Scala for-comprehension type inference - scala

Question

The next code
def f(chars: List[Char]): List[List[Char]] = chars match {
case Nil => List(Nil)
case x :: xs => for {
v <- f(xs)
} yield List(x) :: v
}
gives the error message
- type mismatch; found : List[List[Any]] required: List[List[Char]]
Please help me understand why 'for' chooses the most general Any instead of Char here? What topic in language spec should I read? Thanks.

Answer 1

The result, you are yielding is a mix of List[List[List[Char]]] and List[List[Char]]. Scala upcasts that to List[List[Any]]. For your case either of the following will do the job:
scala> def f(chars: List[Char]): List[List[Char]] = chars match {
| case Nil => List(Nil)
| case x :: xs => for {
| v <- f(xs)
| } yield x :: v
| }
f: (chars: List[Char])List[List[Char]]
scala> def f(chars: List[Char]): List[List[Char]] = chars match {
| case Nil => List(Nil)
| case x :: xs => for {
| v <- f(xs)
| } yield List(x) ++ v
| }
f: (chars: List[Char])List[List[Char]]

Answer 2

The problem is List(x) -- it needs to be x.
First, v iterates over the results of f(xs), and f returns List[List[Char]]. That means the result will be List[X], where X is the type returned by yield.
The type of v is List[Char], since it is iterating over the contents of f(xs). So we have to figure out the type of List(x) :: v, which is prepending a List[Char] on a List[Char]. It is not concatenating them: it is adding a list to a list containing only characters. The resulting list will have both Char and List[Char] in it.
Since the only type that satisfy both is Any, then X will be Any and the result of the for-comprehension List[Any].