[英]F# custom computation workflow
給出結果類型
type Result<'t> = OK of 't | Error of string
我有這些函數都返回Async <Result <'t>>這些函數組合如下:
let a = async { return Result.OK 1000 }
let b = async { return Result.Error "some message" }
let sum x y =
async {
let! r1 = x
match r1 with
| Result.OK v1 ->
let! r2 = y
match r2 with
| Result.OK v2 -> return v1 + v2
| Result.Error msg -> return Result.Error msg
| Result.Error msg -> return Result.Error msg
}
這段代碼看起來很糟糕,所以我希望這樣:
type Result = Ok of int | Error of string
type MyMonadBuilder() =
member x.Bind (v,f) =
async {
let! r = v
match r with
| Ok r' -> return! f r'
| Error msg -> return Error msg
}
member x.Return v = async {return Ok v }
member x.Delay(f) = f()
let mymonad = MyMonadBuilder()
let runMyMonad = Async.RunSynchronously
let a = mymonad { return 10 }
let b = mymonad { return 20 }
let c =
mymonad {
return Result.Error "Some message"
//??? The above doesn't work but how do I return a failure here?
}
let d =
async {
return Ok 1000
}
//how to wrap this async with mymonad such that I can use it together with my other computation expressions?
let sum x y =
mymonad {
let! v1 = x
let! v2 = y
return v1 + v2
}
[<EntryPoint>]
let main argv =
let v = sum a b |> runMyMonad
match v with
| Ok v' -> printfn "Ok: %A" v'
| Error msg -> printf "Error: %s" msg
System.Console.Read() |> ignore
0
所以問題是:
......這樣我就可以寫了
let f (a:MyMonad<int>) (b:MyMonad<string>) = ...
更新:
另外,我想並行運行幾個mymonad操作,然后查看結果數組以查看錯誤和成功之處。 出於這個原因,我認為使用例外不是一個好主意。
另外,關於問題3,我的意思是使我的類型參數化並且不透明,以使調用者不知道/不在乎他們正在處理異步。 我編寫monad的方式是調用方始終可以使用Async.RunSynchronously運行mymonad表達式。
更新2:
到目前為止,我最終得到以下結果:
代碼如下所示:
type MyMonad<'t> = 't Result Async
type MyMonadBuilder() =
member x.Bind<'t> (v,f) : MyMonad<'t>=
async {
let! r = v
match r with
| Ok r' -> return! f r'
| Error msg -> return Error msg
}
member x.Return<'t> v : MyMonad<'t> = async {return Ok v }
member x.ReturnFrom<'t> v : MyMonad<'t> = v
member x.Delay(f) = f()
let failwith<'t> : string -> MyMonad<'t> = Result.Error >> async.Return
對於我的目的,這看起來相當不錯。 謝謝!
異步工作流通過異常自動支持錯誤處理,因此慣用解決方案只是使用異常。 如果要區分某些特殊類型的錯誤,則可以只定義自定義異常類型:
exception MyError of string
// Workflow succeeds and returns 1000
let a = async { return 1000 }
// Workflow throws 'MyError' exception
// (using return! means that it can be treated as a workflow returning int)
let b = async { return! raise (MyError "some message") }
// Exceptions are automatically propagated
let sum = async {
let! r1 = a
let! r2 = b
return r1 + r2 }
如果要處理異常,可以在異步工作流中使用try ... with MyError msg -> ...
您可以定義一個自定義計算生成器,以使用諸如Result
類的代數數據類型重新實現此功能,但是除非您有充分的理由這樣做,否則我不建議您使用這種方法-它不適用於標准庫,很復雜,不適合一般的F#風格。
在您的計算表達式中,值的類型為Async<Result<'T>>
, return
自動將類型'T
的參數包裝在返回Ok
的異步工作流中。 如果要構造表示失敗的值,可以使用return!
並創建一個返回Result.Error
的異步工作流。 你可能需要這樣的東西:
let c = mymonad {
return! async.Return(Result.Error "Some message")
}
let d = mymonad {
return 1000
}
但正如我所說,使用例外是一種更好的方法。
編輯:要在注釋中回答問題-如果您有許多異步計算,您仍然可以將最終結果包裝在您的自定義類型中。 但是,您不需要重建整個異步工作流庫 - 仍可以使用標准異常處理基元操作中的錯誤:
// Primitive async work that may throw an exception
let primitiveAsyncWork = async { ... }
// A wrapped computation that returns standard Option type
let safeWork = async {
try
let! res = primitiveAsyncWork
return Some res
with e -> return None }
// Run 10 instances of safeWork in parallel and filter out failed computations
async { let! results = [ for i in 0 .. 9 -> safeWork ] |> Async.Parallel
return results |> Seq.choose id }
我的ExtCore庫中的asyncChoice
工作流已經實現了-在NuGet上可用,因此您要做的就是添加對項目的引用,在源文件中打開ExtCore.Control
命名空間,並開始編寫如下代碼:
open ExtCore.Control
let asyncDivide100By (x : int) =
asyncChoice {
if x = 0 then
return! AsyncChoice.error "Cannot divide by zero."
else
return (100 / x)
}
let divide100By (x : int) =
let result =
asyncDivide100By x
|> Async.RunSynchronously
match result with
| Choice1Of2 result ->
printfn "100 / %i = %i" x result
| Choice2Of2 errorMsg ->
printfn "An error occurred: %s" errorMsg
[<EntryPoint>]
let main argv =
divide100By 10
divide100By 1
divide100By 0
0 // Exit code
asyncChoice是使用F#Core庫中的標准Async<'T>
和Choice<_,_>
類型構造的,因此您不會出現任何兼容性問題。
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