构建 Nat N 的固定大小列表 [英] Building Fixed Size List of Nat N

问题描述

我试图定义一个函数，给定一个 N <: Nat 类型参数，它会构建一个正好包含 3 个 N 的 List.

I tried to define a function that, given a N <: Nat type parameter, builds a List with exactly 3 N's.

import shapeless._
import shapeless.nat._

scala> def natNOfSize3[N <: Nat](n: Nat): Sized[List[N], _3] = 
     Sized[List, _3](List(n, n, n))
<console>:17: error: wrong number of type parameters for overloaded method value apply with alternatives:
  [CC[_]]()(implicit cbf: scala.collection.generic.CanBuildFrom[Nothing,Nothing,CC[Nothing]], implicit ev: shapeless.AdditiveCollection[CC[Nothing]])shapeless.Sized[CC[Nothing],shapeless._0] <and>
  [CC[_]]=> shapeless.SizedBuilder[CC]

       def natNOfSize3[N <: Nat](n: Nat): Sized[List[N], _3] = Sized[List, _3](List(n, n, n))             ^

但我不明白为什么它失败了.

But I don't understand why it failed.

推荐答案

一个问题是你的 n 被输入为 Nat，而不是 N>——我认为这只是一个错字.一旦你解决了这个问题，你可以像这样编写方法:

One issue is that your n is typed as Nat, not N—I assume that's just a typo. Once you've fixed that, you can write the method like this:

import shapeless._, nat._

def natNOfSize3[N <: Nat](n: N): Sized[List[N], _3] = Sized[List](n, n, n)

请注意，Sized.apply 接受类型为 * -> 的单个类型参数.*，而不是提供集合，而是提供元素.

Note that Sized.apply takes a single type parameter of kind * -> *, and instead of providing a collection, you provide the elements.

如果你真的想传入一个集合，你可以使用wrap:

If you really want to pass in a collection, you could use wrap:

def natNOfSize3[N <: Nat](n: N): Sized[List[N], _3] = Sized.wrap(List(n, n, n))

但是，如果您对元素数量撒了谎，那么编译器将无法帮助您.

But then the compiler isn't going to be able to help you if you've lied about the number of elements.

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