在F#中一次异步一个?

喜欢 this :

type System.Threading.SemaphoreSlim with

    /// Wait for capacity to be available. Returns false if timeout elapsed before this as achieved
    member semaphore.Await(timeout: TimeSpan): Async<bool> = async {
        let! ct = Async.CancellationToken
        return! semaphore.WaitAsync(timeout, ct) |> Async.AwaitTaskCorrect
    }

    /// Wait indefinitely for capacity to be available on the semaphore
    member semaphore.Await(): Async<unit> = async {
        let! ct = Async.CancellationToken
        return! semaphore.WaitAsync(ct) |> Async.AwaitTaskCorrect
    }

    /// Throttling wrapper that waits asynchronously until the semaphore has available capacity
    member semaphore.Throttle(workflow: Async<'T>): Async<'T> = async {
        do! semaphore.Await()
        try return! workflow
        finally semaphore.Release() |> ignore

如果你想要的DOP是1,那么你:

let limiter = new System.Threading.SemaphoreSlim(1)
Seq.replicate 5 asyncThing |> Seq.map limiter.Throttle |> Async.Parallel |> Async.RunSynchronously

^the asyncThing 体执行将被序列化,受限于任何执行都将被获取/释放信号量括起来的事实。如果你只需要一个,当然可以用比信号量更轻的权重锁原语来做同样的事情。

(当然,你也可以通过 Seq.replicate 5 asyncThing |> Async.Sequential |> Async.RunSynchronously ,但我相信这种限制就是你想要的)

(取决于你是什么 真正地 做,还有一个 TaskCell 在里面 Equinox.Core (参见使用示例 in the tests ))

我认为你可以使用for循环:

async {
    for item in 1..10 do
        do! Async.Sleep 1000
        printfn $"item: {item}"
}