在Scala中延迟函数执行的最简单方法是什么,比如JavaScript的setTimeout
?理想情况下,每个延迟执行没有产生线程,即顺序执行。我能找到的最接近的是Akka的Scheduler,但这太过分了。
为了我的测试目的,我打开了数千个连接,然后他们在10秒钟内获得响应。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Scala版本的做法是什么?我不在乎res.end
是在多个线程中执行还是在一个线程中排队。
答案 0 :(得分:20)
厌倦了最简单地回答这个问题的问题,这里是标准的JVM习语:
$ 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 = java.util.Timer@2d9ffd6f
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 = java.util.concurrent.ScheduledThreadPoolExecutor@2c5d529e
scala> x.schedule(new Callable[Int]() { def call = { println("Ran"); 42 }}, 1L, TimeUnit.SECONDS)
res3: java.util.concurrent.ScheduledFuture[Int] = java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask@3ab0f534
scala> Ran
没有用于在标准库中安排延迟任务的API,但您可以使用固定延迟生成ExecutionContext
,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 = $anon$1@40d3ab8b
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))
或者你可以使用Akka的调度程序,这是Scheduled Executor in Scala
的规范答案旧的单行:
最简单的仍然是future { blocking(Thread.sleep(10000L)); "done" }
但是我想为这个家伙制作一个广告,我刚刚遇到过这个广告,它会为您提供进度指示或中间值。我希望它有一个不同的名字,就是全部。
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] = scala.concurrent.DelayedLazyVal@50b56ef3
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
这是一个使用Either的替代公式,用于计算延迟后的值。当然仍有时间Left
时使用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]] = scala.concurrent.DelayedLazyVal@78f9c6f2
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