斯卡拉函數返回對象類型

Question

我有一個定義，看起來像這樣：

sealed trait ServiceResponse 
    case object OK extends ServiceResponse 
    case object SHIT extends ServiceResponse 

    private def remoteService(waitTime: FiniteDuration): ServiceResponse = { 
    val awaitable: Future[Ok.type] = Future.successful { 
     Thread.sleep(waitTime.toMillis) 
     OK 
    } 
    Await.ready(awaitable, waitTime) 
    } 

的remoteService功能抱怨說，它期待一個類型。那麼爲什麼這是一個問題呢？爲什麼我不能簡單地返回一個Object類型？

Answer 1

這是因爲您需要使用Await.result而不是Await.ready。這工作：

import scala.concurrent._ 
import scala.concurrent.duration._ 

sealed trait ServiceResponse { 

    case object OK extends ServiceResponse 
    case object SHIT extends ServiceResponse 

    private def remoteService(waitTime: FiniteDuration): ServiceResponse = { 
    val awaitable: Future[OK.type] = Future.successful { 
     Thread.sleep(waitTime.toMillis) 
     OK 
    } 
    Await.result(awaitable, waitTime) 
    } 

} 

而且你有一個錯字：未來[Ok.Type]應該是未來[OK.type]