Applying an argument list to curried function using foldLeft in Scala

Question

Is it possible to do a foldLeft on a list of arguments, where the initial value supplied to the fold is a fully curried function, the operator is apply

Answer 1

Ok no scalaz and no solution but an explanation. If you use your f.curried.apply with 1 and then with 2 arguments in the REPL observe the return-result types DO actually differ each time! FoldLeft is quite simple. It's fixed in it's type with your starting argument which is f.curried and since that has not the same signature as f.curried.apply(1) it doesn't work. So the starting argument and the result HAVE to be of the same type. The type hast to be consistent for the starting-sum element of foldLeft. And your result would be even Int so that would absolutely not work. Hope this helps.

Answer 2

Your function expects exactly 4 Int arguments. foldLeft is a function that applies to an arbitrary number of elements. You mention List(1,2,3,4) but what if you have List(1,2,3,4,5) or List()?

List.foldLeft[B] also expects a function to return the same type B, but in your case Int and some Function1[Int, _] is not the same type.

Whatever solution you come up with would not be general either. For instance what if your function is of type (Int, Float, Int, String) => Int? You would then need a List[Any]

So it's definitely not a job for List.foldLeft.

With that in mind (warning very un-scala code):

class Acc[T](f: Function1[T, _]) {
  private[this] var ff: Any = f
  def apply(t: T): this.type = {
    ff = ff.asInstanceOf[Function1[T,_]](t)
    this
  }
  def get = ff match { 
    case _: Function1[_,_] => sys.error("not enough arguments")
    case res => res.asInstanceOf[T]
  }
}

List(1,2,3,4).foldLeft(new Acc(f.curried))((acc, i) => acc(i)).get
// res10: Int = 10

Answer 3

This turns out to be quite a bit simpler than I initially expected.

First we need to define a simple HList,

sealed trait HList

final case class HCons[H, T <: HList](head : H, tail : T) extends HList {
  def ::[H1](h : H1) = HCons(h, this)
  override def toString = head+" :: "+tail.toString
}

trait HNil extends HList {
  def ::[H1](h : H1) = HCons(h, this)
  override def toString = "HNil"
}

case object HNil extends HNil
type ::[H, T <: HList] = HCons[H, T]

Then we can define our fold-like function inductively with the aid of a type class,

trait FoldCurry[L <: HList, F, Out] {
  def apply(l : L, f : F) : Out
}

// Base case for HLists of length one
implicit def foldCurry1[H, Out] = new FoldCurry[H :: HNil, H => Out, Out] {
  def apply(l : H :: HNil, f : H => Out) = f(l.head)
}

// Case for HLists of length n+1
implicit def foldCurry2[H, T <: HList, FT, Out]
  (implicit fct : FoldCurry[T, FT, Out]) = new FoldCurry[H :: T, H => FT, Out] {
    def apply(l : H :: T, f : H => FT) = fct(l.tail, f(l.head))
}

// Public interface ... implemented in terms of type class and instances above
def foldCurry[L <: HList, F, Out](l : L, f : F)
  (implicit fc : FoldCurry[L, F, Out]) : Out = fc(l, f)

We can use it like this, first for your original example,

val f1 = (i : Int, j : Int, k : Int, l : Int) => i+j+k+l
val f1c = f1.curried

val l1 = 1 :: 2 :: 3 :: 4 :: HNil

// In the REPL ... note the inferred result type
scala> foldCurry(l1, f1c)
res0: Int = 10

And we can also use the same unmodified foldCurry for functions with different arity's and non-uniform argument types,

val f2 = (i : Int, s : String, d : Double) => (i+1, s.length, d*2)
val f2c = f2.curried

val l2 = 23 :: "foo" :: 2.0 :: HNil

// In the REPL ... again, note the inferred result type
scala> foldCurry(l2, f2c)
res1: (Int, Int, Double) = (24,3,4.0)