我在寻找与下面的OCaml代码相同的行为,其中编译器理解匹配是穷举的,因为我们已经表示了两个scuteine必须具有相同的类型:
type circle
type rectangle
type _ figure =
| Circle : int -> circle figure
| Rectangle : int * int -> rectangle figure
let equal_figure : type a. a figure -> a figure -> bool = fun f1 f2 -> match (f1, f2) with
| Circle r1, Circle r2 -> Int.(r1 = r2)
| Rectangle (x1, y1), Rectangle (x2, y2) -> Int.(x1 = x2 && y1 = y2)
(* the compiler knows this match is exhaustive *)
我可以将这个例子直接移植到Scala,穷举检查器会做正确的事情:
sealed trait CircleMarker
sealed trait RectangleMarker
enum Fig[T]:
case Circle(r: Int) extends Fig[CircleMarker]
case Rectangle(x: Int, y: Int) extends Fig[RectangleMarker]
def equalFig[T](f1: Fig[T], f2: Fig[T]): Boolean = (f1, f2) match
case (Fig.Circle(r1), Fig.Circle(r2)) => r1 == r2
case (Fig.Rectangle(x1, y1), Fig.Rectangle(x2, y2)) => x1 == x2 && y1 == y2
(* the compiler knows this match is exhaustive *)
scastie
在Scala中,有没有更简洁的方式来表达这一点,而不使用虚幻的CircleMarker
和RectangleMarker
特性?
1条答案
按热度按时间t2a7ltrp1#
你可以试试F界限