基本上,我想将以下内容更改为有限的线程解决方案,因为在我的情况下,计算列表太大,产生的线程太多,我想用较少的线程来试验和测量性能。
// the trivial approach (and largely my current situation)
let doWork() =
[1 .. 10]
|> List.map (fun i -> async {
do! Async.Sleep (100 * i) // longest thread will run 1 sec
return i * i // some complex calculation returning a certain type
})
|> Async.Parallel
|> Async.RunSynchronously // works, total wall time 1s
我的新方法,这个代码是由this online snippet from Tomas Petricek借用/启发的(我测试过,它有效,但我需要它来返回一个值,而不是单位)。
type LimitAgentMessage =
| Start of Async<int> * AsyncReplyChannel<int>
| Finished
let threadingLimitAgent limit = MailboxProcessor.Start(fun inbox -> async {
let queue = System.Collections.Generic.Queue<_>()
let count = ref 0
while true do
let! msg = inbox.Receive()
match msg with
| Start (work, reply) -> queue.Enqueue((work, reply))
| Finished -> decr count
if count.Value < limit && queue.Count > 0 then
incr count
let work, reply = queue.Dequeue()
// Start it in a thread pool (on background)
Async.Start(async {
let! x = work
do! async {reply.Reply x }
inbox.Post(Finished)
})
})
// given a synchronous list of tasks, run each task asynchronously,
// return calculated values in original order
let worker lst =
// this doesn't work as expected, it waits for each reply
let agent = threadingLimitAgent 10
lst
|> List.map(fun x ->
agent.PostAndReply(
fun replyChannel -> Start(x, replyChannel)))
现在,有了这个,原始代码将成为:
let doWork() =
[1 .. 10]
|> List.map (fun i -> async {
do! Async.Sleep (100 * i) // longest thread will run 1 sec
return i * i // some complex calculation returning a certain type
})
|> worker // worker is not working (correct output, runs 5.5s)
总而言之,输出是正确的( 计算并传播回复),但在(有限集合)线程中没有这样做。
我一直在玩一下,但是我觉得我错过了显而易见的事情(而且,谁知道,有人可能会喜欢有限线程邮箱处理器按顺序返回计算的想法。)
答案 0 :(得分:3)
问题是对agent.PostAndReply
的调用。 PostAndReply
将阻止工作完成。在List.map
内调用此选项将导致工作按顺序执行。一种解决方案是使用PostAndAsyncReply
,它不会阻塞,也会返回一个异步句柄来获取结果。
let worker lst =
let agent = threadingLimitAgent 10
lst
|> List.map(fun x ->
agent.PostAndAsyncReply(
fun replyChannel -> Start(x, replyChannel)))
|> Async.Parallel
let doWork() =
[1 .. 10]
|> List.map (fun i -> async {
do! Async.Sleep (100 * i)
return i * i
})
|> worker
|> Async.RunSynchronously
这当然只是一种可能的解决方案(让所有异步句柄回来并等待它们并行)。