least become visible to the reader. Writer two returns from the write operation without noticing that the operation failed. To make sure that the write succeeded writer two would have to wait and verify that its message actually becomes visible to the reader. You can even go a step further and make writer two block. Just assume that the next thing it does is waiting for a response from the receiver of its message. Now because writer one failed the message never reaches the receiver, the receiver never sends a response and the writer waits indefinitely for the response, i.e. is blocked. Now the failed writer one is really and the only reason writer two is blocked, even though only indirectly. And you can not really call it a corruption of the data structure because it is the exact same stated that occurs during normal operation. If writer one would resume execution after two weeks suddenly the corruption would be gone.

wflfof 11y ago

Regarding applying the wait/lock/obstruction freedom definition to Vyukov's MPSC implementation you reference. It's clear that the MPSC design is not wait-free, nor is it lock-free. Dmitry identifies this fact in his initial postings about that design: https://groups.google.com/forum/#!topic/lock-free/Vd9xuHrLggE https://groups.google.com/forum/#!topic/lock-free/Vd9xuHrLgg... """ Push function is blocking wrt consumer. I.e. if producer blocked in (*), then consumer is blocked too. """ Between the XCHG and the update of next. If the producing thread completes the XCHG but fails to update next, all threads will fail to make progress. "Progress" in this case being the producers transferring data to the consumer. Put another way, in this algorithm a single misbehaving thread can prevent all threads from making progress. Regarding the idea of "maliciousness", specifically if a participant thread is "killed mid-operation". That idea is the very definition of wait-freedom: that all participants complete the algorithm in a bounded number of their own steps, irrespective of the activities of other threads. The definitions of wait/lock/obstruction freedom are well specified. I suggest the first half of _The Art of Multiprocessor Programming_ (the revised edition!) by Herlihy and Shavit for a deep dive.

mjpt777 11y ago
I've read the _The Art of Multiprocessor Programming_ and am happy to read it again. Maybe I have misinterpreted. So if "killed mid-operation" must be supported then I don't see how many algorithms can be said to make progress. Take for example the Lamport SPSC queue[1], if the producer gets killed mid operation between steps 4 and 5 of the push operation. Then the data is in the queue but the consumer is blocked from ever seeing it with this line of reasoning. The Lamport SPSC queue is considered wait-free by the concurrency community I know. I base my reasoning on this. What if the producer takes a pause for a long time between steps 4 and 5 before continuing? However if to be wait-free an algorithm must allow all other threads to continue using the data structure, not just continue making progress in other ways by being non-blocking and completing in a finite number of steps, then I stand corrected. If wait-free must include coping with any thread being killed mid operation is there a term for being lock-free but also having all threads not block and complete in a finite number of steps for their interaction with the algorithm? [1] - http://arxiv.org/pdf/1012.1824.pdf http://arxiv.org/pdf/1012.1824.pdf