Scala:用两种类型参数化的类型的简写

Question

Scala允许使用以下类型:

=:=[Int, String] 

如:

Int=:=String

此功能似乎不限于此类型,我也可以举例如:

type Or[A,B] = Either[A,B]
val x: Int Or String = Right("value")

这是怎么回事？

Answer 1

它的工作原理如说明书中所述.如果这是过于正式和过于抽象,这里只是简单说明它的内容.

1. 您可以对中缀类型使用任意标识符,因此以下所有定义都是有效的:

   type foobar[X, Y] = (X, Y)
   type <=[X, Y] = Y => X
   type !+?[X, Y] = (X, Y)
   type `or failure`[X, Err] = scala.util.Either[Err, X]
   
   val x: Int foobar String = (42, "hello world")
   val y: String <= Int = n => "#" * n
   val z: Int !+? Float = (42, 3.1415f)
   val w: Int `or failure` String = scala.util.Right(42)
   

2. 中缀类型的参数不仅限于种类*,它也适用于更高级的参数:

   type of[F[_], X] = F[X]
   val l1: List of Int = List(42)
   

3. 没有运算符优先级.中缀类型都向左或向右关联:

   type +[A, B] = scala.util.Either[A, B]
   type *[A, B] = (A, B)
   
   // It's     ((Int * String) + Float) * Double
   // It's NOT (Int * String) + (Float * Double)
   val a: Int * String + Float * Double =
     (scala.util.Left((42, "foo")), 1.0d)
   

4. :以右侧关联的中缀类型:

   type -:[A, B] = B => A
   
   val f: String -: Int -: Double =
     (g: (Double => Int)) => "foo" * g(42d)
   
   // Not: (g: Double) => (i: Int) => "foo"
   

5. 左关联和右关联中缀运算符不能混合:

   // error: left- and right-associative operators
   // with same precedence may not be mixed
   val wontCompile: Int * Int -: Int = i => (i, i)