[英]Semigroup with function in Scala
我正在嘗試將Haskell Semigroup
轉換為Scala。 Haskell代碼可以正常工作,但我無法在Scala中編寫它
Haskell:
import Data.Semigroup
newtype Combine a b = Combine { unCombine :: (a -> b) }
instance Semigroup b => Semigroup (Combine a b) where
Combine f <> Combine g = Combine (f <> g)
f = Combine $ \n -> Sum (n + 1)
g = Combine $ \n -> Sum (n - 1)
print (unCombine (f <> g) $ 0) -- Sum 0
print (unCombine (f <> g) $ 10) -- Sum 20
Scala代碼
import cats.Semigroup
import cats.instances.all._
trait Combine[A, B] {
def unCombine(a: A): B
}
val f = new Combine[Int, Int] {
override def unCombine(n: Int): Int = n + 1
}
val g = new Combine[Int, Int] {
override def unCombine(n: Int): Int = n - 1
}
implicit val mySemigroup: Semigroup[Combine[Int, Int]] = new Semigroup[Combine[Int, Int]] {
def combine(x: Combine[Int, Int], y: Combine[Int, Int]): Combine[Int, Int] = (x,y) match {
// ???
}
}
除了由@KartikSabharwal答案,因為這兩個Semigroup
和Combine
是功能接口,因為Scala的2.12可以定義具體情況是這樣的:
implicit val mySemigroup: Semigroup[Combine[Int, Int]] =
(x, y) => a => x.unCombine(a) + y.unCombine(a)
@KartikSabharwal提到的通用案例在Scala 2.12中看起來像這樣:
// Don't forget to NOT import `cats.instances.all._` together with this import
import cats.implicits._
implicit def combineSemigroup[A, B](
implicit ev: Semigroup[B]
): Semigroup[Combine[A, B]] =
(x, y) => a => x.unCombine(a) combine y.unCombine(a)
像在Scala 2.11中這樣:
import cats.implicits._
implicit def combineSemigroup[A, B](
implicit ev: Semigroup[B]
): Semigroup[Combine[A, B]] =
new Semigroup[Combine[A, B]] {
override def combine(x: Combine[A, B], y: Combine[A, B]): Combine[A, B] =
new Combine[A, B] {
override def unCombine(a: A): B = x.unCombine(a) combine y.unCombine(a)
}
}
這是可以回答您特定問題的代碼。
import cats.Semigroup
import cats.instances.all._
object Main extends App {
trait Combine[A, B] {
def unCombine(a: A): B
}
override def main(args: Array[String]): Unit = {
implicit val mySemigroup: Semigroup[Combine[Int, Int]] =
new Semigroup[Combine[Int, Int]] {
def combine(x: Combine[Int, Int], y: Combine[Int, Int]): Combine[Int, Int] =
new Combine[Int, Int] {
override def unCombine(n: Int): Int =
Semigroup[Int].combine(x.unCombine(n), y.unCombine(n))
}
}
val f = new Combine[Int, Int] {
override def unCombine(n: Int): Int = n + 1
}
val g = new Combine[Int, Int] {
override def unCombine(n: Int): Int = n - 1
}
val example = Semigroup[Combine[Int, Int]].combine(f, g).unCombine(10)
printf("%d\n", example)
}
}
理想情況下,我想在本質上復制Haskell代碼並實現某種形式
// 'a' can be any type
// Semigroup[b] must exist
implicit val mySemigroup: Semigroup[Combine[a, b]] =
def combine(x: Combine[a, b], y: Combine[a, b]): Combine[a, b] =
new Combine[a, b] {
override def unCombine(n: a): b =
Semigroup[b].combine(x.unCombine(n), y.unCombine(n))
}
但是我對Scala的了解還不夠多。 找出答案后,我將對其進行更新,否則其他人可以跟進並編輯此答案/發布更好的答案。
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