如何从Async[IO]创建Async[Future]



我正在尝试在我的代码中隐式添加异步和同步 doobie 存储库。同步和异步[F]工作正常IO。我想将它们转换为未来并面临问题

我尝试从IO创建自己的Aync

def futureAsync(implicit F: MonadError[Future, Throwable]): Async[Future] = new Async[Future] {
override def async[A](k: (Either[Throwable, A] => Unit) => Unit): Future[A] = IO.async(k).unsafeToFuture()
override def asyncF[A](k: (Either[Throwable, A] => Unit) => Future[Unit]): Future[A] =
throw new Exception("Not implemented Future.asyncF")
override def suspend[A](thunk: => Future[A]): Future[A] = thunk
override def bracketCase[A, B](acquire: Future[A])(use: A => Future[B])(release: (A, ExitCase[Throwable]) => Future[Unit]): Future[B] =
throw new Exception("Not implemented Future.bracketCase")
override def raiseError[A](e: Throwable): Future[A] = F.raiseError(e)
override def handleErrorWith[A](fa: Future[A])(f: Throwable => Future[A]): Future[A] = F.handleErrorWith(fa)(_ => f(new Exception("")))
override def pure[A](x: A): Future[A] = F.pure(x)
override def flatMap[A, B](fa: Future[A])(f: A => Future[B]): Future[B] = F.flatMap(fa)(f)
override def tailRecM[A, B](a: A)(f: A => Future[Either[A, B]]): Future[B] = F.tailRecM(a)(f)
}

我对在 asyncF 和括号中实现两个函数感到震惊 有人可以帮忙吗?

正如Reactormonk在上面的评论中所说,不可能为具有正确语义的Future编写Async实例,因为Async扩展了Sync,并且Sync需要可以重复运行的计算的表示,而Scala的未来在定义时开始运行并且无法重新运行。

非法实例

不过,亲眼看到这一点很有启发性,我鼓励您尝试编写自己的可编译但(必然)非法的Async[Future]实例,而无需查看下一个代码块。不过,为了这个例子,这里有一个快速的草图:

import scala.concurrent.{ExecutionContext, Future, Promise}
import scala.util.{Failure, Success}
import cats.effect.{Async, ExitCase, IO}
def futureAsync(implicit c: ExecutionContext): Async[Future] = new Async[Future] {
def async[A](k: (Either[Throwable, A] => Unit) => Unit): Future[A] =
IO.async(k).unsafeToFuture()
def asyncF[A](k: (Either[Throwable, A] => Unit) => Future[Unit]): Future[A] = {
val p = Promise[A]()
val f = k {
case Right(a) => p.success(a)
case Left(e) => p.failure(e)
}
f.flatMap(_ => p.future)
}
def suspend[A](thunk: => Future[A]): Future[A] = Future(thunk).flatten
def bracketCase[A, B](acquire: Future[A])(use: A => Future[B])(
release: (A, ExitCase[Throwable]) => Future[Unit]
): Future[B] = acquire.flatMap { a =>
use(a).transformWith {
case Success(b) => release(a, ExitCase.Completed).map(_ => b)
case Failure(e) => release(a, ExitCase.Error(e)).flatMap(_ => Future.failed(e))
}
}
def raiseError[A](e: Throwable): Future[A] = Future.failed(e)
def handleErrorWith[A](fa: Future[A])(f: Throwable => Future[A]): Future[A] =
fa.recoverWith { case t => f(t) }
def pure[A](x: A): Future[A] = Future.successful(x)
def flatMap[A, B](fa: Future[A])(f: A => Future[B]): Future[B] = fa.flatMap(f)
def tailRecM[A, B](a: A)(f: A => Future[Either[A, B]]): Future[B] = f(a).flatMap {
case Right(b) => Future.successful(b)
case Left(a) => tailRecM(a)(f)
}
}

这将编译得很好,并且可能适用于某些情况(但请不要实际使用它!不过,我们已经说过它不能有正确的语义,我们可以通过使用猫效应定律模块来证明这一点。

检查法律

首先,我们需要一些你不需要担心的样板内容:

import cats.kernel.Eq, cats.implicits._
import org.scalacheck.Arbitrary
implicit val throwableEq: Eq[Throwable] =  Eq.by[Throwable, String](_.toString)
implicit val nonFatalArbitrary: Arbitrary[Throwable] =
Arbitrary(Arbitrary.arbitrary[Exception].map(identity))
implicit def futureEq[A](implicit A: Eq[A], ec: ExecutionContext): Eq[Future[A]] =
new Eq[Future[A]] {
private def liftToEither(f: Future[A]): Future[Either[Throwable, A]] =
f.map(Right(_)).recover { case e => Left(e) }
def eqv(fx: Future[A], fy: Future[A]): Boolean =
scala.concurrent.Await.result(
liftToEither(fx).zip(liftToEither(fy)).map {
case (rx, ry) => rx === ry
},
scala.concurrent.duration.Duration(1, "second")
)
}

然后我们可以定义一个测试来检查实例的Async定律:

import cats.effect.laws.discipline.{AsyncTests, Parameters}
import org.scalatest.FunSuite
import org.typelevel.discipline.scalatest.Discipline
object FutureAsyncSuite extends FunSuite with Discipline {
implicit val ec: ExecutionContext = ExecutionContext.global
implicit val params: Parameters =
Parameters.default.copy(allowNonTerminationLaws = false)
checkAll(
"Async",
AsyncTests[Future](futureAsync).async[String, String, String]
)
}

然后我们可以运行法律测试:

scala> FutureAsyncSuite.execute()
FutureAsyncSuite:
- Async.async.acquire and release of bracket are uncancelable
- Async.async.ap consistent with product + map
- Async.async.applicative homomorphism
...

你会看到大多数测试都是绿色的;这个实例做对了很多事情。

违反法律的地方

不过,它确实显示了三个失败的测试,包括以下内容:

- Async.async.repeated sync evaluation not memoized *** FAILED ***
GeneratorDrivenPropertyCheckFailedException was thrown during property evaluation.
(Discipline.scala:14)
Falsified after 1 successful property evaluations.
Location: (Discipline.scala:14)
Occurred when passed generated values (
arg0 = "淳칇멀",
arg1 = org.scalacheck.GenArities$$Lambda$7154/1834868832@1624ea25
)
Label of failing property:
Expected: Future(Success(驅ṇ숆㽝珅뢈矉))
Received: Future(Success(淳칇멀))

如果您查看定律定义,您会发现这是一个测试,它使用delay定义一个Future值,然后对其进行多次排序,如下所示:

val change = F.delay { /* observable side effect here */ }
val read = F.delay(cur)
change *> change *> read

另外两个失败是类似的"未记忆"违规。这些测试应该看到副作用发生两次,但在我们的例子中,不可能以这样一种方式为Future编写delaysuspend(不过,值得尝试说服自己是这种情况)。

你应该做什么

总结一下:你可以编写一个Async[Future]实例,它将通过 78 个Async定律测试中的 75 个,但不可能编写一个通过所有测试的实例,并且使用非法实例是一个非常糟糕的主意:你的代码的潜在用户和像 Doobie 这样的库都会假设你的实例是合法的, 如果你不辜负这个假设,你就为复杂而烦人的错误打开了大门。

值得注意的是,为具有合法Async实例的Future编写最小包装器并不难(例如,我在我的 catbird 库中有一个名为Rerunnable的 Twitter 未来的包装器)。不过,您真的应该坚持使用cats.effect.IO,并使用提供的转换在代码的任何部分与 future(使用传统的基于Future的 API )进行转换。

最新更新