我有一个需要观察的事物序列,一个C#任务。这些C#任务不应同时运行,而应在之后另一个。基本上,我需要实现与此C#等效的F#问题:
天真地翻译此C#代码会得到如下内容:
let run (idx:int) (delay:int) =
async {
sprintf "start: %i (%i)" idx delay |> System.Diagnostics.Trace.WriteLine
let! t = System.Threading.Tasks.Task.Delay(delay) |> Async.AwaitTask
sprintf "finish: %i" idx |> System.Diagnostics.Trace.WriteLine
t
}
Observable.generate (new Random()) (fun _ -> true) id (fun s -> s.Next(250, 500))
|> Observable.take 20
|> Observable.mapi(fun idx delay -> idx, delay)
|> Observable.bind(fun (idx, delay) -> Observable.ofAsync (run idx delay))
|> Observable.subscribe ignore
|> ignore
这无法按预期工作,因为我不在任何地方等待结果。甚至有一种在F#中执行此操作而不阻塞线程的方法,就像C#的等待会一样吗?
F#中存在一个方便的库,称为AsyncSeq:https://www.nuget.org/packages/FSharp.Control.AsyncSeq/
与添加到C#8.0中的IAsyncEnumerable<T>
非常相似,这为您提供了一个不错的解决方案。
let run (idx:int) (delay:int) =
async {
sprintf "start: %i (%i)" idx delay |> System.Diagnostics.Trace.WriteLine
do! System.Threading.Tasks.Task.Delay(delay) |> Async.AwaitTask
sprintf "finish: %i" idx |> System.Diagnostics.Trace.WriteLine
}
Observable.generate (new Random()) (fun _ -> true) id (fun s -> s.Next(250, 500))
|> Observable.take 20
|> Observable.mapi(fun idx delay -> idx, delay)
|> AsyncSeq.ofObservableBuffered
|> AsyncSeq.iterAsync (fun (idx,delay) -> run idx delay)
|> Async.RunSynchronously