listf#value-restriction

F# value restriction in empty list


I have a F# function:

let removeEven (listToGoUnder : _ list) =
    let rec listRec list x =
        match list with
        | [] -> []
        | head::tail when (x%2 = 0) -> head :: listRec (tail) (x+1)
        | head::tail -> listRec (tail) (x+1)

     listRec listToGoUnder 0

It removes all elements at an even index in a list. It works if I give the list some imput, like removeEven ['1';'2';'3'] I get ['1';'3'] which I am supposed to. But when I insert a empty list as parameter, I get this error:

stdin(78,1): error FS0030: Value restriction. The value 'it' has been inferred to have generic type

val it : '_a list Either define 'it' as a simple data term, make it a function with explicit arguments or, if you do not intend for it to be generic, add a type annotation.

Help, anybody?


Solution

  • The empty list ([]) is quite special; it can be a list of any type. Therefore, the compiler complains that you don't have a specific type for []. Adding type annotation on the argument helps to solve the problem:

    let results = removeEven ([]: int list)
    

    or more idiomatic type annotation as suggested by @kvb:

    let results: int list = removeEven []
    

    This is probably beyond the question, but your function should be named as removeOdd since indices often start from 0 and your function removes all elements with odd indices. Moreover, things are much more clear if you use pattern matching on first two elements of the list rather than keep a counter x for checking indices:

    let rec removeOdd = function
        | [] -> []
        | [x] -> [x]
        | x::_::xs -> x::removeOdd xs