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(希望)关于 Scalaz Zio 的简单问题。

我有一些旧代码重构为 Zio。我希望该代码的一条路径保持原样运行:

  • 同步
  • 阻塞
  • 在当前线程上(这是一个硬性要求)

我怎样才能运行IO它以使其表现得像旧的阻塞代码?

我目前使用:

  private lazy val blockingRts = new RTS {}
  def runBlocking[E, A](io: IO[E, A]): Either[E, A] = {
    blockingRts.unsafeRun(io.attempt)
  }

它似乎可以解决问题,但我不确定这是正确的。这是否与旧代码 100% 向后兼容?

4

1 回答 1

1

好的,我终于深入了解并实现了一些似乎满足我要求的东西:

  /**
    * Executes the IO synchronous and blocking on the current thread, thus running an IO
    * without any of the advantages of IO. This can be useful for maintaining backwards compatibility.
    * Rethrows any exception that was not handled by the IO's error handling.
    */
  @throws
  def runLegacy[E, A](io: IO[E, A]): Either[E, A] = {
    syncBlockingRunTimeSystem.unsafeRunSync[Nothing, Either[E, A]](io.either) match {
      case Exit.Success(v) => v
      case Exit.Failure(Cause.Die(exception)) => throw exception
      case Exit.Failure(Cause.Interrupt) => throw new InterruptedException
      case Exit.Failure(fail) => throw FiberFailure(fail)
    }
  }

  private lazy val syncBlockingRunTimeSystem = Runtime(
    (),
    PlatformLive.fromExecutor(new Executor {
      override def yieldOpCount: Int = Int.MaxValue
      override def metrics: Option[ExecutionMetrics] = None
      override def submit(runnable: Runnable): Boolean = {
        runnable.run()
        true
      }
      override def here: Boolean = true
    })
  )

我还写了几个测试:

  "runLegacy" should {
    "run synchronous code in blocking fashion on current thread" in {
      var runCount = 0
      val io = IO.succeedLazy { runCount += 1 }
        .map { _ => runCount +=1 }
        .flatMap { _ =>
          runCount += 1
          IO.effect {
            runCount += 1
            Thread.currentThread()
          }
        }

      runCount shouldBe 0
      runLegacy(io) shouldBe Right(Thread.currentThread())
      runCount shouldBe 4
    }

    "run parallel code sequentially on current thread" in {
      val ios = (1 to 500).map { i => IO.succeedLazy { i } }
      runLegacy(IO.reduceAll(IO.succeed(0), ios) {
        case (a, b) => a + b
      }) shouldBe Right((500 * 501) / 2)
    }

    "run many flatMaps without overflowing" in {
      var runCount = 0
      val io = IO.succeedLazy { runCount += 1 }
      val manyIo = (1 to 9999).foldLeft(io) { case (acc, _) => acc.flatMap { _ => io } }
      runLegacy(manyIo)
      runCount shouldBe 10000
    }

    case object TestException extends Throwable

    "handle sync blocking errors" in {
      case object TestException extends Throwable
      runLegacy(IO.effect(throw TestException)) shouldBe Left(TestException)
    }

    "rethrow unhandled exceptions" in {
      assertThrows[TestException.type] {
        runLegacy(IO.succeedLazy(throw TestException))
      }
    }
  }
于 2019-04-01T20:52:55.787 回答