在scala中使用foldLeft反转

Question

def foldLeft[A, B] (as: List[A], z: B) (f: (B, A) => B) : B = as match {
  case Nil => z
  case Cons(x, xs) => foldLeft(xs, f(z, x))(f)
}

def reverse[A] (as: List[A]): List[A] =
  foldLeft(as, List[A]())((h, acc) => Cons(acc, h))

I am not sure how List[A] in foldLeft is of type B. Can anyone clear the process happening in this functions?

Answer 1

This reverse implementation is calling foldLeft with A as it's first type argument ( foldLeft#A = A ) and List[A] as it's second type argument ( foldLeft#B = List[A] ). Here is a type annotated version that makes this very explicit:

def reverse[A] (as: List[A]): List[A] =
  foldLeft[A, List[A]](as = as: List[A], z = List[A]())(
    (h: List[A], acc: A) => Cons(acc, h): List[A]
  )

Answer 2

Also Cons (if it is a Cons from standard library) creates a stream instead of list. Probably You want to use :: instead:

def reverse[A] (as: List[A]): List[A] =
    foldLeft(as, List[A]())((acc, h) => h :: acc)