从scalaz中的队列创建进程
我试着按照第一个例子 在以下代码中填充一些空白并添加一些调试打印:从scalaz中的队列创建进程,scala,scalaz,scalaz-stream,Scala,Scalaz,Scalaz Stream,我试着按照第一个例子 在以下代码中填充一些空白并添加一些调试打印: import java.util.concurrent.ScheduledExecutorService import scala.concurrent.{Await, Future} import scala.concurrent.duration._ import scalaz.concurrent.Task import scalaz.stream.async.mutable.Queue import scalaz.str
import java.util.concurrent.ScheduledExecutorService
import scala.concurrent.{Await, Future}
import scala.concurrent.duration._
import scalaz.concurrent.Task
import scalaz.stream.async.mutable.Queue
import scalaz.stream.{Process, Sink}
object ProcessTest {
def main(args: Array[String]): Unit = {
import scala.concurrent.ExecutionContext.Implicits.global
import scalaz.stream.async
val q: Queue[Int] = async.unboundedQueue[Int]
val src: Process[Task, Int] = q.dequeue
// Thread 1
val f1 = Future {
for (i <- 0 to 10) {
println(s"enqueueOne $i")
Thread.sleep(100)
q.enqueueOne(i)
}
println("closing")
q.close
println("closed")
}
// Thread 2
val f2 = Future {
val buf = new collection.mutable.ArrayBuffer[Int]
val snk: Sink[Task, Int] = scalaz.stream.io.fillBuffer(buf)
val run: Task[Unit] = src.map(x => {
println(s"map $x")
x
}).to(snk).run
println("running")
run.get.runFor(3.seconds)
println(s"result = ${buf.toList}")
}
Await.result(f1, 10.seconds)
Await.result(f2, 10.seconds)
}
}
我做错了什么?
这个街区在哪里
(我正在使用scalaz stream 0.8.6)好的,我发现了问题:
enqueueOne
和close
返回任务,这些任务必须运行:
// Thread 1
val f1 = Future {
for (i <- 0 to 10) {
println(s"enqueueOne $i")
Thread.sleep(100)
q.enqueueOne(i).run
}
println("closing")
q.close.run
println("closed")
}
//线程1
val f1=未来{
为了
// Thread 1
val f1 = Future {
for (i <- 0 to 10) {
println(s"enqueueOne $i")
Thread.sleep(100)
q.enqueueOne(i).run
}
println("closing")
q.close.run
println("closed")
}