如何在Scala中使用编码器类型类处理Option [英] How to handle Option with an encoder typeclass in scala

问题描述

我有一个经典的 Encoder 类型类.

I have a classic Encoder typeclass.

trait Encoder[A] {
  def encode(a: A): String
}

我有两个问题

问题1:差异从何而来:

Question 1 : where does the divergence comes from :

[error] … diverging implicit expansion for type sbopt.Test.Encoder[None.type]
[error] starting with value stringEncoder in object Test
[error]   show(None)

implicit val stringEncoder = new Encoder[String] {
  override def encode(a: String): String = a
}
implicit def optionEncoder[A: Encoder]: Encoder[Option[A]] =
  (a: Option[A]) => {
    val encoderA = implicitly[Encoder[A]]
    a.fold("")(encoderA.encode)
  }
implicit def someEncoder[A: Encoder]: Encoder[Some[A]] =
  (a: Some[A]) => {
    val encoderA = implicitly[Encoder[A]]
    encoderA.encode(a.get)
  }
implicit def noneEncoder[A: Encoder]: Encoder[None.type] =
  (_: None.type) => ""

def show[A: Encoder](a: A) = println(implicitly[Encoder[A]].encode(a))

show(None)

问题2:我大概看到了编码器不是协变的.优点和缺点是什么?

Question 2 : I saw in circe that the Encoder is not contravariant. What are the pros and cons ?

trait Encoder[-A] {
  def encode(a: A): String
}

implicit val stringEncoder: Encoder[String] = (a: String) => a

implicit def optionEncoder[A: Encoder]: Encoder[Option[A]] =
  (a: Option[A]) => {
    val encoderA = implicitly[Encoder[A]]
    a.fold("")(encoderA.encode)
  }

def show[A: Encoder](a: A) = println(implicitly[Encoder[A]].encode(a))

show(Option("value"))
show(Some("value"))
show(None)

推荐答案

关于1.

您对 noneEncoder 的定义不好.您还有一个额外的上下文绑定(甚至还有额外的类型参数).

Your definition of noneEncoder is not good. You have an extra context bound (and even extra type parameter).

使用

implicit def noneEncoder/*[A: Encoder]*/: Encoder[None.type] =
  (_: None.type) => ""

它编译:

show[Option[String]](None)
show[None.type](None)
show(None)

您最初对 noneEncoder 的定义意味着您为 None.type 拥有一个 Encoder 实例，前提是您有一些实例.> A (不受限制，即可以推断).通常，如果您具有唯一的隐式(或至少只有唯一的更高优先级的隐式)，则此方法有效.例如，如果只有 stringEncoder 和原始的 noneEncoder ，则 show [None.type](None)和 show(None)即可编译.

Your original definition of noneEncoder meant that you had an instance of Encoder for None.type provided you had an instance for some A (not constrained i.e. to be inferred). Normally this works if you have the only implicit (or at least the only higher-priority implicit). For example if there were only stringEncoder and original noneEncoder then show[None.type](None) and show(None) would compile.

关于2.

PROS .使用反变的 Encoder

trait Encoder[-A] {
  def encode(a: A): String
}

您可以删除 someEncoder 和 noneEncoder ， optionEncoder 就足够了

you can remove someEncoder and noneEncoder, optionEncoder will be enough

show(Some("a"))
show[Option[String]](Some("a"))
show[Option[String]](None)
show[None.type](None)
show(None)

CONS..有人认为协变类型类的行为与直觉相反:

CONS. Some people believe that contravariant type classes behave counterintuitively:

https://github.com/scala/bug/issues/2509

https://groups.google.com/g/scala-language/c/ZE83TvSWpT4/m/YiwJJLZRmlcJ

也可能相关:在Scala 2.13中，如何隐式使用[值单例类型]?

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