Sure, I'll get to it this weekend probably. I don't know what jira is so some information of how to do this would be very helpful.
Thank you, semih On Tue, Mar 8, 2011 at 8:31 AM, Patrick Hunt <[email protected]> wrote: > On Tue, Mar 8, 2011 at 5:59 AM, Flavio Junqueira <[email protected]>wrote: > >> I believe the goal of the examples was never to be a complete solutions to >> barriers or queues, but just to give a quick bootstrap to beginners. It is >> true, though, that the documentation page does not make that claim, and can >> be misleading. >> >> I see two possible action points out of this discussion: >> 1- State clearly in the beginning that the example discussed is not >> correct under the assumption that a process may finish the computation >> before another has started, and the example is there for illustration >> purposes; >> 2- Have another example following the current one that discusses the >> problem and shows how to fix it. This is an interesting option that >> illustrates how one could reason about a solution when developing with >> zookeeper. >> >> > This (2) sounds much better to me. Semih, would you like to give that a > try? (updating the docs I mean) > > Patrick > > >> If you are interested in helping us fix it, Semih, then you could perhaps >> create a jira and assign yourself to fix it. I can help you out. >> >> -Flavio >> >> On Mar 7, 2011, at 11:23 AM, Semih Salihoglu wrote: >> >> Hi Mahadev, >> >> Sorry for the late response. I agree, actually in this other documentation >> http://hadoop.apache.org/zookeeper/docs/r3.0.0/recipes.html, where there >> is >> only the pseudo-code, I think this situation is avoided. Here there is >> another znode /ready that all nodes have a watch on. And after each node >> writes their own ephemeral child, they don't wait. They read how many of >> has >> been written and the last one writes the /ready znode and everyone wakes >> up. >> The only race condition in this one is that there can be two nodes trying >> to >> write /ready and only one of them will succeed but this is ok. >> >> Thank you again, >> >> semih >> >> On Sat, Mar 5, 2011 at 6:41 PM, Mahadev Konar <[email protected]> wrote: >> >> Semih, >> >> You pointed it out right. It is possible ot enter into a situation >> >> like that. The recipe does have a bug. It can be fixed with the last >> >> client creating a special znode and every node in the list watching >> >> for that (so itll be an indication for entering the barrier). no? >> >> >> thanks >> >> mahadev >> >> >> On Sat, Mar 5, 2011 at 5:06 PM, Semih Salihoglu <[email protected]> >> >> wrote: >> >> Hi All, >> >> >> I am new to this group and to ZooKeeper. I was readin the Barrier >> >> tutorial >> >> in one of the ZooKeeper documentations. >> >> http://hadoop.apache.org/zookeeper/docs/current/zookeeperTutorial.html . >> >> A >> >> barrier primitive is exactly how I want to use ZooKeeper. I have a >> >> question >> >> about this example. It's not really a ZooKeeper question, it's more a >> >> question about the Barrier primitive I think. Here it is: In the enter >> >> method of this Barrier implementation below >> >> >> boolean enter() throws KeeperException, InterruptedException{ >> >> zk.create(root + "/" + name, new byte[0], Ids.OPEN_ACL_UNSAFE, >> >> CreateMode.EPHEMERAL_SEQUENTIAL); >> >> while (true) { >> >> synchronized (mutex) { >> >> List<String> list = zk.getChildren(root, true); >> >> >> if (list.size() < size) { >> >> mutex.wait(); >> >> } else { >> >> return true; >> >> } >> >> } >> >> } >> >> } >> >> >> could there be a race condition? Let's say there are two >> >> machines/nodes: node1 and node2 that will use this code to synchronize >> >> over ZK. Let's say the following steps take place: >> >> >> >> 1. node1 calls the zk.create method and then reads the number of >> >> children, and sees that it's 1 and starts waiting. >> >> 2. node2 calls the zk.create method (doesn't call the >> >> zk.getChildren method yet, let's say it's very slow) >> >> 3. node1 is notified that the number of children on the znode >> >> changed, it checks that the size is 2 so it leaves the barrier, it >> >> does its work and then leaves the barrier, deleting its node. >> >> 4. node2 calls zk.getChildren and because node1 has already left, >> >> it sees that the number of children is equal to 1. Since node1 will >> >> never enter the barrier again, it will keep waiting. >> >> >> Could this scenario happen? If not, what is preventing this? I haven't >> >> copied the code piece that enters barrier-does work-leaves barrier. >> >> But in the link I pasted above, it's the barrierTest(String args[]) >> >> method. >> >> >> Thank you very much in advance, >> >> >> semih >> >> >> >> >> *flavio* >> *junqueira* >> >> research scientist >> >> [email protected] >> direct +34 93-183-8828 >> >> avinguda diagonal 177, 8th floor, barcelona, 08018, es >> phone (408) 349 3300 fax (408) 349 3301 >> >> >> >
