實現IList <T> 在F＃中

Question

我已經遍及SO和Internet，但仍然無法弄清我所缺少的內容以及如何在F＃中為這種類型實現通用IList接口：

type OrCondition() as self =
    inherit Condition()
    member val Conditions: List<Condition> = new List<Condition>() with get, set
    interface IList<Condition> with
        member this.Item 
            with get(index) = self.Conditions.[index]
            and  set(index)(value) = self.Conditions.[index] <- value
        member this.IndexOf item = self.Conditions.IndexOf(item)
        member this.Insert(index, item) = self.Conditions.Insert(index, item)
        member this.RemoveAt(index) = self.Conditions.RemoveAt(index)
        member this.Count with get() = self.Conditions.Count
        member this.IsReadOnly with get() = false
        member this.Add(item) = self.Conditions.Add(item)
        member this.Clear() = self.Conditions.Clear()
        member this.Contains(item) = self.Conditions.Contains(item)
        member this.CopyTo(conditions, index) = self.Conditions.CopyTo(conditions, index)
        member this.Remove(item) = self.Conditions.Remove(item)
        member this.GetEnumerator() = (Seq.cast<Condition> self.Conditions).GetEnumerator()

現在，編譯器抱怨我還沒有實現Collections.IEnumerable.GetEnumerator（）並且我知道它，但是我真的不知道該怎么做。

更新：最終結果如下。 非常感謝加蘭先生。 另外，值得指出的是，我僅引用System.Collections.Generic，而當非IEnumerable非常規版本駐留時，卻忘記了引用System.Collection。 很遺憾，ReSharper不支持F＃。

open System.Collections
open System.Collections.Generic

type OrCondition() as self =
    inherit Condition()
    member val Conditions = new List<Condition>() with get, set
    interface IList<Condition> with
        member this.Item 
            with get(index) = self.Conditions.[index]
            and  set(index)(value) = self.Conditions.[index] <- value
        member this.IndexOf item = self.Conditions.IndexOf(item)
        member this.Insert(index, item) = self.Conditions.Insert(index, item)
        member this.RemoveAt(index) = self.Conditions.RemoveAt(index)
        member this.Count with get() = self.Conditions.Count
        member this.IsReadOnly with get() = false
        member this.Add(item) = self.Conditions.Add(item)
        member this.Clear() = self.Conditions.Clear()
        member this.Contains(item) = self.Conditions.Contains(item)
        member this.CopyTo(conditions, index) = self.Conditions.CopyTo(conditions, index)
        member this.Remove(item) = self.Conditions.Remove(item)
        member this.GetEnumerator() = self.Conditions.GetEnumerator() :> IEnumerator<Condition>
        member this.GetEnumerator() = self.Conditions.GetEnumerator() :> IEnumerator 

Answer 1

這是因為IList具有兩種GetEnumerator方法，一種用於IEnumerable<T>接口，另一種用於非通用IEnumerable接口。

您可以添加此成員以實現IEnumerable

member this.GetEnumerator() = (this.Conditions :> IEnumerable).GetEnumerator()

Answer 2

最簡單的方法是

interface System.Collections.IEnumerable with
    member this.GetEnumerator() = this.Conditions.GetEnumerator() :> _