为何F#中使用function定义的函数无法标记inline？

先看两个功能完全一致的F#函数：

let episodeDict1 (a: obj) = 
    match a with
    | :? (Episode -> _) as generator ->
        let episodes = [D; W; M]
        let values = episodes |> List.map generator
        List.zip episodes values
    | :? _ as value -> [D, value; W, value; M, value]
    |> dict |> Dictionary


let episodeDict2: obj -> Dictionary<Episode, _> = function
    | :? (Episode -> _) as generator ->
        let episodes = [D; W; M]
        let values = episodes |> List.map generator
        List.zip episodes values |> dict |> Dictionary
    | :? _ as value -> [D, value; W, value; M, value] |> dict |> Dictionary

尝试给第二个函数添加inline修饰符时，会触发错误：

FS0832 Only functions may be marked 'inline'

原因解释 #

function是F#里的语法糖，本质等价于fun x -> match x with ...的lambda表达式。当你写成let episodeDict2: ... = function ...时，语法上是把一个lambda表达式赋值给一个绑定，而非直接定义一个带参数的函数。

而F#的inline修饰符有严格的语法限制：它只能直接作用于用let显式声明参数的函数定义（比如第一个函数let episodeDict1 (a: obj) = ...这种形式），不能用于赋值lambda表达式的绑定——哪怕这个绑定的类型是函数，语法上也不属于“直接定义的函数”范畴。

解决方法 #

如果要给这类函数添加inline，需要改成显式声明参数的形式，或者把function展开为带参数的fun表达式：

方式1：显式声明参数

let inline episodeDict2 (a: obj) : Dictionary<Episode, _> = 
    match a with
    | :? (Episode -> _) as generator ->
        let episodes = [D; W; M]
        let values = episodes |> List.map generator
        List.zip episodes values |> dict |> Dictionary
    | :? _ as value -> [D, value; W, value; M, value] |> dict |> Dictionary

方式2：展开function为fun表达式

let inline episodeDict2: obj -> Dictionary<Episode, _> = fun a -> 
    match a with
    | :? (Episode -> _) as generator ->
        let episodes = [D; W; M]
        let values = episodes |> List.map generator
        List.zip episodes values |> dict |> Dictionary
    | :? _ as value -> [D, value; W, value; M, value] |> dict |> Dictionary

这两种写法都符合inline的语法要求，可以正常编译。

内容的提问来源于stack exchange，提问作者Franco Tiveron