What is the Enum.GetName equivalent for F# union member?

三世轮回 提交于 2019-11-30 08:13:23

You need to use the classes in the Microsoft.FSharp.Reflection namespace so:

open Microsoft.FSharp.Reflection

///Returns the case name of the object with union type 'ty.
let GetUnionCaseName (x:'a) = 
    match FSharpValue.GetUnionFields(x, typeof<'a>) with
    | case, _ -> case.Name  

///Returns the case names of union type 'ty.
let GetUnionCaseNames <'ty> () = 
    FSharpType.GetUnionCases(typeof<'ty>) |> Array.map (fun info -> info.Name)

// Example
type Beverage =
    | Coffee
    | Tea

let t = Tea
> val t : Beverage = Tea

GetUnionCaseName(t)
> val it : string = "Tea"

GetUnionCaseNames<Beverage>()
> val it : string array = [|"Coffee"; "Tea"|]
ympostor

@DanielAsher's answer works, but to make it more elegant (and fast? because of the lack of reflection for one of the methods), I would do it this way:

type Beverage =
    | Coffee
    | Tea
    static member ToStrings() =
        Microsoft.FSharp.Reflection.FSharpType.GetUnionCases(typeof<Beverage>)
            |> Array.map (fun info -> info.Name)
    override self.ToString() =
        sprintf "%A" self

(Inspired by this and this.)

I would like to propose something even more concise:

open Microsoft.FSharp.Reflection

type Coffee = { Country: string; Intensity: int }

type Beverage =
    | Tea
    | Coffee of Coffee

    member x.GetName() = 
        match FSharpValue.GetUnionFields(x, x.GetType()) with
        | (case, _) -> case.Name  

When union case is simple, GetName() may bring the same as ToString():

> let tea = Tea
val tea : Beverage = Tea

> tea.GetName()
val it : string = "Tea"

> tea.ToString()
val it : string = "Tea"

However, if union case is fancier, there will be a difference:.

> let coffee = Coffee ({ Country = "Kenya"; Intensity = 42 })
val coffee : Beverage = Coffee {Country = "Kenya"; Intensity = 42;}

> coffee.GetName()
val it : string = "Coffee"

> coffee.ToString()
val it : string = "Coffee {Country = "Kenya";        Intensity = 42;}"
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