在Scala中处理高级对象时,类型信息会丢失

trait Builder[M[_]] {
    def build[A]: M[A]
    def buildPair[A,B]: (M[A],M[B]) = (build[A],build[B])
}

class List[A]

class BuilderList extends Builder[List] {
    def build[A] = new List[A]
}

val l: List[String] = (new BuilderList).build[String]

val ll: (List[String],List[Double]) = (new BuilderList).buildPair[String,Double]

defined trait Builder
defined class List
defined class BuilderList
l: List[String] = List@5c7754a7
ll: (List[String],List[Double]) = (List@62874648,List@7b0f6537)

如果我现在想将它应用于具有两个类型参数的类型,比如说

class Map[K,V]

我想能写

trait BuilderMap[K] extends Builder[Map[K,_]] {...}

但当然这不起作用,因为Scala中的类型参数不是咖喱.

我发现以下技巧允许我通过编译：

trait PartialApplier[F[_,_],K] {
    type PartiallyApplied[_] = F[K,_]
}

class BuilderMap[K] extends Builder[PartialApplier[Map,K]#PartiallyApplied] {
    def build[V] = new Map[K,V]
}

但后来发生了一些奇怪的效果,我无法弄清楚原因：

scala> val m: Map[Int,String] = (new BuilderMap[Int]).build[String]
m: Map[Int,String] = Map@71da0c64

scala> val mm: (Map[Int,String],Map[Int,Double]) = (new BuilderMap[Int]).buildPair[String,Double]
<console>:21: error: type mismatch;
 found   : (Map[Int,_])
 required: (Map[Int,Double])
   val mm: (Map[Int,Double]

当我使用PartialApplier技巧时,似乎在较高知名度的特性生成器中定义的函数会丢失一些类型信息.

有没有办法让所有这些工作顺利进行？也许PartialApplier技巧不是正确的方法！