获取“单身类型"类型. [英] Get type of a "singleton type"

问题描述

我们可以通过无形创建文字类型:

We can create a literal types via shapeless:

import shapeless.syntax.singleton._
var x = 42.narrow
// x: Int(42) = 42

但是如果什至不可能创建类型别名，该如何使用Int(42)作为类型呢?

But how can I operate with Int(42) as a type if it's even impossible to create type alias

type Answ = Int(42) // won't compile
// or
def doSmth(value: Int(42)) = ... // won't compile

推荐答案

1)在 Typelevel Scala 中，您可以只写

1) In Typelevel Scala you can write just

val x: 42 = 42

type Answ = 42

def doSmth(value: 42) = ???

2)在 Dotty Scala中，您可以编写同样的文字.

2) In Dotty Scala you can write the same.

3)在 Lightbend Scala (即标准Scala)+ Shapeless中，您可以编写

3) In Lightbend Scala (i.e. standard Scala) + Shapeless you can write

import shapeless.Witness
import shapeless.syntax.singleton._

val x: Witness.`42`.T = 42.narrow

type Answ = Witness.`42`.T

def doSmth(value: Witness.`42`.T) = ???

在情况1)build.sbt应该是

In case 1) build.sbt should be

scalaOrganization := "org.typelevel"
scalaVersion := "2.12.3-bin-typelevel-4"
scalacOptions += "-Yliteral-types"

在情况2)build.sbt应该是

In case 2) build.sbt should be

scalaOrganization := "ch.epfl.lamp"
scalaVersion := "0.3.0-RC2"

和plugins.sbt

and plugins.sbt

addSbtPlugin("ch.epfl.lamp" % "sbt-dotty" % "0.1.5")

在情况3)build.sbt应该是

In case 3) build.sbt should be

scalaOrganization := "org.scala-lang"
scalaVersion := "2.12.3"
libraryDependencies += "com.chuusai" %% "shapeless" % "2.3.2"

4)，或者您可以同时使用Typelevel Scala和Shapeless.

4) or you can use Typelevel Scala and Shapeless at the same time.

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