scala – 如何在此示例中修复类型推断错误？

sealed trait Status
object Error1 extends Status
case class Ok(x: Int) extends Status  

def foo(opt: Option[Int]): Status = opt.fold(Error1)(x => Ok(x))

当我在REPL中尝试时,我收到一个错误：

scala> def foo(opt: Option[Int]): Status = opt.fold(Error1)(x => Ok(x))
<console>:11: error: type mismatch;
found   : Ok
required: Error1.type
   def foo(opt: Option[Int]): Status = opt.fold(Error1)(x => Ok(x))
                                                               ^

我可以解决它,但它看起来不是特别优雅：

// work around the type error above
 val error1: Status = Error1
 def ok(x: Int): Status = Ok(x)

 def foo(opt: Option[Int]): Status = opt.fold(error1)(x => ok(x))

您如何建议解决此类问题？