如何在 F# 中声明只有一个 Literal 属性的多个 Literals？

Question

我正在尝试找到一种优雅的方式来为符号分配键，而无需执行以下操作。

let [<Literal>] North = ConsoleKey.UpArrow // etc.

我宁愿做这样的事情，只使用一个属性。有什么办法可以做到吗？

[<Literal>]
type Direction =
    | North of ConsoleKey.UpArrow
    | East of ConsoleKey.RightArrow
    | South of ConsoleKey.DownArrow
    | West of ConsoleKey.LeftArrow

Answer 1

假设您的目标是在模式匹配中使用这些，这是一种方法：

// Use a type alias to shorten the name for ConsoleKey
type Key = ConsoleKey

// Create a general purpose active pattern that simply tests for equality
let (|Is|_|) a b = if a = b then Some () else None

// This is how you would use it
let describeMovement key =
    match key with
    | Is Key.UpArrow -> "up"
    | Is Key.RightArrow -> "right"
    | Is Key.DownArrow -> "down"
    | Is Key.LeftArrow -> "left"
    | _ -> "invalid"

如何在 F# 中声明只有一个 Literal 属性的多个 Literals？

How to declare multiple Literals with only one Literal attribute in F#?

f#

literals

discriminated-union