有没有办法保证 Scala 中的类型类存在 case 类复制方法? [英] Is there a way to guarantee case class copy methods exist with type classes in Scala?

问题描述

在下面的示例中，我有一个类型类 Foo，并且想以某种方式保证所有成员都符合 Foo(例如 Bar> 通过 barFoo) 有一个复制方法，例如通过 case 类 生成的方法.我还没有想到这样做的方法.在这种情况下，复制签名可能类似于 copy(foo: F, aa: List[T] = foo.aa,maybe: Option[T] = foo.maybe): F.

In the example below I have a type class Foo, and would like to somehow guarantee that all members conforming to Foo (such as Bar via barFoo) have a copy method such as the one generated by way of being a case class. I haven't thought of a way to do this. In this case the copy signature would be possibly be something like copy(foo: F, aa: List[T] = foo.aa, maybe: Option[T] = foo.maybe): F.

trait Foo[F] {
  type T
  def aa(foo: F): List[T]
  def maybe(foo: F): Option[T]
}

final case class Bar(aa: List[String], maybe: Option[String])
object Bar {
  implicit val barFoo = new Foo[Bar] {
    type T = String
    def aa(foo: Bar): List[String] = foo.aa
    def maybe(foo: Bar): Option[T] = foo.maybe
  }
}

推荐答案

我不能用类型成员来做，但这里有一个带有类型参数的版本.另外需要给Foo添加一个方法来构造对象.

I couldn't do it with type member, but here you are a version with type parameters. Also it is necessary to add a method to Foo to construct the object.

trait Foo[F, T] {
  def aa(foo: F): List[T]
  def maybe(foo: F): Option[T]
  def instance(aa: List[T], maybe: Option[T]): F
}

class Bar(val aa: List[String], val maybe: Option[String]) {
  override def toString = s"Bar($aa, $maybe)"
}

object Bar {
  implicit val barFoo = new Foo[Bar, String] {
    def aa(foo: Bar): List[String] = foo.aa
    def maybe(foo: Bar): Option[String] = foo.maybe
    def instance(aa: List[String], maybe:Option[String]):Bar = new Bar(aa, maybe)
  }
}

implicit class FooOps[A, T](fooable:A)(implicit foo:Foo[A, T]) {

  def copy(aa: List[T] = foo.aa(fooable), maybe: Option[T] = foo.maybe(fooable)) = {
    foo.instance(aa, maybe)
  }

}

val r = new Bar(List(""), Option("")).copy(aa = List("asd"))

println(r)

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