Talk:Producer–consumer problem

message: Am I the only one who's confused by the two contrasting implementations (inadequate vs semaphore-based)? They seem to be written by two different people. I think we all agree that the inadequate implementation is incorrect. However, I don't see how using two semaphores fillCount and emptyCount makes a difference. Rather what seems to fix the problem is the use of atomic down() and up() procedures. Why does the use of down() and up() in the "inadequate" implementation with a single semaphore not fix the problem with that? Houseofwealth (talk) 18:10, 8 April 2019 (UTC)
replies: text: I don't think the described deadlock is related to non-atomic increments and decrements at all; you'd need critical sections around whole pieces of code which handle the count. It's the testing of the values of the counter that's racy in the deadlock, not the changes to it. It is already noted that non-atomic increments and decrements (called "concurrent accesses to shared variables" there) would make it "unsatisfactory" even without the deadlock. That said, I don't think the author of the "good" implementation wanted to imply you can't write a solution with just a single semaphore, it's just that the code is shorter by using two. The difference in the implementations is explained by the fact that all the checking, sleeping and waking is moved into the implementation of the semaphore functions down() and up() rather than directly in the producer and consumer. There is a balance to find here: you could expand the examples to include more of the code, choose other methods to deal with the concurrency, show two inadequate solutions, one with a deadlock and the other with racy shared updates, etcetera, but does this improve the article or just make it larger? Maybe it should be noted that the checking, sleeping and waking is now handled in up() and down(), though. Digital Brains (talk) 10:46, 10 April 2019 (UTC)
thread_title: The good vs. the bad (implementations)

message: Any one else thinks that the code implementation in "The Producer Consumer Design Pattern created with C++ and the Boost Thread Library" is not a good implementation and does not merit a link at WikiPedia?  —Preceding unsigned comment added by 24.5.131.68 (talk) 04:41, 4 January 2010 (UTC)
thread_title: Untitled

message: It seems like there may be an error in the first semaphore solution with just one producer and consumer, although I'm not sure I understand how the buffer data is supposed to be structured.
What if the consumer does:
determine last slot in the buffer to read from
and the producer does:
determine next slot and write there
then consumer completes the remove of the OLD slot
isn't there now a "hole" in the buffer. I guess this depends on whether the buffer data is meant to be sequential and is only a problem unless the buffer knows whether the consumer has read from a specific spot.
Can you help me with this? Thank you.  —Preceding unsigned comment added by 132.239.217.4 (talk) 19:43, 19 August 2009 (UTC)
thread_title:  Question about first semaphore case 

message: The article defines the 'producer-consumer' as a producer and a consumer sharing a fixed-size buffer. If one would like to use a non-fixed-size buffer (such as a linked list), would the situation no longer be a 'producer-consumer' situation? That seems rather counter-intuitive to me.
My Operating Systems textbook (Stallings, just added the reference) defines producer-consumer as follows:
There are one or more producers generating some type of data (records, characters) and placing these in a buffer. There is a single consumer that is taking items out of the buffer one at a time. The system is to be constrained to prevent the overlap of buffer operations. That is, only one agent (producer or consumer) may access the buffer at any one time.
It then proceeds to give examples of implementations, first some with a hypothetical infinite buffer, then finally one adding the 'new and realistic restriction' of a finite buffer. --Raboof (talk) 18:06, 2 December 2009 (UTC)
replies: text: The infinite size buffer is a simple thought exercise to lead to the final solution. In any case, one can argue that "infinite" is still "fixed-size". If one is free to enlarge the buffer through linked lists or memory reallocation, there would be no problem for the producer, as it can always place new data in an enlarged buffer. Only a fixed size buffer introduces the complete producer-consumer problem. C xong (talk) 06:32, 26 May 2010 (UTC)
thread_title:  fixed-size buffer requirement? 

message: After numerous contacs about this issue, and because most IT persons are not aware that semaphores/monitors/mutexes are not needed for the producer-consumer, I moved this section to the main article.
Jvanehv
I am having some trouble understanding your addition. You don't use the notation used in the previous examples (Token instead of Item?), you introduce concepts without explaining what they are (volatile types, sched_yield). Furthermore, a reader that comes from the previous sections will identify all problems identified in the previous sections, i.e.: deadlock on reading the produceCount - consumeCount, multiple producer issue both on the producer and consumer. Apparently, these are magically solved by the sched_yield, and I think that your precondition is that there is only one procuder and one consumer thread. I would try to fix the section myself but what you are trying to say is very unclear.Dtheodorou (talk) 14:19, 30 March 2013 (UTC)
thread_title:  Without semaphores or monitors 

message: This sentences says to me that the code shown in the Using monitors section does have data races. I have no idea how to fix this, I am not even sure I understand the intentions behind the quoted sentence. Ali Baharev (talk) 08:42, 4 January 2013 (UTC)
replies: text: I wrote a new monitor in C++ implementation. It is very standard, that is there should be no data race in my implementation. But as always, I am sure to 99.x%, not to 100% AndreAdrian (talk) 14:28, 16 May 2022 (UTC)
thread_title:  Data race in the Using monitors section? 

message: I appreciate the inclusion of the C++11 example! Unfortunately, this example currently contains a data race that might result in a deadlock situation:
In principle, it might occur that some consumer thread only starts-up after that all producers  finished. Such consumer will forever wait for producers to register themselves, and cause the t.join() in main() to wait forever.  — Preceding unsigned comment added by Jvanehv (talk • contribs) 09:23, 2 April 2013 (UTC)
thread_title:  C++11 example 

message: I believe that there must be a memory barrier added to the examples between writing the item to the shared buffer and declaring to the other thread that it is readable.  It seems to me that, as written, the "without semaphores or monitors" must ensure that the sharedBuffer memory write has completed before the produceCount is incremented.  This would typically be done with a memory barrier between the write to the buffer and the increment, as well as a memory barrier between the read of the produceCount and the read of the sharedBuffer.  All that C guarantees is that the single-threaded observable behavior is correct, not that all threads will observe the same state at the same time.  Am I wrong? Skrapple (talk) 20:05, 16 May 2019 (UTC)
thread_title:  Memory Barriers Needed 

message: The original C.A.R. Hoare solution is faulty. There is an increment count statement in procedure append, but no decrement count statement in procedure remove. See C.A.R. Hoare; 1974; Monitors: An Operating System Structuring Concept, 4. Example: Bounded Buffer:
bounded buffer: monitor
  begin buffer:array 0..N-1 of portion;
    lastpointer: 0..N-1;
    count: 0..N;
    nonempty, nonfull: condition;
  procedure append(x: portion);
    begin if count = N then nonfull.wait;
      note 0 <= count < N;
      buffer[lastpointer] := x;
      lastpointer := lastpointer (+) 1;
      count := count + 1;
      nonempty.signal
    end append;
  procedure remove(result x: portion);
    begin if count = 0 then nonempty.wait;
      note 0 < count <= N;
      x := buffer[lastpointer (-) count];
      nonfull.signal
    end remove;
  count := 0; lastpointer := 0;
end bounded buffer;
The circled operations (+), (-) are taken modulo N. Question: Shall Wikipedia show the faulty original solution?  AndreAdrian (talk) 15:03, 6 January 2022 (UTC)
thread_title:  Display original but faulty solution? 

message: The Lamport solution for producer-consumer without semaphores or monitors is faulty, too. This is the published solution, see Lamport, Leslie; 1977; Proving the Correctness of Multiprocess Programs, The Producer/Consumer Example:
Producer:
  L:  if s - r mod k = b then goto L fi;
      put message in buffer;
      s := s + 1 mod k;
      goto L;
Consumer:
  L:  if s - r mod k = 0 then goto L fi;
      take message from buffer;
      r := r + 1 mod k;
      goto L;
The published solution does assume the modulo operation has less precedence then subtraction. This is very uncommon. The second mistake is the remainder calculation. If the term s - r becomes negative, because of a wrap-around, the modulo operation has a negative result. Again the question: shall Wikipedia show the faulty original? AndreAdrian (talk) 19:58, 6 January 2022 (UTC)
replies: text: I don't think s-r can become negative even with wraparound... we always have s <= r <= s + b, so 0 <= s-r <= b. Mtzguido (talk) 15:01, 10 April 2022 (UTC)
replies: text: Nevermind, I see it now, it can become negative after doing the mod k. In any case, I think this algorithm is correct just with the clarification that `mod` gives the positive remainder. Mtzguido (talk) 23:30, 10 April 2022 (UTC)
thread_title:  Even more faulty original solution 

message: The article says it does, and provides a C++ implementation that's supposed to fix that, but I don't see why the original has a problem. Since only the producer modifies s, and only the consumer modifies r, then the lack of atomicity is not a problem. There's also no need of making the taking/writing of the message atomic with the update of the respective variable.  — Preceding unsigned comment added by Mtzguido (talk • contribs) 14:51, 10 April 2022 (UTC)
replies: text: The Lamport solution has a problem. You are correct: every function changes only one variable. But this neglects the term (s - r). The value of this term changes if s or r change. This is the weakness of the Lamport solution. Can you make the term (s - r) atomic? If not the following can happen: function Producer has copied value r to a register and will later calculate (s - r) with this register value. Now the scheduler switches to function Consumer and changes variable r. And the scheduler switches back. We have now a value of r in memory and a different value of r in a register. Do you think that this will work well? Because of this, I used the classic circular buffer solution that needs the variable count to be atomic. And this variable needs only count up and down. Again: you need some kind of thread synchronization. It can be the Dekker algorithm, it can be atomic variable or it can be something else. AndreAdrian (talk) 14:21, 16 May 2022 (UTC)
thread_title:  Lamport's solution does not assume atomicity? 

message: Proposal to revert to early 2022 version. I think current version is not clear in why the producer consumer is a problem and why synchronization barriers are necessary.  — Preceding unsigned comment added by 181.230.113.80 (talk) 02:54, 5 January 2023 (UTC)
thread_title:  What's the Problem? 

message: I came here from Stack Overflow to help a developer classify their problem but feel a bit alienated. Where's the introduction to the problem itself? The article starts with explaining the solution. From the link given above  I could at least get to the former intro, but it seems no good idea to refer to an older revision for explaining something that is even in the title and what the article is initially written for. Please put it back. --Buckiboy (talk) 13:00, 15 September 2023 (UTC)
thread_title:  Reinstate the Intro