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我正在使用反射枚举 F# 中的一些可区分联合中的值（例如，如何在 F# 中枚举可区分联合？）。我想使用从使用反射获得的值来生成不同的记录类型，这些记录类型由我正在枚举的有区别的联合组成，但我不确定如何将类型 UnionCaseInfo 转换为实际的联合案例。有可能进行这样的演员吗？下面的代码正是我想要做的（区分联合中的值不同，变量名也不同）。我知道我可以使用枚举，但我宁愿不使用它们而不是区分联合。

open System
open Microsoft.FSharp.Collections
open Microsoft.FSharp.Reflection

let GetUnionCaseName (x:'a) = 
    match FSharpValue.GetUnionFields(x, typeof<'a>) with
    | case, _ -> case.Name  

type shape =
    | Square
    | Circle
    | Triangle
    | Other

type color =
    | Black
    | Red
    | Blue
    | Green
    | White

type coloredShape = { Shape: shape; Color: color }

let shapeCases = FSharpType.GetUnionCases typeof<shape>
let colorCases = FSharpType.GetUnionCases typeof<color>

let boardOfRelevantPossibilities = Microsoft.FSharp.Collections.Array2D.init<coloredShape> 5 3 (fun x y -> {Shape = Other; Color = Black})

let OtherShape = GetUnionCaseName(shape.Other)
let rand = Random()

for shapeCase in shapeCases do
    // Is there a way to do the following comparison this without using string comparisons
    if not (shapeCase.Name.Equals OtherShape) then
        for colorCase in colorCases do
            let mutable addedToBoard = false

            while not addedToBoard do
                let boardRowIndex = rand.Next(0,4)
                let boardColumnIndex = rand.Next(0,2)

                if boardOfRelevantPossibilities.[boardRowIndex,boardColumnIndex].Shape.Equals shape.Other then
                    addedToBoard <- true

                    // I want to utilize colorCase instead of other and shapeCase instead of black
                    boardOfRelevantPossibilities.[boardRowIndex,boardColumnIndex] <- {Shape = Other; // Shape should be determined by shapeCase instead of Other  
                        Color = White } // Color should be determined by colorCase instead of White

Console.ReadKey() |> ignore

我将我的代码重新分解为以下内容：

open System
open Microsoft.FSharp.Collections
open Microsoft.FSharp.Reflection

let allUnionCases<'T>() =
    FSharpType.GetUnionCases(typeof<'T>)
    |> Array.map (fun case -> FSharpValue.MakeUnion(case, [||]) :?> 'T)

type shape =
    | Square
    | Circle
    | Triangle
    | Other

type color =
    | Black
    | Red
    | Blue
    | Green
    | White

type coloredShape = { Shape: shape; Color: color }

let shapeCases = FSharpType.GetUnionCases typeof<shape>
let colorCases = FSharpType.GetUnionCases typeof<color>

let numberOfRows = 5
let numberOfColumns = 3
let boardOfRelevantPossibilities = Microsoft.FSharp.Collections.Array2D.init<coloredShape> numberOfRows numberOfColumns (fun x y -> {Shape = Other; Color = Black})

let rand = Random()

for shapeCase in allUnionCases<shape>() do
    // No string comparison anymore
    if shapeCase <> shape.Other then
        for colorCase in allUnionCases<color>() do
            let mutable addedToBoard = false

            while not addedToBoard do
                let boardRowIndex = rand.Next(0,numberOfRows)
                let boardColumnIndex = rand.Next(0,numberOfColumns)

                if boardOfRelevantPossibilities.[boardRowIndex,boardColumnIndex].Shape.Equals shape.Other then
                    addedToBoard <- true
                    // utilizing colorCase and shapeCase to create records to fill array
                    boardOfRelevantPossibilities.[boardRowIndex,boardColumnIndex] <- {Shape = shapeCase; Color = colorCase } 


printfn "%A" boardOfRelevantPossibilities
Console.ReadKey() |> ignore

这种新的重构结合了反射以枚举可区分联合，同时允许我生成由这些可区分联合组成的不同记录类型。我还发现了两个错误，它们在重构代码中得到修复。

4

1 回答 1

11

老实说，我更喜欢在静态方法中手动枚举所有联合案例，而不是通过反射创建它们。

正如@John 所说，您还需要使用FSharpValue.MakeUnion一步：

let allUnionCases<'T>() =
    FSharpType.GetUnionCases(typeof<'T>)
    |> Array.map (fun case -> FSharpValue.MakeUnion(case, [||]) :?> 'T)

在while循环中，您应该使用=而不是使用没有完全限定名称的Equals联合案例 ( )。shape.Other

/// Use type params for clarity; type inference should work fine without them
for shapeCase in allUnionCases<shape>() do
    for colorCase in allUnionCases<color>() do
        let mutable addedToBoard = false
        while not addedToBoard do
            let r = rand.Next(0,4)
            let c = rand.Next(0,2)
            if boardOfRelevantPossibilities.[r, c].Shape = Other then
                addedToBoard <- true
                boardOfRelevantPossibilities.[r, c] <- { Shape = shapeCase; 
                                                         Color = colorCase }
于 2012-12-27T09:52:18.097 回答