scala - '1.narrow' 的类型

Question

查看 Singleton Types :

import shapeless._, syntax.singleton._

scala> 1.narrow
res3: Int(1) = 1

我试图编写一个函数，给定一个单例1，即根据上述内容，返回???:

scala> def f(a: Int(1)): Unit = ???
<console>:1: error: ')' expected but '(' found.
def f(a: Int(1)): Unit = ???
            ^
<console>:1: error: '=' expected but ')' found.
def f(a: Int(1)): Unit = ???
               ^

但是编译失败。

1.narrow 的类型是什么以及如何在函数中使用它？

Answer 1

您可以通过 shapeless.Witness 语法使用此类型:

def f(a: Witness.`1`.T): Unit = ???

val a = 1.narrow
f(a) // compiles

val b = 2.narrow
f(b) // type mismatch; found: Int(2) required: Int(1)

val c = 1
f(c) // type mismatch; found: c.type (with underlying type Int) required: Int(1)

还有一个Typelevel Scala编译器分支，支持类型中的文字(使用适当的 compiler flag ):

def f(a: 1): Unit = ???