F# 如何将接口(interface)传递给函数？

Question

我的操作数据如下所示:

type IHasShow =
  abstract member show:bool
type ShowHideNotCompletedData       = {show:bool}
type ShowHideCompletedData          = {show:bool}
[<Pojo>]
type ActionData =
  | ShowHideNotCompleted of ShowHideNotCompletedData
  | ShowHideCompleted of ShowHideCompletedData

后来我尝试将 ShowHideNotCompletedData 或 ShowHideCompletedData 传递给函数，该函数只关心 bool “show”成员，但无法弄清楚如何传递/转换它:

let setShowItem (data:IHasShow) negate item =
  if data.show && (negate item.completed) then
    { item with show = true}
  else if (negate item.completed) then
    { item with show = false}
  else
    item

但是如何调用这个函数呢？

let setShowFn = setShowItem (data :> IHasShow) not

错误:

Type constraint mismatch. The type 
    'ShowHideNotCompletedData'    
is not compatible with type
    'IHasShow'

尝试过

let setShowFn = setShowItem data not

错误:

The type 'ShowHideNotCompletedData' is not compatible with the type 'IHasShow'

除了复制粘贴 setShowItem 并采用 ShowHideNotCompletedData 和 ShowHideCompleted 之外，还有其他方法吗？

如果有帮助；完整的源代码在这里:https://github.com/amsterdamharu/riot_redux_fable

最简单的解决方案是不传递数据，而只传递 bool 值:

let setShowItem show negate item =
  if (negate item.completed) then//simplified if statement
    { item with show = show}
  else
    item
//...

| ShowHideCompleted data ->
  let setShowFn = setShowItem data.show id
  { state with
      showCompleted = data.show
      items = state.items
      |> Array.map setShowFn}

我仍然想知道如何定义泛型类型并传递它。

Answer 1

在您当前的解决方案中，您的两种类型 ShowHideNotCompletedData 和 ShowHideCompletedData 是记录。它们具有接口(interface)的所有字段，但没有显式实现它们。解决方案是使接口(interface)显式化:

type ShowHideNotCompletedData(show) =
    interface IHasShow with
        member this.show = show
type ShowHideCompletedData(show) = 
    interface IHasShow with
        member this.show = show

实例化为ShowHideNotCompletedData true。对于替代解决方案，您可能需要咨询一些有关鸭子类型的 SO 问题，例如 this

话虽如此:我有一种预感，您的数据类型定义有点太复杂了。 @robkuz 发布了一个没有界面的答案。您自己的建议是将 bool 传递到函数中，这在模块化和可测试性方面似乎更好。