Question

我有以下界面：

type IFactory<'TIn, 'TOut> =
  abstract Create: 'TIn -> 'TOut

我正在尝试编写ComposedFactory。以下内容似乎是正确的语法，因为VS没有抱怨它：

type ComposedFactory<'TIn, 'TMid, 'TOut>
  (midFactory: IFactory<'TIn, 'TMid>,
   outFactory: IFactory<'TMid, 'TOut>) =

    let Create' =
        midFactory.Create >> outFactory.Create

    interface IFactory<'TIn, 'TOut> with
        member __.Create x = Create' x

但是我两次定义“创建”的事实让人感到愚蠢。我只想要一个接口。我该怎么办？

Answer 1

您可以在运动中这样做：

type ComposedFactory<'TIn, 'TMid, 'TOut>
   (midFactory: IFactory<'TIn, 'TMid>,
    outFactory: IFactory<'TMid, 'TOut>) =

interface IFactory<'TIn, 'TOut> with
    member __.Create x = (midFactory.Create >> outFactory.Create) x

但是我不能说它比您以前拥有的更可取。

Answer 2

恐怕我不知道该怎么做。我猜想接口函数的实现显式要求在其定义中指定参数。

我可以向您建议的最佳替代方法是管道输送

member __.Create x = x |> midFactory.Create |> outFactory.Create

F＃在接口成员实现中使用合成符号

2 个答案: