在具有子类型的类上创建Scalaz equal实例
我有以下简单的ADT,如何实现equality typeclass的实例而不诉诸于显式模式匹配所有可能的组合在具有子类型的类上创建Scalaz equal实例,scala,scalaz,Scala,Scalaz,我有以下简单的ADT,如何实现equality typeclass的实例而不诉诸于显式模式匹配所有可能的组合 import scalaz._ import Scalaz._ sealed trait Billinginfo case class CreditCard(number: Int, holder: String, Address: String) extends Billinginfo case object COD extends Billinginfo case class In
import scalaz._
import Scalaz._
sealed trait Billinginfo
case class CreditCard(number: Int, holder: String, Address: String) extends Billinginfo
case object COD extends Billinginfo
case class Invoice(cId: String) extends Billinginfo
object Billinginfo{
implicit val BillingEqual = Equal.equal[Billinginfo]{(b1,b2) =>
(b1,b2) match {
case (Invoice(c1), Invoice(c2)) => c1 === c2
case (CreditCard(a,b,c), CreditCard(d,e,f)) =>
a === d &&
b === e &&
c === f //writing exhaustive match would be tedious
}
}
你(至少)有两个选择。一是使用“自然”平等。如果您没有案例类成员的任何自定义类型,这应该可以正常工作:
implicit val BillingEqual: Equal[Billinginfo] = Equal.equalA[Billinginfo]
或者可以使用Shapess的类型类实例派生:
import shapeless._
import scalaz.{ Coproduct => _, :+: => _, _ }, Scalaz._
object EqualDerivedOrphans extends TypeClassCompanion[Equal] {
object typeClass extends TypeClass[Equal] {
def product[H, T <: HList](eh: Equal[H], et: Equal[T]): Equal[H :: T] =
tuple2Equal(eh, et).contramap {
case h :: t => (h, t)
}
def project[A, B](b: => Equal[B], ab: A => B, ba: B => A): Equal[A] =
b.contramap(ab)
def coproduct[L, R <: Coproduct](
el: => Equal[L],
er: => Equal[R]
): Equal[L :+: R] = eitherEqual(el, er).contramap {
case Inl(l) => Left(l)
case Inr(r) => Right(r)
}
val emptyProduct: Equal[HNil] = Equal.equal((_, _) => true)
val emptyCoproduct: Equal[CNil] = Equal.equal((_, _) => true)
}
}
import EqualDerivedOrphans._
请注意,您不需要在元组上进行额外级别的匹配。应该向Billinginfo伴随对象添加“自然”相等(隐式val BillingEqual:Equal[Billinginfo]=Equal.equalA[Billinginfo])以启用Invoice(1)==Invoice(1)?@user2726995是的
=
语法将起作用,但是我们只能在幕后使用=
。啊,必须显式地键入发票(“a”)作为Billinginfo才能获得它。谢谢
implicit val BillingEqual = Equal.equal[Billinginfo] {
case (Invoice(c1), Invoice(c2)) => c1 === c2
case (CreditCard(a, b, c), CreditCard(d, e, f)) =>
a === d && b === e && c === f
case (COD, COD) => true
case _ => false
}