我想要一个A|B类型作为A|B|C. 这可以在 Scala 中编码吗？如果是，如何？
我希望我可以在implicitly[¬¬[IF] <:< T]下面进行编译（这里的原始代码），但事实并非如此。有没有办法修复此代码以允许子类型化？
object NUnion{
type ¬¬[A] = ¬[¬[A]]
type ¬[A] = A => Nothing
trait Disj[T] {
type or[S] = Disj[T with ¬[S]]
type apply = ¬[T]
}
// for convenience
type disj[T] = { type or[S] = Disj[¬[T]]#or[S] }
type T = disj[Int]#or[Float]#or[String]#apply
type IF = disj[Int]#or[Float]#apply
implicitly[¬¬[Int] <:< T] // works
// implicitly[¬¬[Double] <:< T] // doesn't work
// implicitly[¬¬[IF] <:< T] // doesn't work - but it should
}
我也试过这个（从这里）：
object Kerr{
def f[A](a: A)(implicit ev: (Int with String with Boolean) <:< A) = a match {
case i: Int => i + 1
case s: String => s.length
}
f(1) //works
f("bla") // works
def g[R]()(implicit ev: (Int with String with Boolean) <:< R):R = "go" // does not work
}
但在这里我不能将联合类型设为“一等”，它们只能作为参数类型存在，不能作为返回类型存在。
这种方法同样的问题：
object Map{
object Union {
import scala.language.higherKinds
sealed trait ¬[-A]
sealed trait TSet {
type Compound[A]
type Map[F[_]] <: TSet
}
sealed trait ∅ extends TSet {
type Compound[A] = A
type Map[F[_]] = ∅
}
// Note that this type is left-associative for the sake of concision.
sealed trait ∨[T <: TSet, H] extends TSet {
// Given a type of the form `∅ ∨ A ∨ B ∨ ...` and parameter `X`, we want to produce the type
// `¬[A] with ¬[B] with ... <:< ¬[X]`.
type Member[X] = T#Map[¬]#Compound[¬[H]] <:< ¬[X]
// This could be generalized as a fold, but for concision we leave it as is.
type Compound[A] = T#Compound[H with A]
type Map[F[_]] = T#Map[F] ∨ F[H]
}
def foo[A : (∅ ∨ String ∨ Int ∨ List[Int])#Member](a: A): String = a match {
case s: String => "String"
case i: Int => "Int"
case l: List[_] => "List[Int]"
}
def geza[A : (∅ ∨ String ∨ Int ∨ List[Int])#Member] : A = "45" // does not work
foo(geza)
foo(42)
foo("bar")
foo(List(1, 2, 3))
// foo(42d) // error
// foo[Any](???) // error
}
}