将示例从 Haskell 翻译成 Scala(HKT,类型约束)

Translate a sample from Haskell to Scala (HKT, type constraints)

我正在学习函数式编程,我在 Haskell 中制作了这个示例,它按我想要的方式工作,但是当我不知道如何在 Scala 中进行此类约束时,我不明白我是如何做到的可以在 Scala atm 中使用 HKT 和约束。

{-# LANGUAGE GADTs #-}

module Complex
  ( Complex
  , add
  ) where

data Complex a where
  Complex :: (Show a, Fractional a) => a -> a -> Complex a

instance Show (Complex a) where
  show (Complex a b) = "z = " ++ show a ++ " + i * " ++ show b

add :: Complex a -> Complex a -> Complex a
add (Complex a b) (Complex c d) = Complex (a + c) (b + d)

提前致谢:)

所以,我做到了,但似乎有点不对,我可以做得更好吗?

  case class ComplexNumber[T](realPart: T, imagPart: T){
    override def toString: String = s"z= $realPart + $imagPart i"
  }

  object ComplexNumber {
    def add[T](a: ComplexNumber[T], b: ComplexNumber[T])(implicit evidence: Numeric[T]): ComplexNumber[T] = {
      ComplexNumber(evidence.plus(a.realPart, b.realPart), evidence.plus(a.imagPart, b.imagPart))
    }
  }

我不得不使用这个证据助手似乎很奇怪..

希望这有帮助,但不能让对象泛型感到遗憾

case class Complex[T](real: T, img: T) {
    override def toString: String = s"z= $real + $img i"
}

object Complex {
    def add(cmplx: Complex[Double], cmplx2: Complex[Double]): Complex[Double]
        = new Complex[Double](cmplx.real + cmplx2.real, cmplx.img + cmplx2.img)
}

val c1 = Complex[Double](3, 4)
val c2 = Complex[Double](5, 1)

println(Complex.add(c1, c2))

"evidence helper" 保证类型参数 T 仅限于可以加在一起(或任何其他算术运算)的数字类型。

它可以隐藏在一些语法糖后面,使 T 成为 "context bound" 类型参数。您还可以引入一些额外的隐式以使加法语法更自然。

object ComplexNumber {
  import Numeric.Implicits._  //can be placed elsewhere in the file
  def add[T: Numeric](a: ComplexNumber[T], b: ComplexNumber[T]):ComplexNumber[T] =
    ComplexNumber(a.realPart + b.realPart, a.imagPart + b.imagPart)
}

您还可以使用更自然的客户端语法。

import Numeric.Implicits._
case class ComplexNumber[T: Numeric](realPart: T, imagPart: T){
  override def toString: String = s"z= $realPart + $imagPart i"
  def +(that: ComplexNumber[T]):ComplexNumber[T] =
    ComplexNumber(this.realPart + that.realPart, this.imagPart + that.imagPart)
}

用法:

ComplexNumber(8,1) + ComplexNumber(4,4)  //res0: ComplexNumber[Int] = z= 12 + 5 i