是否可以通过单个HList定义多个隐式证据？

Question

我有一段代码，概念上类似于以下代码：

//library code
trait Support[K, V]

def partialHandler[K, V](key: K, value: V)(implicit ev: Support[K, V]) = ???

//user code
implicit val intIntSupport = new Support[Int, Int] {}
implicit val intStringSupport = new Support[Int, String] {}
...

partialHandler(1, "foo)
partialHandler(1, 1)

[我想知道是否有一种方法可以让该库的用户更优雅地定义受支持的(K, V)类型，例如：

val supportedTypes = new Support[Int, Int] {} :: new Support[Int, String] {} :: HNil

（（本质上，我正在寻找从几乎未知的HList到Support[K, V]的隐式转换。这看起来不可行，但也许我遗漏了某些东西。）]

Answer 1

尝试隐含supportedTypes

import shapeless.ops.hlist.Selector
import shapeless.{HList, HNil}

// library code
trait Support[K, V]

object Support {
  implicit def mkSupport[L <: HList, K, V](implicit l: L, sel: Selector[L, Support[K, V]]): Support[K, V] = null
}

def partialHandler[K, V](key: K, value: V)(implicit ev: Support[K, V]) = ???

//user code
implicit val supportedTypes = new Support[Int, Int] {} :: new Support[Int, String] {} :: new Support[Long, Double] {} :: HNil

partialHandler(1, "foo")
partialHandler(1, 1)
partialHandler(1L, 1.0)
// partialHandler("foo", "bar") // doesn't compile