Scala中的通用伴随对象超类型

Question

我有两个班,Guid和UserGuid.Guid有一个类型参数.UserGuid是Guid的一个特例,代表一个没有类的实体(User),所以我把它实现为Guid [Any].

我有几种Guid的应用方法,我想在两种类型之间共享,所以我将它们放在一个超类(GuidFactory)中.但是,由于Guid是参数化的,我必须参数化GuidFactory特征,否则生成的Guid将被参数化为Guid [_].

因此,我的伴随对象UserGuid无法编译,抱怨:

错误:com.wixpress.framework.scala.UserGuid不带任何类型参数,预期:一个对象UserGuid扩展GuidFactory [UserGuid]

有什么方法可以在Guid和UserGuid之间共享应用方法,还是我必须复制它们或使用转换？

代码如下.

abstract class TypeSafeId[I, T](val id: I) extends Serializable    
class Guid[T](override val id: String) extends TypeSafeId[String, T](id)
class UserGuid(override val id: String) extends Guid[Any](id)

trait GuidFactory[I[A] <: Guid[A]] {
  def apply[T](id: String): I[T]
  def apply[T](id: UUID): I[T] = apply(id.toString)
  def apply[T](ms: Long, ls: Long): I[T] = apply(new UUID(ms, ls))
  def apply[T](bytes: Array[Byte]):I[T] = apply(UUID.nameUUIDFromBytes(bytes))
  def random[T] = apply[T](UUID.randomUUID())
}

object Guid extends GuidFactory[Guid] {
  override def apply[T](id: String) = new Guid[T](id)
}

object UserGuid extends GuidFactory[UserGuid] {
  override def apply(id: String) = new UserGuid(id)
}

Answer 1

这是我能建议的最好的：

import java.util.UUID
abstract class TypeSafeId[I, T](val id: I) extends Serializable
class Guid[T](override val id: String) extends TypeSafeId[String, T](id)
class UserGuid(override val id: String) extends Guid[Any](id)

trait GuidFactory[G] {
  def apply(id: String): G
  def apply(id: UUID): G = apply(id.toString)
  def apply(ms: Long, ls: Long): G = apply(new UUID(ms, ls))
  def apply(bytes: Array[Byte]): G = apply(UUID.nameUUIDFromBytes(bytes))
  def random = apply(UUID.randomUUID())
}

object Guid {
  def apply[T] = new GuidFactory[Guid[T]] {
    def apply(id: String) = new Guid[T](id)
  }
}

object UserGuid extends GuidFactory[UserGuid] {
  override def apply(id: String) = new UserGuid(id)
}

val guid1 = Guid[String]("123")