猫与验证的mapN

问题描述 投票:5回答:1

我与猫初学者。我有一个经验证的猫一个错误。我用一个列表蓄电池这样的:

case class Type(
 name: String,
 pattern: String,
 primitiveType: PrimitiveType = PrimitiveType.string,
 sample: Option[String] = None,
 comment: Option[String] = None,
 stat: Option[Stat] = None
) {
 type ValidationResult[A] = Validated[List[String], A]

 def checkValidity(): ValidationResult[Boolean] = {
  val errorList: mutable.MutableList[String] = mutable.MutableList.empty

  val patternIsValid = Try {
   primitiveType match {
    case PrimitiveType.struct =>
    case PrimitiveType.date =>
      new SimpleDateFormat(pattern)
    case PrimitiveType.timestamp =>
      pattern match {
        case "epoch_second" | "epoch_milli" =>
        case _ if PrimitiveType.formatters.keys.toList.contains(pattern) =>
        case _ =>
          DateTimeFormatter.ofPattern(pattern)
      }
    case _ =>
      Pattern.compile(pattern)
   }
 }
 if (patternIsValid.isFailure)
  errorList += s"Invalid Pattern $pattern in type $name"
 val ok = sample.forall(this.matches)
 if (!ok)
  errorList += s"Sample $sample does not match pattern $pattern in type $name"
 if (errorList.nonEmpty)
  Invalid(errorList.toList)
 else
  Valid(true)
}
}

当我使用此功能与我的情况类类型:

case class Types(types: List[Type]) {

 type ValidationResult[A] = Validated[List[String], A]

 def checkValidity(): ValidationResult[Boolean] = {
   val typeNames = types.map(_.name)
   val dup: ValidationResult[Boolean] =
   duplicates(typeNames, s"%s is defined %d times. A type can only be defined once.")
  (dup,types.map(_.checkValidity()).sequence).mapN((_,_) => true)
 }
}

我有这样的错误

Error:(29, 39) Cannot prove that cats.data.Validated[List[String],Boolean] <:< G[A].
(dup,types.map(_.checkValidity()).sequence: _*).mapN((_,_) => true)

你能不能帮我解决这个问题?

谢谢你的帮助。

scala scala-cats
1个回答
8
投票

很多年前,我写了一个关于潜在问题long blog post你正在运行到这里,如果你有兴趣在历史上还是一些老的解决方法,但幸运的是,现在的解决方案是很容易(假设你在斯卡拉2.11或2.12) :只需添加-Ypartial-unification到您的Scala编译器选项。如果您正在使用SBT,例如,可能是这样的:

scalacOptions += "-Ypartial-unification"

就大功告成了。

如果由于某种原因,你不能添加编译器选项,你必须明确地提供某种类型的参数。这里有一个快速简化版本:

import cats.data.Validated, cats.implicits._


case class Foo(i: Int) {
  type ValidationResult[A] = Validated[List[String], A]

  def check: ValidationResult[Boolean] =
    if (i < 0) Validated.invalid(List("bad")) else Validated.valid(true)
}

case class Foos(values: List[Foo]) {
  type ValidationResult[A] = Validated[List[String], A]

  def dup: ValidationResult[Unit] = Validated.valid(())
  def check: ValidationResult[Boolean] =
    (dup, values.map(_.check).sequence).mapN((_, _) => true)
}

这将失败,你看到的错误(假设你没有添加-Ypartial-unification):

<console>:22: error: Cannot prove that cats.data.Validated[List[String],Boolean] <:< G[A].
           (dup, values.map(_.check).sequence).mapN((_, _) => true)
                                     ^

要解决它,你可以写:

case class Foos(values: List[Foo]) {
  type ValidationResult[A] = Validated[List[String], A]

  def dup: ValidationResult[Unit] = Validated.valid(())
  def check: ValidationResult[Boolean] =
    (dup, values.map(_.check).sequence[ValidationResult, Boolean]).mapN((_, _) => true)
}

我想你也很可能只是移动式别名包级别,但我不是100%肯定没有,我不上进检查,对不起。

一个注脚:你有map然后sequence,您可以通过使用traverse,而不是使事情更快一点任何时候:

  def check: ValidationResult[Boolean] =
    (dup, values.traverse[ValidationResult, Boolean](_.check)).mapN((_, _) => true)

同样,你可以删除该类型的参数,并让类型推断推测出来,如果你有-Ypartial-unification启用。

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