Future with Timeout in Scala

You can go for a slightly easier approach using Await. The Await.result method takes timeout duration as a second parameter and throws a TimeoutException on timeout.

try {
  import scala.concurrent.duration._
  Await.result(aref, 10 seconds);
} catch {
    case e: TimeoutException => // whatever you want to do.
}

I needed the same behavior as well, so this is how I solved it. I basically created an object that creates a timer and fails the promise with a TimeoutException if the future hasn't completed in the specified duration.

package mypackage

import scala.concurrent.{Promise, Future}
import scala.concurrent.duration.FiniteDuration
import akka.actor.ActorSystem
import scala.concurrent.ExecutionContext.Implicits.global

object TimeoutFuture {

  val actorSystem = ActorSystem("myActorSystem")
  def apply[A](timeout: FiniteDuration)(block: => A): Future[A] = {
    val promise = Promise[A]()
    actorSystem.scheduler.scheduleOnce(timeout) {
      promise tryFailure new java.util.concurrent.TimeoutException
    }

    Future {
      try {
        promise success block
      }
      catch {
        case e:Throwable => promise failure e
      } 
    }

    promise.future
  }
}

Although you already got some answers on how to achieve it with blocking the additional thread to handle the timeout, I would suggest you to try a different way, for the reason Rex Kerr already gave. I don't exactly know, what you are doing in f(), but if it is I/O bound, I would suggest you to just use an asynchronous I/O library instead. If it is some kind of loop, you could pass the timeout value directly into that function and throw a TimeoutException there, if it exceeds the timeout. Example:

import scala.concurrent.duration._
import java.util.concurrent.TimeoutException

def doSth(timeout: Deadline) = {
  for {
    i <- 0 to 10
  } yield {
    Thread.sleep(1000)
    if (timeout.isOverdue)
      throw new TimeoutException("Operation timed out.")

    i
  }
}

scala> future { doSth(12.seconds.fromNow) }
res3: scala.concurrent.Future[scala.collection.immutable.IndexedSeq[Int]] = 
  scala.concurrent.impl.Promise$DefaultPromise@3d104456

scala> Await.result(res3, Duration.Inf)
res6: scala.collection.immutable.IndexedSeq[Int] =
  Vector(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

scala> future { doSth(2.seconds.fromNow) }
res7: scala.concurrent.Future[scala.collection.immutable.IndexedSeq[Int]] = 
  scala.concurrent.impl.Promise$DefaultPromise@f7dd680

scala> Await.result(res7, Duration.Inf)
java.util.concurrent.TimeoutException: Operation timed out.
    at $anonfun$doSth$1.apply$mcII$sp(<console>:17)
    at $anonfun$doSth$1.apply(<console>:13)
    at $anonfun$doSth$1.apply(<console>:13)
    at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:244)
    ...

scala> res7.value
res10: Option[scala.util.Try[scala.collection.immutable.IndexedSeq[Int]]] =
  Some(Failure(java.util.concurrent.TimeoutException: Operation timed out.))

This will only use only 1 thread, that will be terminated after timeout + execution time of a single step.

You could also try using a CountDownLatch like this:

def launch(f: () => Unit, timeout: Int): Future[Try[Unit]] = {

  val aref = new java.util.concurrent.atomic.AtomicReference[Thread]() 

  import ExecutionContext.Implicits.global
  val latch = new CountDownLatch(1)
  Future {
    latch.await()
    aref.get().interrupt
  }

  Future {
    aref.set(Thread.currentThread) 
    latch.countDown()
    Try(f())
  }
}

Now I'm waiting forever with my call to latch.await(), but you could certainly change that to:

latch.await(1, TimeUnit.SECONDS)

and then wrap it with a Try to handle if when/if it times out.