将无形的可扩展记录传递给函数 [英] Passing a Shapeless Extensible Record to a Function
问题描述
我正在尝试学习 Shapeless(使用 2.10.2 版).我创建了一个非常简单的可扩展记录:
I am trying to learn Shapeless (using version 2.10.2). I have created a very simple extensible record:
val rec1 = ("foo" ->> 42) :: HNil
根据 REPL，这有类型
According to the REPL, this has type
shapeless.::[Int with shapeless.record.KeyTag[String("foo"),Int],shapeless.HNil]
我正在尝试定义一个简单的函数:
I am trying to define a simple function:
def fun(x: ::[Int with KeyTag[String("foo"), Int], HNil]) = x("foo")
但它甚至无法编译.我不能在类型声明中使用 String("foo") 并得到一个错误.
but it does not even compile. I cannot use a String("foo") in the type declaration, and get an error.
我有两个问题:
- 如何在代码中指定可扩展记录的类型?
- 处理具有更多字段的记录时，类型声明的长度和复杂性将难以管理.有没有办法为该类型创建别名(给定记录的特定实例)或其他一些解决方法?
编辑
我发现:
val rec1 = ("foo" ->> 42) :: HNil
val rec2 = ("foo" ->> 43) :: HNil
var x = rec1
x = rec2
效果很好.我得出的结论是 rec1、rec2 和 x 属于同一类型.我只是不知道如何在代码中表达这种类型！
works well. I conclude rec1, rec2, and x are of the same type. I just don't know how to express that type in code!
推荐答案
这里有一些更笼统的东西，我认为可以回答您的问题.假设我们要编写一个方法，该方法可以处理带有 "foo" 键的任何记录.我们可以结合使用见证和选择器:
Here's something a little more general that I think might answer your question. Suppose we want to write a method that will work on any record with a "foo" key. We can use a combination of a witness and a selector:
import shapeless._, record._, syntax.singleton._
val w = Witness("foo")
def fun[L <: HList](xs: L)(implicit sel: ops.record.Selector[L, w.T]) = xs("foo")
然后:
scala> fun(("foo" ->> 42) :: HNil)
res0: Int = 42
或者:
scala> fun(("bar" ->> 'a) :: ("foo" ->> 42) :: HNil)
res1: Int = 42
如果我们真的只想允许没有其他字段的记录，我们可以这样写:
If we really wanted to only allow records with no other fields, we could write the following:
def fun(l: Int with KeyTag[w.T, Int] :: HNil) = l("foo")
但这与通常使用记录的方式有些矛盾.
But that's somewhat at odds with the way records are generally used.
我们必须精确地定义见证，因为 Scala 2.10 没有提供任何直接引用单例类型的方法——例如参见 my fork Alois Cochard 的 Shona 项目进行了一些讨论.
We have to define the witness precisely because Scala 2.10 doesn't provide any way to refer to a singleton type directly—see for example my fork of Alois Cochard's Shona project for some discussion.
作为最后的免责声明，我要补充一点，我自己才刚刚熟悉 Shapeless 2.0，但我认为即使是 Miles 也没有足够的魔力来解决这个限制.
I will add as a final disclaimer that I'm only just now getting familiar with Shapeless 2.0 myself, but I don't think even Miles is magical enough to get around this limitation.
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