sweetyBaby sweetyBaby - 1 year ago 116
Scala Question

What 's the difference between foldRight and foldLeft in concat

Why I cannot use fold Left in the following code:

def concatList[T](xs: List[T],ys:List[T]): List[T]=
(xs foldLeft ys)(_::_)

Actually it is hard for me to understand the differences between foldRight and foldLeft, it there any examples to illustrate the real differences?


Answer Source

Well, you can,

scala> def concatList[T](xs: List[T],ys:List[T]) = 
         (xs foldLeft ys)( (a, b) => b :: a )
concatList: [T](xs: List[T], ys: List[T])List[T]

scala> concatList(List(1,2,3), List(6,7,8))
res0: List[Int] = List(3, 2, 1, 6, 7, 8)

Was that the result you were expecting? I don't think so.

First let's look at the signature of the folds and :: (only a simplification for illustrative purposes, but fits perfectly in our case) :

given a List[T]
  def ::(v:T): List[T] // This is a right associative method, more below
  def foldLeft[R](r:R)(f: (R,T) => R):R
  def foldRight[R](r:R)(f: (T,R) => R):R

Now, apply one argument list in foldLeft we xs.foldLeft(ys) and unifying the types from our signature from foldLeft sample call:

List[T] : List[Int], therefore T : Int, and R : List[Int], that applied to foldLeft signature gives

foldLeft[List[Int]](r:List[Int])( f:(List[Int],Int) => List[Int] )

Now, for the usage of ::, a :: b compiles to b.::(a), Scala often refers to it as a right associative method. This a special syntax sugar for methods ending in : and quite convenient when defining a list: 1 :: 2 :: Nil is like writing Nil.::(2).::(1).

Continuing our instantiation of foldLeft, the function we need to pass has to look like this: (List[Int],Int) => List[Int]. Consider (a,b) => a :: b, if we unify that with the type of we f get:

a : List[Int], and b : Int, comparing that with the signature of a2 :: b2, a2 : Int, b2 : List[Int]. For this to compile, a and a2 in conjunction with b and b2 must have the same types each. Which they don't!

Notice in my example, I inverted the arguments, making a match the type of b2 and b match the type of a2.

I will offer yet another version that compiles:

def concatList[T](xs: List[T],ys:List[T]) = (xs foldLeft ys)( _.::(_) )

To cut the story short, look at foldRight signature

def foldRight[R](r:R)(f: (T,R) => R):R

The arguments are already inverted, so making f = _ :: _ gives us the right types.

Wow, that was a lot of explanation about type inference and I am sort on time, but I still owe a explanation on difference between the meaning of fold left and right. For now have a look at https://wiki.haskell.org/Fold, in special these two imagines:

foldLeft foldRight

Notice, the argument to foldl and foldr are inverted, it first takes the function and them the inital arguments, r in the signatures, and instead of :: for list construction it uses just :. Two very small details.

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