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inheritance - 我可以避免为从接口派生的类编写样板代码吗？

问题描述

我有许多实现接口的类。每当我定义一个类时，我必须在类定义中编写类似这样的内容：

interface Model.IModel with

        member this.Assets          = modelArg.Assets
        member this.AuxData         = modelArg.AuxData
        member this.Benchmark       = modelArg.Benchmark
        member this.CalcPeriods     = modelArg.CalcPeriods
        member this.CashTicker      = modelArg.CashTicker
        member this.Description     = modelArg.Description
        member this.FirstValidDate  = modelArg.FirstValidDate
        member this.Name            = modelArg.Name
        member this.ParamData       = modelArg.ParamData
        member this.ParamList       = modelArg.ParamList
        member this.Strategy        = modelArg.Strategy

        member this.CalcStartTime with get() = calcStartTime and
                                       set v = calcStartTime <- v

        member this.Counter with get() = counter and
                                 set v = counter <- v

有没有办法避免对从接口 IModel 派生的每个类重复此操作？代码在所有情况下都是相同的。

标签： inheritanceinterfacef#boilerplate

解决方案

为了使讨论更加具体，并且由于评论空间有限，以下内容对您有用还是不合适？

 type ModelArg = {
     Number: int
     Name: string
 }

 type IModel(modelArg: ModelArg) = 
    abstract member Number: int
    abstract member Name: string
    default __.Number = modelArg.Number
    default __.Name = modelArg.Name

type ConcreteModel1(modelArg: ModelArg) =
    inherit IModel(modelArg)

type ConcreteModel2(modelArg: ModelArg) =
    inherit IModel(modelArg)

let modelArg1 = {Number=2; Name ="Joe"}
let modelArg2 = {Number=3; Name = "Jim"}

let getNumberAndName(x: IModel) =
    (x.Number, x.Name)

let model1 = ConcreteModel1(modelArg1)
let model2 = ConcreteModel2(modelArg2)

getNumberAndName(model1)
//val it : int * string = (2, "Joe")
getNumberAndName(model2)
//val it : int * string = (3, "Jim")

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