scala – 如何“提取”类型参数来实例化另一个类
以下Scala代码有效:
object ReducerTestMain extends App { type MapOutput = KeyVal[String,Int] def mapFun(s:String): MapOutput = KeyVal(s,1) val red = new ReducerComponent[String,Int]((a: Int,b: Int) => a + b) val data = List[String]("a","b","c","b") data foreach {s => red(mapFun(s))} println(red.mem) // OUTPUT: Map(a -> 1,b -> 3,c -> 2) } class ReducerComponent[K,V](f: (V,V) => V) { var mem = Map[K,V]() def apply(kv: KeyVal[K,V]) = { val KeyVal(k,v) = kv mem += (k -> (if (mem contains k) f(mem(k),v) else v)) } } case class KeyVal[K,V](key: K,value:V) 我的问题是我想像这样实例化ReducerComponent: val red = new ReducerComponent[MapOutput,b: Int) => a + b) 甚至更好: val red = new ReducerComponent[MapOutput](_ + _) 这意味着很多事情: >我想键入 – 检查MapOutput的类型是KeyVal [K,C], 要问很多吗? :)我想写点类似的东西 class ReducerComponent[KeyVal[K,V]](f: (V,V) => V) {...} 当我将实例化ReducerComponent时,我只有f和MapOutput,所以推断V是可以的.但后来我只将KeyVal [K,V]作为类中的类型参数,它可以与KeyVal [_,_]不同. 我知道如果你理解类型推理是如何工作的,我问的可能是疯了,但我不知道!我甚至不知道什么是一个好的方法继续 – 除了在我的高级代码中一直提出明确的类型声明.我应该改变所有的架构吗? 解决方法
写一个简单的工厂:
case class RC[M <: KeyVal[_,_]](){ def apply[K,V) => V)(implicit ev: KeyVal[K,V] =:= M) = new ReducerComponent[K,V](f) } def plus(x: Double,y: Double) = x + y scala> RC[KeyVal[Int,Double]].apply(plus) res12: ReducerComponent[Int,Double] = ReducerComponent@7229d116 scala> RC[KeyVal[Int,Double]]()(plus) res16: ReducerComponent[Int,Double] = ReducerComponent@389f65fe 如您所见,ReducerComponent具有适当的类型.这里使用隐式证据从M<:KeyVal [_,_]中捕获K和V. 附:上面的版本要求为f明确指定参数类型,如(_:Double)(_:Double).如果你想避免这种情况: case class RC[M <: KeyVal[_,_]](){ def factory[K,V](implicit ev: KeyVal[K,V] =:= M) = new { def apply(f: (V,V) => V) = new ReducerComponent[K,V](f) } } scala> RC[KeyVal[Int,Double]].factory.apply(_ + _) res5: ReducerComponent[Int,Double] = ReducerComponent@3dc04400 scala> val f = RC[KeyVal[Int,Double]].factory f: AnyRef{def apply(f: (Double,Double) => Double): ReducerComponent[Int,Double]} = RC$$anon$1@19388ff6 scala> f(_ + _) res13: ReducerComponent[Int,Double] = ReducerComponent@24d8ae83 更新.如果你想generelize keyval – 使用类型函数: type KV[K,V] = KeyVal[K,V] //may be anything,may implement `type KV[K,V]` from some supertrait case class RC[M <: KV[_,_]](){ def factory[K,V](implicit ev: KV[K,V] =:= M) = new { def apply(f: (V,V](f) } } 但请记住,从您的问题中应用仍然需要KeyVal [K,V]. 您还可以将KV传递到某个类: class Builder[KV[_,_]] { case class RC[M <: KV[_,_]](){ def factory[K,V](f) } } } scala> val b = new Builder[KeyVal] scala> val f = b.RC[KeyVal[Int,Double]].factory scala> f(_ + _) res2: ReducerComponent[Int,Double] = ReducerComponent@54d9c993 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |