實施的IList <T>在F＃

Question

我到處都找過SO和互聯網，但仍無法找出什麼我失蹤，如何實現在F＃泛型IList接口這樣的類型：

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() 

眼下編譯器抱怨我沒有實現Collections.IEnumerable.GetEnumerator（），我知道它，但我真的不知道該怎麼做。

更新：下面的最終結果。非常感謝Garland先生。另外，值得指出的是，當引用IEnumerable的非泛型版本時，我只引用了System.Collections.Generic並忘記引用System.Collection。所以傷心的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

這是因爲IList有兩個GetEnumerator方法，一個用於IEnumerable<T>接口和一個用於非通用IEnumerable接口。

可以添加這些成員來實現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() :> _