Scala - foldLeft type inference fail
fold, scala, types
Solution
The error occurs because you wrote `xs foldLeft (init) (f)` instead of `(xs foldLeft init)(f)` or `xs.foldLeft(init)(f)`.
The former does not work, because Scalas operator notation rules allow only to leave the dot and parentheses if a call occurs in form `obj method param`, which is not the case with `foldLeft` because it has two parameter lists.
Problem
In my code, I have the following: ``` type Occurrences = List[(Char, Int)] def subtract(x: Occurrences, y: Occurrences): Occurrences = { val yMap = y.toMap x foldLeft (List[(Char, Int)]()) { // ERROR case (a: List[(Char, Int)], xe: (Char, Int)) => if(yMap.contains(xe._1)) (xe._1 -> (xe._2 - yMap(xe._1))) :: a else a } } ``` It fails on compile-time, at the `{` that is right before the error mark in the code. The error message is the following: missing parameter type for expanded function The argument types of an anonymous function must be fully known. (SLS 8.5) Expected type was: Int 1) How could that be? As far as I can see, there is no room for misinterpretation of type information here, and I find a lot of such examples over the internet. How can I fix that? 2) Why is it thinking that the expected type is `Int` after all?