F#将对象转换为接口

Question

我有一个名为'Pane'的课程(想想玻璃窗格),它实现了IPane:

type IPane =
    abstract PaneNumber : int with get, set
    abstract Thickness : float<m> with get, set
    abstract ComponentNumber : int with get, set
    abstract Spectra : IGlassDataValues with get, set
    ...

type Pane(paneNumber, componentNumber, spectra, ...) =
    let mutable p = paneNumber
    let mutable n = componentNumber
    let mutable s = spectra
    ...

    interface IPane with
        member this.PaneNumber
            with get() = p
            and set(value) = p <- value
        member this.ComponentNumber
            with get() = n
            and set(value) = n <- value
        member this.Spectra
            with get() = s
            and set(value) = s <- value
            ...

我创建了一个窗格列表(窗格列表):

let p = [ p1; p2 ]

但是我需要将它转换为IPane列表,因为这是另一个函数中的参数类型.以下代码生成错误:

let p = [ p1; p2 ] :> IPane list
'Type constraint mismatch. The type
  Pane list
is not compatible with type
  IPane list
The type 'IPane' does not match the type 'Pane'

当Pane实现IPane时,这很令人困惑.只需将Pane list对象作为参数传递给所需的函数也会产生错误.

如何将窗格列表转换为IPane列表？

Answer 1

F#不允许以你想要的方式继承.

更好的方法是使用:

[p1 ;p2] |> List.map (fun x -> x:> IPane)

或者,您可以将功能更改为使用此类功能

let f (t:#IPane list) = ()

在这里你可以做f [p1;p2]的#告诉编译器,从继承的任何类型IPane的罚款.