F#这是Option.map中的错误

Question

给出以下代码:

let mapOption (f : ('a -> 'b)) (x : 'a option) =
    match x with
    | Some x -> Some(f(x))
    | None -> None

let mapOptions (f : ('a -> 'b)) (xs : 'a option list) : 'b option list =
    xs
    |> List.map (fun (x : 'a option) -> mapOption f x)

let myList = [None; Some 1; Some 2; None]

let a = myList |> mapOptions (fun x -> x + 2)

let b = myList |> List.map(fun x-> x |> Option.map(fun y -> y + 2))

为什么a和b的结果等于:

[null; Some 3; Some 4; null] 要么 val it : int option list = [null; Some 3; Some 4; null]

不应该是:

[None; Some 3; Some 4; None]

Answer 1

Nonenull在CLR中表示.你可以通过试验FSI来看到:

> [Some 3; None];;
val it : int option list = [Some 3; null]

它仍然有效,但::

> [Some 3; None] |> List.choose id;;
val it : int list = [3]

所以[null; Some 3; Some 4; null]是这样的[None; Some 3; Some 4; None]:

> a = [None; Some 3; Some 4; None];;
val it : bool = true

aOP的值在哪里.