scala - LabelledGeneric 具有不同类型

Question

鉴于这些 ADT：

case class SourceRecord(name: String, number: Int)
case class TargetRecord(number: Int, name: String)

我可以轻松地从 a 映射SourceRecord到 a TargetRecord：

val sourceGen = LabelledGeneric[SourceRecord]
val targetGen = LabelledGeneric[TargetRecord]
val sourceRecord = SourceRecord("record_1", 1)
val targetRecord = targetGen.from(sourceGen.to(sourceRecord).align[targetGen.Repr])

但是，如果我引入一种新类型，我很难理解如何执行转换：

case class MyNumber(value: Int) extends AnyVal
case class SourceRecord(name: String, number: Int)
case class TargetRecord(number: MyNumber, name: String)

Error: could not find implicit value for parameter align: shapeless.ops.hlist.Align[sourceGen.Repr,targetGen.Repr]

我猜我需要以某种方式提供一个类型类来对齐，Int但MyNumber我不知道如何。谁能告诉我？

Answer 1

您可以手动定义它，如下所示：

import shapeless.labelled.FieldType
import shapeless.ops.hlist.Align
import shapeless.record.Record
import shapeless.{::, HNil, Witness}
import shapeless.syntax.singleton._

implicit val align: Align[Record.`'name -> String, 'number -> Int`.T, Record.`'number -> MyNumber, 'name -> String`.T] = {
  case s :: (i: FieldType[Witness.`'number`.T, Int]) :: HNil => 'number ->> MyNumber(i) :: 'name ->> s :: HNil
}