U.Mutlu wrote:
Peter Dimov wrote, On 2011-11-15 01:54:
U.Mutlu wrote:
Hi, in my current project I'm confronted with an IMO challenging real-world problem, and am looking for an optimal or near-optimal solution for it:
There is a standard vector (std::vector) of structs (ie. data records), and 3 threads working on that vector: thread 1: about every 10 seconds appends a new record to the vector, or updates an existing record, there are no deletions done. thread 2: walks over all records in a read-only manner and generates a list; it takes about 60 seconds. thread 3: walks over all records in a read-only manner and generates a different list; it takes about 90 seconds.
Of course all threads are running simultanously, but thread 2 fires its job every 3 minutes, thread 3 every 5 minutes, and thread 1 is permanently working (reacting on external events).
The problem is this: when thread 2 or 3 are running (remember 60 or 90 seconds) then using the usual shared locking schemes thread 1 cannot do its job, although it is the most important thread and its job is time-critical (recording external events).
Is there a better locking solution to this problem?
It depends. How long does it take to make a copy of the vector?
Hmm. let's say it takes 35 seconds, but this method consumes very much memory, eventually tripling the memory consumption (when thread2.job and thread3.job run in parallel). Other, resource-efficient, alternatives?
// thread 1 lock mutex; update vector; unlock mutex;
// thread 2 lock mutex; make a copy of the vector; unlock mutex; walk over copy;
// thread 3: see thread 2
Hi, If you have just a writer thread and two (or more) reader threads, you could split your application in two parts. The writer part writes directly to his vector without using any mutex and in addition log some updates that need to be done on the reader part (using a mutex or any lock-free available algorithm). Any thread in the reader part starts by updating the reader vector by taking in account the updates in the log. The space required is 2 copies of the data and the temporary queue of commands to be updated. The synchronization time is reduced to the writing/reading of a command in the queue command. This is a little bit more complex, but scales well when you need to add more readers. Best, Vicente -- View this message in context: http://boost.2283326.n4.nabble.com/thread-Locking-Problem-App-Level-tp404131... Sent from the Boost - Users mailing list archive at Nabble.com.