当基类实现IEnumerable时实现IEnumerable <'T>

Question

是否有一种简洁的方法IEnumerable<'T>在基类型实现非泛型的类型中实现IEnumerable？我正在使用mkEnumerator下面的功能来做这件事,但似乎过分了.

open System
open System.Collections.Generic
open System.Configuration

let mkEnumerator<'T> (e : System.Collections.IEnumerator) =
  { new IEnumerator<'T> with
      member __.Current = e.Current |> unbox
    interface System.Collections.IEnumerator with
      member __.Current = e.Current
      member __.MoveNext() = e.MoveNext()
      member __.Reset() = e.Reset()
    interface IDisposable with
      member __.Dispose() =
        match e with
        | :? IDisposable as d -> d.Dispose()
        | _ -> () }

type MyConfigElement() = 
  inherit ConfigurationElement()

type MyConfigCollection() =
  inherit ConfigurationElementCollection()
  override __.CreateNewElement() = Unchecked.defaultof<_>
  override __.GetElementKey(_) = Unchecked.defaultof<_>
  interface IEnumerable<MyConfigElement> with
    member __.GetEnumerator() = mkEnumerator<MyConfigElement> (base.GetEnumerator())

更新

根据desco的建议,我现在正在使用:

type MyConfigCollection() =
  inherit ConfigurationElementCollection()
  ...
  interface IEnumerable<MyConfigElement> with
    member __.GetEnumerator() = (Seq.cast<MyConfigElement> this).GetEnumerator()

Answer 1

布莱恩建议的简短版本：

let mkEnum<'T> (ie : System.Collections.IEnumerable) = (Seq.cast<'T> ie).GetEnumerator()