执行延迟功能

在Scala中延迟函数执行的最简单方法是什么，例如JavaScript的setTimeout？理想情况下，每延迟执行时间不产生线程，即顺序执行。我能找到的最接近的是Akka的Scheduler，但这是一种矫枉过正。执行延迟功能

为了我的测试目的，我打开了数千个连接，然后他们在10秒内得到响应。在node.js中，它看起来像：

http.createServer(function (req, res) { 
    res.writeHead(200, {'Content-Type': 'text/plain'}); 
    setTimeout(function() {res.end('Hello World\n');}, 10000); 
}).listen(8080, '127.0.0.1');

但是，最接近的Scala版本也会这样做吗？我不在乎res.end是要在多个线程中执行还是在单个线程中排队。

来源

2013-06-22 Oleg Mikheev

可能的复制，我不知道这样的回答可以帮助：http://stackoverflow.com/a/16629357/1296806但也许如果你想要一个毫无价值一次性进行测试。 –

你也可以看看这个其他问题：http://stackoverflow.com/questions/16359849/scala-scheduledfuture –

厌倦了越来越高射炮回答的简单问题太简单，这里有标准的JVM成语：

$ scala 
Welcome to Scala 2.11.8 (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_65). 
Type in expressions for evaluation. Or try :help. 

scala> import java.util.{Timer,TimerTask} 
import java.util.{Timer, TimerTask} 

scala> val timer = new Timer 
timer: java.util.Timer = [email protected] 

scala> def delay(f:() => Unit, n: Long) = timer.schedule(new TimerTask() { def run = f() }, n) 
delay: (f:() => Unit, n: Long)Unit 

scala> delay(() => println("Done"), 1000L) 

scala> Done 


scala> import java.util.concurrent._ 
import java.util.concurrent._ 

scala> val x = Executors.newScheduledThreadPool(2) 
x: java.util.concurrent.ScheduledExecutorService = [email protected] 

scala> x.schedule(new Callable[Int]() { def call = { println("Ran"); 42 }}, 1L, TimeUnit.SECONDS) 
res3: java.util.concurrent.ScheduledFuture[Int] = java.[email protected]3ab0f534 

scala> Ran

没有在标准库调度延迟的任务没有API，但你可以做的ExecutionContext具有固定的延迟，以便使用Scala Future。

scala> import scala.concurrent._ 
import scala.concurrent._ 

scala> implicit val xx = new ExecutionContext() { 
    | def reportFailure(t: Throwable) = t.printStackTrace() 
    | def execute(r: Runnable) = x.schedule(new Callable[Unit]() { def call = r.run() }, 1L, TimeUnit.SECONDS) 
    | } 
xx: scala.concurrent.ExecutionContext = [email protected] 

scala> Future(println("hello")) 
res4: scala.concurrent.Future[Unit] = List() 

scala> hello 

scala> Future(42) 
res5: scala.concurrent.Future[Int] = List()     

scala> .value 
res6: Option[scala.util.Try[Int]] = Some(Success(42))

或者你可以用阿卡的调度，这是在Scheduled Executor in Scala

老一班轮规范答案：

最简单的仍然只是future { blocking(Thread.sleep(10000L)); "done" }

，但我想把它放到一个我刚刚遇到的这个人的广告，它给你一个进度指标或中间值。我有点希望它有一个不同的名字，就是这样。

scala> import concurrent._ 
import concurrent._ 

scala> import ExecutionContext.Implicits._ 
import ExecutionContext.Implicits._ 

scala> import duration._ 
import duration._ 

scala> val deadline = 60.seconds.fromNow 
deadline: scala.concurrent.duration.Deadline = Deadline(38794983852399 nanoseconds) 

scala> new DelayedLazyVal(() => deadline.timeLeft.max(Duration.Zero), blocking { 
    | Thread.sleep(deadline.timeLeft.toMillis) 
    | Console println "Working!" 
    | }) 
res9: scala.concurrent.DelayedLazyVal[scala.concurrent.duration.FiniteDuration] = [email protected] 

scala> res9() 
res10: scala.concurrent.duration.FiniteDuration = 23137149130 nanoseconds 

scala> res9.isDone 
res11: Boolean = false 

scala> res9() 
res12: scala.concurrent.duration.FiniteDuration = 12499910694 nanoseconds 

scala> res9() 
res13: scala.concurrent.duration.FiniteDuration = 5232807506 nanoseconds 

scala> Working! 


scala> res9.isDone 
res14: Boolean = true 

scala> res9() 
res15: scala.concurrent.duration.FiniteDuration = 0 days

下面是一个替代公式，用来计算延迟后的值。当然使用Left的时候仍然有时间Left。

scala> new DelayedLazyVal(()=> if (deadline.hasTimeLeft) Left(deadline.timeLeft) else 
    | Right("Working!"), blocking(Thread.sleep(deadline.timeLeft.toMillis))) 
res21: scala.concurrent.DelayedLazyVal[Product with Serializable with scala.util.Either[scala.concurrent.duration.FiniteDuration,String]] = [email protected] 

scala> res21() 
res22: Product with Serializable with scala.util.Either[scala.concurrent.duration.FiniteDuration,String] = Left(28553649064 nanoseconds) 

scala> res21() 
res23: Product with Serializable with scala.util.Either[scala.concurrent.duration.FiniteDuration,String] = Left(9378334087 nanoseconds) 

scala> res21.isDone 
res24: Boolean = false 

scala> res21() 
res25: Product with Serializable with scala.util.Either[scala.concurrent.duration.FiniteDuration,String] = Right(Working!) 

scala> res21.isDone 
res26: Boolean = true

来源

2013-06-23 19:01:00

不会导致成千上万的线程睡10秒？ –

是否有非阻塞解决方案？ –

@OlegMikheev是的。 –

执行延迟功能

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