1 :: List[Nothing] in foldLeft - scala

Question

If:
scala> val l = List() // List() same as List[Nothing]()
l: List[Nothing] = List()
scala> 1 :: l
res0: List[Int] = List(1)
or:
scala> 1 :: List[Nothing]()
res6: List[Int] = List(1)
Why then this does not work out:
scala> List(1,2,3). foldLeft( List() ) ((acc,x) => x :: acc)
So I have to type this explicitly List[Int]():
scala> List(1,2,3). foldLeft( List[Int]() ) ((acc,x) => x :: acc)
res3: List[Int] = List(3, 2, 1)
?
Though it does in Haskell, for example:
foldl (\acc x -> x:acc) [] [1,2,3]

Answer 1

Let's look at scala's foldLeft signature:
List[+A].foldLeft[B](z: B)(f: (B, A) ⇒ B): B
and haskell's signature:
foldl :: (b -> a -> b) -> b -> [a] -> b
They pretty much same, but:
1) scala has problem with type inference between [pseudo-]curried parameter lists, just compare:
scala> def aaa[A](a: A)(b: A) = {}
aaa: [A](a: A)(b: A)Unit
scala> aaa(null: Any)(5)
scala> aaa(5)(null: Any)
<console>:21: error: type mismatch;
found : Any
required: Int
aaa(5)(null: Any)
^
So scala can choose bigger type from left to right only.
More than that, this is a problem only for [pseudo-]curried functions:
scala> def aaa[T](a: T, b: T) = a
aaa: [T](a: T, b: T)T
scala> aaa(List("a"), List(6.0))
res26: List[Any] = List(a)
scala> aaa(List(6.0), List("a"))
res27: List[Any] = List(6.0)
Here scala not only picked a bigger type - it found a common supertype of both T's. So, it's trying to choose bigger type (if it's in the left part) by default, but looking for a common supertype inside one parameter list.
Note: I'm talking about [pseudo-] currying, as method with multiple parameter lists (B)((B, A) => B)B becoming curried function only after eta-expansion: foldLeft _ gives B => (B,A) => B
2) haskell uses "object" of List itself as function's parameter, which allows you to do even:
Prelude> let f = foldl (\acc x -> x:acc) []
:: [a] -> [a] //here is the polymorphic function
Prelude> f [1,2,3]
[3,2,1]
Prelude> f ["1","2","3"]
["3","2","1"]
In scala you need:
scala> def f[T](x: List[T]) = x.foldLeft(List[T]()) ((acc,x) => x :: acc)
f: [T](x: List[T])List[T]
scala> f(List(1,2,3))
res3: List[Int] = List(3, 2, 1)
scala> f(List("1","2","3"))
res3: List[String] = List(3, 2, 1)
3) Finally, let's rewrite foldLeft and place monoid's 'add' and 'identity' to the same parameter list (to avoid separate inference from p.1):
def foldLeft[T, U](l: List[T])(identity: U, add: (U,T) => U) = l.foldLeft(identity)(add)
and define polymorphic add operation:
scala> def add[A](x: List[A], y: A) = y :: x
add: [A](x: List[A], y: A)List[A]
So you can:
scala> foldLeft(List(1,2,3))(Nil, add)
res63: List[Int] = List(3, 2, 1)
in comparision with:
scala> List(1,2,3).foldLeft(Nil)(add)
<console>:9: error: polymorphic expression cannot be instantiated to expected type;
found : [A, B](x: List[A], y: A)List[A]
required: (scala.collection.immutable.Nil.type, Int) => scala.collection.immutable.Nil.type
List(1,2,3).foldLeft(Nil)(add)
^
Unfortunately, scala can't infer generic type for lambdas, so you can't:
scala> foldLeft(List(1,2,3))(Nil, (acc,x) => x :: acc)
<console>:10: error: missing parameter type
foldLeft(List(1,2,3))(Nil, (acc,x) => x :: acc)
as you can't:
scala> val a = (acc,x) => x :: acc
<console>:7: error: missing parameter type
val a = (acc,x) => x :: acc
^
2 & 3) Because scala has no polymorphic lambdas at all. Can't infer A => List[A] => A (where A is a type parameter) from (acc,x) => x :: acc (even A => A from val a = (a) => a), but Haskell can:
Prelude> let lambda = \acc x -> x:acc
:: [a] -> a -> [a]
Prelude> let f = foldl(lambda) []
Prelude> f [1,2,3]
[3,2,1]
Here is an eta-expansion of perviously defined add generic method in scala:
scala> add _
res2: (List[Nothing], Nothing) => List[Nothing] = <function2>

Answer 2

def foldLeft[B](z: B)(f: (B, A) => B): B
Nothing is a sub-type of every other type, in this case Int. Since List() is inferred to have type List[Nothing], then f is expected to be (List[Nothing], A) => List[Nothing]
But in the function (acc, x) => x :: acc) , A is an Int, which means you should have:
(List[Nothing], Int) => List[Nothing]
When really you have:
(List[Nothing], Int) => List[Int]
And thus the type mismatch, because List[Int] can't be a List[Nothing].
This is similar to:
class A
class B extends A
scala> List.fill(5)(new A).foldLeft(List.empty[B])((acc, x) => x :: acc)
<console>:10: error: type mismatch;
found : A
required: B
List.fill(5)(new A).foldLeft(List.empty[B])((acc, x) => x :: acc)
^