从 HList 中获取元素 [英] Getting elements from an HList

问题描述

我玩弄了 HList 并按预期工作:

I toyed around with HList and the following works as expected:

val hl = 1 :: "foo" :: HNil
val i: Int = hl(_0)
val s: String = hl(_1)

但是，我无法让以下代码工作(让我们暂时假设随机访问列表是一个聪明的主意;-)):

However, I can't get the following piece of code working (let's assume for a moment random access on lists is a smart idea ;-)):

class Container(hl: HList) {
    def get(n: Nat) = hl(n)
}

val container = new Container(1 :: "foo" :: HNil)
val i: Int = container.get(_0)
val s: String = container.get(_1)

我想让 get 根据它的参数返回一个 Int 和 String.我假设，如果可能的话，我必须使用 Aux 或 at 但我不知道如何做到这一点.

I'd like to have get return an Int and String according to it's parameter. I assume, if possible at all, I have to use Aux or at but I'm not sure how to do this.

推荐答案

尝试这些方法，

scala> import shapeless._, nat._, ops.hlist._
import shapeless._
import nat._
import ops.hlist._

scala> class Container[L <: HList](hl: L) {
     |   def get(n: Nat)(implicit at: At[L, n.N]): at.Out = hl[n.N]
     | }
defined class Container

scala> val container = new Container(1 :: "foo" :: HNil)
container: Container[shapeless.::[Int,shapeless.::[String,shapeless.HNil]]] = ...

scala> container.get(_0)
res1: Int = 1

scala> container.get(_1)
res2: String = foo

这里的第一个重要区别是，我们没有将 hl 输入为普通的 HList，这会丢失有关元素类型的所有特定信息，而是对精确类型进行参数化并保留其结构为 L.第二个区别是我们使用L来索引隐含的At类型的类实例，用于在get中执行索引.

The first crucial difference here is that rather than typing hl as plain HList, which loses all specific information about the types of the elements, we parametrize over the precise type of the argument and preserve its structure as L. The second difference is that we use L to index the implicit At type class instance which is used to perform the indexing in get.

另请注意，由于存在从 Int 文字到 Nat 的隐式转换，您可以编写，

Also note that because there is an implicit conversion from Int literals to Nat's you can write,

scala> container.get(0)
res3: Int = 1

scala> container.get(1)
res4: String = foo

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