如何为此Func调用编写延续传递样式?

时间:2015-10-13 13:28:59

标签: f# c#-to-f# nbuilder

给出:

open System
open System.Linq.Expressions
open Microsoft.FSharp.Quotations
open Microsoft.FSharp.Linq.RuntimeHelpers
open FizzWare.NBuilder

let toLinq (expr: Expr<'a -> 'b>) =
    let linq = LeafExpressionConverter.QuotationToExpression expr
    let call = linq :?> MethodCallExpression
    let lambda = call.Arguments.[0] :?> LambdaExpression
    Expression.Lambda<Func<'a,'b>>(lambda.Body, lambda.Parameters)

let inline with'<'a,'b> (f:Expr<'a->'b>) (value:'b) (operable:IOperable<'a>) = 
    let f = toLinq f
    operable.With(f,value)

let size = 20
    let builderList =
        Builder<dbEncounter.ServiceTypes.Patients>.CreateListOfSize(size).All()
        |> with' <@ fun x -> x.PatientID @> 0
        |> with' <@ fun x -> x.ForeignEHRID @> (Nullable 0)
        |> with' <@ fun x -> x.PatientInfoID @> (Nullable 0)
        |> (fun b -> b.With(fun x-> x.PatientGUID <- Nullable (Guid.NewGuid()); x.PatientGUID ))
        |> withf (fun x-> x.PatientGUID <- Nullable (Guid.NewGuid()); x.PatientGUID) // this line doesn't compile as a replacement for the previous line

我尝试撰写withf

let inline withf<'a,'b> (f:Func<'a,_>) (operable:IOperable<'a>) = 
operable.With(f)

尝试使用withf替换其他选项的错误是

Error This function takes too many arguments, or is used in a context where a function is not expected

1 个答案:

答案 0 :(得分:1)

感谢@kvb对另一个问题的其他答案

,找到了答案

Interop between F# and C# lambdas

我只需要像这样调用Func构造函数:

let inline withf<'a,'b> (f:'a->'b) (operable:IOperable<'a>) = 
    operable.With(Func<'a,'b>(f))

现在这样可行:

let makePatients size = 
    let builderList =
        Builder<dbEncounter.ServiceTypes.Patients>.CreateListOfSize(size).All()
        |> with' <@ fun x -> x.PatientID @> 0
        |> with' <@ fun x -> x.ForeignEHRID @> (Nullable 0)
        |> with' <@ fun x -> x.PatientInfoID @> (Nullable 0)
        //|> (fun b -> b.With(fun x-> x.PatientGUID <- Nullable (Guid.NewGuid()); x.PatientGUID ))
        |> withf (fun x-> x.PatientGUID <- Nullable (Guid.NewGuid()); x.PatientGUID)