F#“​​有状态”计算表达式

时间:2014-12-12 16:29:19

标签: f# functional-programming computation-expression

我目前正在学习F#并遇到一些绊脚石;我认为很多都是学习在功能上思考。

我目前正在学习的一件事是计算表达式,我希望能够定义一个处理某些跟踪状态的计算表达式,例如:

let myOptions = optionListBuilder {
    let! opt1 = {name="a";value=10}
    let! opt2 = {name="b";value=12}
}

我希望能够拥有myOptionsOption<'T> list,因此每个let!绑定操作都会有效地使构建器跟踪&#34;跟踪&#34;随之而来的定义选项。

我不想使用可变状态来做这件事 - 例如拥有由构建器维护的列表,并使用每个bind调用进行更新。

有没有办法让它成为可能?


更新:结果Option<'T> list类型只是代表性的,实际上我可能会有OptionGroup<'T>类型来包含列表以及一些其他信息 - 正如丹尼尔在下面提到的,我可以使用列表理解来获得一个简单的列表。

3 个答案:

答案 0 :(得分:5)

我写了一个字符串构建器计算表达式here

open System.Text

type StringBuilderUnion =
| Builder of StringBuilder
| StringItem of string

let build sb =
    sb.ToString()

type StringBuilderCE () =
    member __.Yield (txt : string) = StringItem(txt)
    member __.Yield (c : char) = StringItem(c.ToString())
    member __.Combine(f,g) = Builder(match f,g with
                                     | Builder(F),   Builder(G)   ->F.Append(G.ToString())
                                     | Builder(F),   StringItem(G)->F.Append(G)
                                     | StringItem(F),Builder(G)   ->G.Append(F)
                                     | StringItem(F),StringItem(G)->StringBuilder(F).Append(G))
    member __.Delay f = f()
    member __.Zero () = StringItem("")
    member __.For (xs : 'a seq, f : 'a -> StringBuilderUnion) =
                    let sb = StringBuilder()
                    for item in xs do
                        match f item with
                        | StringItem(s)-> sb.Append(s)|>ignore
                        | Builder(b)-> sb.Append(b.ToString())|>ignore
                    Builder(sb)

let builder1 = new StringBuilderCE ()

注意到底层类型是不可变的(包含的StringBuilder是可变的,但它不一定是)。每个产量组合当前状态和传入输入,而不是更新现有数据,从而产生StringBuilderUnion的新实例。您可以使用F#列表执行此操作,因为向列表头部添加元素仅仅是构造一个新的价值,而不是改变现有的价值。

使用StringBuilderCE看起来像这样:

//Create a function which builds a string from an list of bytes
let bytes2hex (bytes : byte []) =
    string {
        for byte in bytes -> sprintf "%02x" byte
    } |> build

//builds a string from four strings
string {
        yield "one"
        yield "two"
        yield "three"
        yield "four"
    } |> build

注意到yield而不是let!,因为我实际上并不想使用计算表达式中的值。

答案 1 :(得分:3)

<强>解

使用mydogisbox提供的基线StringBuilder CE构建器,我能够生成以下具有魅力的解决方案:

type Option<'T> = {Name:string;Item:'T}

type OptionBuilderUnion<'T> =
    | OptionItems of Option<'T> list
    | OptionItem of Option<'T>

type OptionBuilder () =
    member this.Yield (opt: Option<'t>) = OptionItem(opt)
    member this.Yield (tup: string * 't) = OptionItem({Name=fst tup;Item=snd tup})
    member this.Combine (f,g) = 
        OptionItems(
            match f,g with
            | OptionItem(F), OptionItem(G) -> [F;G]
            | OptionItems(F), OptionItem(G) -> G :: F
            | OptionItem(F), OptionItems(G) -> F :: G
            | OptionItems(F), OptionItems(G) -> F @ G
        )
    member this.Delay f = f()
    member this.Run (f) = match f with |OptionItems items -> items |OptionItem item -> [item]

let options = OptionBuilder()

let opts = options {
        yield ("a",12)
        yield ("b",10)
        yield {Name = "k"; Item = 20}
    }

opts |> Dump

答案 2 :(得分:2)

F#支持开箱即用的列表推导。

let myOptions =
    [
        yield computeOptionValue()
        yield computeOptionValue()
    ]