scala - 类型依赖隐式

Question

函数中需要两个隐式，但我不能在相同的参数列表，因为我得到了 dependent method type。所以我考虑再次柯里化(Currying)，但这给了我一个语法错误。执行此操作的正确方法是什么？

def add[A](newAnnotations: Seq[A])
    (implicit maybeAdd: MaybeAdd[L, Seq[A]])
    (implicit mod: Modifier[maybeAdd.Out, Seq[A], Seq[A]]):
    Slab[Content, maybeAdd.Out] = {
  val l = maybeAdd(annotations, Seq[A]())
  l.updateWith(_ ++ newAnnotations)
}

Answer 1

我编辑了 MaybeAdd 以具有 Aux 类型，正如@milessabin 所建议的那样。

def add[A, Out0](newAnnotations: Seq[A])(implicit maybeAdd: MaybeAdd.Aux[L, Seq[A], Out0], mod: Modifier[Out0, Seq[A], Seq[A]]): Slab[Content, mod.Out] = {
  val l = maybeAdd(annotations, Seq[A]())
  new Slab(content, mod(l, _ ++ newAnnotations))
}