斯卡拉函数返回对象类型

Question

我有一个定义，看起来像这样：

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功能抱怨说，它期待一个类型。那么为什么这是一个问题呢？为什么我不能简单地返回一个Object类型？

Answer 1

这是因为您需要使用Await.result而不是Await.ready。这工作：

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) 
    } 

} 

而且你有一个错字：未来[Ok.Type]应该是未来[OK.type]