如何正确定义lambda类型？

Question

我使用=:=示例类型lambda来制作简单的最小例子.

=:= 类型带两个参数,我想在类型级别讨好一个.

我采取天真的实现,type Curry[G] = {type l[L] = L =:= G}但在实际使用中它会导致错误:

type X = Int
type Y = Int

type CurryInt[T] = T =:= Int
type Curry[G] = {type l[L] = L =:= G}
type CurrStatic = {type l[L] = L =:= Int}
object CurryObj {type l[L] = L =:= Int}

trait Apply[P[_], T]
implicit def liftApply[P[_], T](implicit ev : P[T]) = new Apply[P,T] {}

implicitly[Apply[CurryInt, Y]] // ok
implicitly[Apply[Curry[X]#l, Y]] // fails
implicitly[Apply[Curry[X]#l, Y]](liftApply) // fails
implicitly[Apply[Curry[X]#l, Y]](liftApply[Curry[X]#l, Y]) // ok
implicitly[Apply[CurrStatic#l, Y]] // fails
implicitly[Apply[CurryObj.l, Y]] // ok

类型推断在这里打破.我应该如何定义类型lambdas以使其工作？

Answer 1

稍微详细一些，但是可以编译:) (scala 2.12.3)

  type X = Int
  type Y = Int

  type CurryInt[T] = T =:= Int
  type Curry[G] = {type l[L] = =:=[L, G]}
  type CurrStatic = {type l[L] = L =:= Int}
  object CurryObj {type l[L] = L =:= Int}

  trait Apply[P[_], T]
  implicit def liftApply[P[_], T](implicit ev : P[T]) = new Apply[P,T] {}


  type L1[R] = =:=[R, X]
  type L2[R] = =:=[R, Int]
  implicitly[Apply[CurryInt, Y]] // ok
  implicitly[Apply[L1, Y]] // ok
  implicitly[Apply[L1, Y]](liftApply[L1, Y]) // ok
  implicitly[Apply[Curry[X]#l, Y]](liftApply[Curry[X]#l, Y]) // ok
  implicitly[Apply[L2, Y]] // ok
  implicitly[Apply[CurryObj.l, Y]] // ok