inheritance - 我可以避免为派生自接口(interface)的类编写样板代码吗？

Question

我有很多实现接口(interface)的类。每当我定义一个类时，我都必须在类定义中写这样的东西:

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(interface) 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。

Answer 1

为了使讨论更具体一些，并且由于评论中的空间有限，下面的内容是否适合您？

 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")