Scala中foldLeft的实现

Question

TraversableOnce使用可变var result实现foldLeft 。

def foldLeft[B](z: B)(op: (B, A) => B): B = {
   var result = z
   this foreach (x => result = op(result, x))
   result
}

我知道递归地实现foldLeft是不切实际的。 现在我想知道是否有可能在没有可变变量的情况下有效地实现foldLeft 。

可以吗？ 为什么不能呢？

Answer 1

尾递归是你的朋友：

def foldLeft[A, B](xs: Seq[A], z: B)(op: (B, A) => B): B = {
  def f(xs: Seq[A], acc: B): B = xs match {
    case Seq()   => acc
    case x +: xs => f(xs, op(acc, x))
  }
  f(xs, z)
}

顺便说一句， TraversableOnce不实现head或tail ，访问元素的唯一方法是使用foreach 。

Answer 2

object FoldImplement:
  def myFoldLeft(lst: List[Int])(acc: Int)(f: (Int, Int)=>Int): Int =
    lst match
      case List() => acc
      case hd::tl => myFoldLeft(tl)(f(hd,acc))(f)

  @main def runFoldImpl =
    println(myFoldLeft(List(1,3,5))(0)((acc,elem)=>acc+elem))

Answer 3

def foldLeft[B](z: B)(op: (B, A) => B): B = {
  val thislist = this.toList
  @tailrec
  def myFold(result: B, list: List[A]): B = list match {
    case Nil => result
    case head :: tail => myFold(op(result,head), tail)
  }
  myFold(z, thislist)
}