从并行任务向单个输出流写入Monix

Question

如何使用Monix任务从一系列任务(可以并行运行)编写单个输出流:例如，我有N个任务可以并行运行，并从HTTP call/server,获得一些响应，我正在将响应写回文件流(就像有限制内存一样)，但我得到stream closed error,想知道是否有一种方法可以使用Monix任务或任何其他API来实现这一目标。

java.io.IOException: Stream closed  

示例代码片段：

 import java.io.{FileOutputStream, OutputStream, OutputStreamWriter}

    import monix.eval.Task
    import monix.execution.ExecutionModel.AlwaysAsyncExecution
    import monix.execution.Scheduler

    import scala.util.Random

    object TaskTest extends App {

      implicit val scheduler = Scheduler(
        AlwaysAsyncExecution
      )
      val list = List(1 to 100)

      val filePath = System.currentTimeMillis() + "_test.txt"
      val outputStreamWriter = new OutputStreamWriter(new FileOutputStream(filePath))
      val futureTask = autoClose(outputStreamWriter) {
        outputStream =>
          val totalCount = SeqOfTasks(outputStream)
          totalCount.map(_.sum)
      }

      futureTask.runToFuture.onComplete {
        case scala.util.Success(value) =>
          println(s"fetched a total of $value")

        case scala.util.Failure(exception) =>

          println("there is some error occurred ")
          println(exception)
      }

//close this stream when all tasks completed.
      def autoClose[A <: AutoCloseable, B](resource: A)(code: A ⇒ B): B = {
        try code(resource)
        finally resource.close()
      }


      def SeqOfTasks(outPutStream: OutputStreamWriter): Task[Seq[Int]] = {

        val list = List(0 to 10: _*)
        Task.wanderUnordered(list)(l => makeHttpCall(outPutStream, l))
      }

      // each taks will fetch some data and write to single out put stream that is shared by other tasks.
      def makeHttpCall(outputStream: OutputStreamWriter, value: Int): Task[Int] = {

        Task {
          val result = getDummyValue(value)
          val bytes = result.mkString("\t") + "\n"
          outputStream.write(bytes)
          //return some value
          result.size
        }
      }

      //return a dummy response for http
      def getDummyValue(n: Int): Seq[Int] = {
        println(s"dummy value : $n")
        Seq.fill(n)(Random.nextInt)
      }

Answer 1

这个问题是由于Task是延迟和异步的，但是autoClose两者都不是。

解决方案是使autoClose任务意识到：

def autoClose[A <: AutoCloseable, B](resource: A)(code: A ⇒ Task[B]): Task[B] = {
    code(resource).guarantee(Task(resource.close()))
  }

注意:这是由 奥列格·皮日科夫 on Monix Gitter聊天回答的。