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Scala 推断类型参数

scala parameters

Scala 推断类型参数,scala,parameters,self-type,inferred-type,Scala,Parameters,Self Type,Inferred Type,为什么我在编译此代码段时出错 trait ID[R <: Record[R] with KeyedRecord[Long]] { this: R => val idField = new LongField(this) } trait ID[R] val idField=新朗菲尔德（本） } 错误： inferred type arguments [ID[R] with R] do not conform to class LongField's type param

为什么我在编译此代码段时出错

trait ID[R <: Record[R] with KeyedRecord[Long]] {

  this: R =>

  val idField = new LongField(this)
}

trait ID[R]
val idField=新朗菲尔德（本）
}
错误：

inferred type arguments [ID[R] with R] do not conform to class LongField's
type parameter bounds [OwnerType <: net.liftweb.record.Record[OwnerType]]

推断的类型参数[ID[R]和R]不符合LongField类的
类型参数边界[OwnerTypeConvert


val idField = new LongField(this)


到

如果未指定类型
R
，则LongField无法检查该类型是否为
记录[OwnerType]
的共同变体。明确提及该类型可以解决此问题

PS：我不知道需要重新确认的其他类声明，但以下声明有效：


case class Record[R]
class KeyedRecord[R] extends Record[R]
class LongField[OwnerType <: Record[OwnerType]](rec: OwnerType)
trait ID[R <: Record[R] with KeyedRecord[Long]] {
  this: R =>
  val idField = new LongField[R](this)
}



案例类记录[R]
类KeyedRecord[R]扩展记录[R]
类LongField[OwnerType

val idField = new LongField[R](this)



case class Record[R]
class KeyedRecord[R] extends Record[R]
class LongField[OwnerType <: Record[OwnerType]](rec: OwnerType)
trait ID[R <: Record[R] with KeyedRecord[Long]] {
  this: R =>
  val idField = new LongField[R](this)
}