Question

（希望如此）关于Scalaz Zio的简单问题。

我有一些重构到Zio的旧代码。我希望该代码的一个路径保持其原来的状态：

同步
阻止
在当前线程上（这是一个硬要求）

如何运行IO使其表现得像旧的阻止代码？

我目前正在使用：

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

这似乎可以解决问题，但我不确定这是正确的。这是100％向后兼容旧代码吗？

Answer 1

好吧，我终于在后台进行了检查，并执行了一些似乎可以满足我要求的事情：

  /**
    * 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
    })
  )

我还写了一些测试：

  "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))
      }
    }
  }

Zio运行阻止向后兼容代码

1 个答案: