实施IList<T> 在F#

Question

我已经看遍了SO和internet，但仍然无法弄清楚我缺少什么以及如何在f＃中为这种类型实现泛型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()

现在编译器抱怨我没有实现集合。不可计数。GetEnumerator()和我知道它，但我真的不知道该怎么做。

更新资料: 最终结果如下。非常感谢先生。加兰。另外，值得指出的是，我只引用了系统。收藏品。通用的，忘了参考系统。Ienumerable的非通用版本驻留时的集合。所以sad 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

这是因为 IList 有两个 GetEnumerator 方法，一个用于 IEnumerable<T> 接口和一个用于非泛型 IEnumerable 界面。

您可以添加此成员来实现 IEnumerable

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

Answer 2

最简单的方式是

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

.