scala 为什么重写Bootstrap，而不是Run方法？

2ic8powd  于 2022-11-09  发布在  Scala
关注(0)|答案(2)|浏览(200)

我正在学习ZIO 2.x，当使用bootstrap层配置Runtime时，它不工作。

  1. object RuntimeCustom extends ZIOAppDefault {
  3.   // It's not work, And I don't know why?
  4.   override val bootstrap = EmailService.live
  6.   def run = (for {
  7.     _ <- ZIO.debug("Start...")
  8.     _ <- EmailService.send("God", "Hi")
  9.     _ <- ZIO.debug("End...")
  10.   } yield ())
  11. }

出现一个错误：

  1. [error] /Users/changzhi/github-repo/zio-start/src/main/scala/zio/reference/experiment/core/RuntimeCustom.scala:35:7: 
  2. [error] 
  3. [error] ──── ZIO APP ERROR ───────────────────────────────────────────────────
  4. [error] 
  5. [error]  Your effect requires a service that is not in the environment.
  6. [error]  Please provide a layer for the following type:
  7. [error] 
  8. [error]    1. example.EmailService
  9. [error] 
  10. [error]  Call your effect's provide method with the layers you need.
  11. [error]  You can read more about layers and providing services here:
  12. [error]  
  13. [error]    https://zio.dev/next/datatypes/contextual/
  14. [error] 
  15. [error] ──────────────────────────────────────────────────────────────────────
  16. [error] 
  17. [error]     _ <- EmailService.send("God", "Hi")
  18. [error]       ^
  19. [error] one error found
  20. [error] (Compile / compileIncremental) Compilation failed

如果我用provide替换，它可以工作。

  1. object RuntimeCustom extends ZIOAppDefault {
  3.   def run = (for {
  4.     _ <- ZIO.debug("Start...")
  5.     _ <- EmailService.send("God", "Hi")
  6.     _ <- ZIO.debug("End...")
  7.   } yield ())
  8.     // This can works, have no doubt
  9.     .provide(EmailService.live)
  10. }

完整版程序在此

  1. package example
  3. import zio._
  4. trait EmailService {
  5.   def send(user: String, content: String): Task[Unit]
  6. }
  7. object EmailService {
  8.   def send(user: String, content: String): ZIO[EmailService, Throwable, Unit] =
  9.     ZIO.serviceWithZIO[EmailService](_.send(user, content))
  11.   val live: ZLayer[Any, Nothing, EmailService] = 
  12.     ZLayer.fromZIO( ZIO.succeed(EmailServiceFake()) <* Console.printLine("Init EmailService") ).orDie
  13. }
  15. case class EmailServiceFake() extends EmailService {
  16.   override def send(user: String, content: String): Task[Unit] =
  17.     Console.printLine(s"sending email to $user")
  18. }
  20. object RuntimeCustom extends ZIOAppDefault {
  22.   // It's not work, And I don't know why?
  23.   //override val bootstrap = EmailService.live
  25.   def run = (for {
  26.     _ <- ZIO.debug("Start...")
  27.     _ <- EmailService.send("God", "Hi")
  28.     _ <- ZIO.debug("End...")
  29.   } yield ())
  30.     // This can works, have no doubt
  31.     .provide(EmailService.live)
  32. }
scala

2条答案

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ohfgkhjo

ohfgkhjo1#

ZIOAppDefault是为不需要额外服务的应用程序设计的，假设它们将只“需要”内置服务，因为您已经用provide方法提供了其他要求。
如果要使用bootstrap方法，则应该扩展ZIOApp并覆盖environmentTag和Environment字段，以便run机制知道如何构造最终环境。

  1. import zio._
  3. object App extends ZIOApp {
  5.   // Tell ZIO how the environment is constructed
  6.   override val environmentTag: EnvironmentTag[Environment] = EnvironmentTag[Environment]
  8.   // Tell the app which layers will be leftover from the `run`
  9.   override type Environment = FooService
  11.   // The app how to construct those remaining layers
  12.   override val bootstrap = FooService.layer
  14.   val run = FooService.doFoo
  16. }
  18. class FooService {
  19.   def doFoo: UIO[Unit] = ZIO.unit
  20. }
  22. object FooService {
  23.   val layer = ZLayer.succeed(new FooService)
  25.   def doFoo = ZIO.serviceWithZIO[FooService](_.doFoo)
  26. }

编辑

使用原始示例来说明这是可行的：

  1. import zio._
  3. trait EmailService {
  4.   def send(user: String, content: String): Task[Unit]
  5. }
  6. object EmailService {
  7.   def send(user: String, content: String): ZIO[EmailService, Throwable, Unit] =
  8.     ZIO.serviceWithZIO[EmailService](_.send(user, content))
  10.   val live: ZLayer[Any, Nothing, EmailService] = 
  11.     ZLayer.fromZIO( ZIO.succeed(EmailServiceFake()) <* Console.printLine("Init EmailService") ).orDie
  12. }
  14. case class EmailServiceFake() extends EmailService {
  15.   override def send(user: String, content: String): Task[Unit] =
  16.     Console.printLine(s"sending email to $user")
  17. }
  19. object RuntimeCustom extends ZIOApp {
  21.   // It's not work, And I don't know why?
  22.   override val bootstrap = EmailService.live 
  24.   override type Environment = EmailService
  26.   override val environmentTag: EnvironmentTag[Environment] = EnvironmentTag[Environment]
  28.   def run = (for {
  29.     _ <- ZIO.debug("Start...")
  30.     _ <- EmailService.send("God", "Hi")
  31.     _ <- ZIO.debug("End...")
  32.   } yield ())
  33. }

https://scastie.scala-lang.org/02MMaWS2S6Wwe55a24H7Sg

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liwlm1x9

liwlm1x92#

像这样的事情应该会奏效：

  1. object YourApp extends ZIOAppDefault {
  3. def run = {
  4.     val app = for {
  5.         _ <- ZIO.debug("Start...")
  6.         _ <- EmailService.send("God", "Hi")
  7.         _ <- ZIO.debug("End...")
  8.     } yield ()
  10.     app.provideLayer(bootstrap)
  11. }
  13. override val bootstrap = {
  14.     //    val confLayer = ZLayer(...getArgs...) construct Layer from getArgs
  15.     //    confLayer >>> EmailService.live
  16.     EmailService.live
  17. }
  18. }
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