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按名称调用参数和函数类型不匹配

callbyname lazy-evaluation scala function

0

0__ · 技术社区 · 14 年前

在以下场景中,按名称参数会导致与函数发生冲突。

给定一些序列化基础结构:

trait Tx {
   def readSource[A](implicit ser: Serializer[A]) : Source[A] = 
      new Source[A] { 
         def get(implicit tx: Tx): A = ser.read(new In {})
      }
}    
trait In
trait Source[A] { def get(implicit tx: Tx): A }
trait Serializer[A] { def read(in: In)(implicit tx: Tx): A }

以及一个示例类型及其序列化程序:

// needs recursive access to itself. for reasons
// beyond the scope of this questions, `self` must
// be a by-name parameter
class Transport(self: => Source[Transport])

// again: self is required to be by-name
def transportSer(self: => Source[Transport]) : Serializer[Transport] = 
   new Serializer[Transport] { 
      def read(in: In)(implicit tx: Tx): Transport = new Transport(self)
   }

现在想象一个名为 Hook 其处理递归/相互连通性:

trait Hook[A] {
   def source: Source[A]
}

及其序列化程序:

def hookSer[A](peerSelf: Source[A] => Serializer[A]) : Serializer[Hook[A]] = 
   new Serializer[Hook[A]] {
      def read(in: In)(implicit tx: Tx) : Hook[A] =
         new Hook[A] with Serializer[A] {
            val source: Source[A] = tx.readSource[A](this)
            def read(in: In)(implicit tx: Tx) : A = peerSelf(source).read(in)
         }
   }

则以下操作失败:

val hs = hookSer[Transport](transportSer)

<console>:15: error: type mismatch;
 found   : => Source[Transport] => Serializer[Transport]
 required: Source[Transport] => Serializer[Transport]
          val hs = hookSer[Transport](transportSer)
                                      ^

如何在不按名称将参数更改为函数的情况下(尽可能)解决此问题?

1 回复 | 直到 14 年前

1

0

0__ 14 年前

似乎一个人可以写类型 (=> Source[A]) => Serializer[A] 以下为:

def hookSer[A](peerSelf: (=> Source[A]) => Serializer[A]) : Serializer[Hook[A]] = ...

val hs = hookSer[Transport](transportSer)
val h  = hs.read(new In {})(new Tx {})
val t  = h.source.get(new Tx {})