Question

我正在努力了解我对F＃工作流程构建器的理解。因此，我想出了以下示例以进行实验。

type failSafeSeq() =
    member this.For(seq, mapFunction) = seq |> Seq.map mapFunction
    member this.Yield(yieldExpr) = yieldExpr
    member this.YieldFrom(yieldBang) = yieldBang
    member this.Combine(a, b) = Seq.append a b
    member this.Delay(delayFun) = delayFun()
    member this.Zero() = Seq.empty 

let failSafe = new failSafeSeq();


let rec allFilesSeq dir =
    failSafe { for file in Directory.EnumerateFiles(dir) do yield file
               for subdir in Directory.EnumerateDirectories dir do yield! (allFilesSeq subdir) }

使用上面的代码，编译在“（allFilesSeq subdir）”处失败，并显示错误消息：

This expression was expected to have type     string     but here has type     seq<string>

我已经尝试了许多不同的咒语来修复此错误，但没有成功。我错过了什么？

Answer 1

type failSafeSeq() =
    member this.For(seq, mapFunction) = Seq.collect mapFunction seq
    member this.Yield(yieldExpr) = Seq.singleton yieldExpr
    member this.YieldFrom(yieldBang) = yieldBang
    member this.Combine(a, b) = Seq.append a b
    member this.Delay(delayFun) = delayFun()
    member this.Zero() = Seq.empty

用于生成一系列序列，因此您需要将它们展平
Yield应该在计算类型中包装值 - 在您的情况下生成具有一个元素的序列

实现F＃Workflow构建器：Yield，For，YieldFrom

1 个答案: