What is the best approach to guarantee that a message will be processed once? You’re asking for a guarantee – you won’t get one. You can reduce probability of a message being processed more than once to a very small amount, but you won’t get a guarantee. I’ll explain why, along with strategies for reducing … Read more
ConcurrentLinkedQueue means no locks are taken (i.e. no synchronized(this) or Lock.lock calls). It will use a CAS – Compare and Swap operation during modifications to see if the head/tail node is still the same as when it started. If so, the operation succeeds. If the head/tail node is different, it will spin around and try … Read more
All functions are actually variables, so it’s actually pretty easy to store all your functions in array (by referencing them without the ()): // Create your functions, in a variety of manners… // (The second method is preferable, but I show the first for reference.) function fun1() { alert(“Message 1”); }; var fun2 = function() … Read more
You can simplify that code a lot: you don’t net to use new Promise with d3.json, since d3.json will itself create the promise. So, you can just do: var files = [“data1.json”, “data2.json”, “data3.json”]; var promises = []; files.forEach(function(url) { promises.push(d3.json(url)) }); Promise.all(promises).then(function(values) { console.log(values) }); Or, if you’re into the code golf, even shorter: … Read more
Well, it’s hard to be sure without a lot of testing, but it certainly looks like this is a bug, either in TThreadedQueue or in TMonitor. Either way it’s in the RTL and not your code. You ought to file this as a QC report and use your example above as the “how to reproduce” … Read more
The following code seems to work: import multiprocessing, time def task(args): count = args[0] queue = args[1] for i in xrange(count): queue.put(“%d mississippi” % i) return “Done” def main(): manager = multiprocessing.Manager() q = manager.Queue() pool = multiprocessing.Pool() result = pool.map_async(task, [(x, q) for x in range(10)]) time.sleep(1) while not q.empty(): print q.get() print result.get() … Read more
priority_queue doesn’t allow iteration through all the members, presumably because it would be too easy in invalidate the priority ordering of the queue (by modifying the elements you traverse) or maybe it’s a “not my job” rationale. The official work-around is to use a vector instead and manage the priority-ness yourself with make_heap, push_heap and … Read more
There are no such feature like retry attempts in RabbitMQ (as well as in AMQP protocol). Possible solution to implement retry attempts limit behavior: Redeliver message if it was not previously redelivered (check redelivered parameter on basic.deliver method – your library should have some interface for this) and drop it and then catch in dead … Read more
Ben’s version is a stack instead of a queue, so i tweaked it a bit: NSMutableArray+QueueAdditions.h @interface NSMutableArray (QueueAdditions) – (id) dequeue; – (void) enqueue:(id)obj; @end NSMutableArray+QueueAdditions.m @implementation NSMutableArray (QueueAdditions) // Queues are first-in-first-out, so we remove objects from the head – (id) dequeue { // if ([self count] == 0) return nil; // to … Read more
I don’t know of a lock-free solution, but you can take a look at the new Dataflow library, part of the Async CTP. A simple BufferBlock<T> should suffice, e.g.: BufferBlock<int> buffer = new BufferBlock<int>(); Production and consumption are most easily done via extension methods on the dataflow block types. Production is as simple as: buffer.Post(13); … Read more