Question

我有以下代码，我需要在我自己的类上调用我通过相互类的引用引用（希望有意义，如果不是代码示例更好地解释它）。

不幸的是它确实有效，因为我在...之前得到了不确定类型对象的＆＃34;查找...＆＃34; ＆＃39; x.Connections。[0] .Dest.Func＆＃39;一部分。

任何人都可以想办法解决这个问题，可能我只想采用一种不同的方法吗？

type Node() =
    member x.Connections = new System.Collections.Generic.List<Connection>()
    member x.Connect other = x.Connections.Add(Connection(x, other))
    member x.Func i : unit = x.Connections.[0].Dest.Func i
and Connection(source : Node, dest : Node) =
    member val Source = source
    member val Dest = dest

忽略在Node上调用Func方法没有任何意义的事实，我已经从实际代码中简化了这一点，试图只显示导致问题的原因。提前谢谢。

Answer 1

尝试在最后声明该方法：

type Node() =
    member x.Connections = new System.Collections.Generic.List<Connection>()
    member x.Connect other = x.Connections.Add(Connection(x, other))

and Connection(source : Node, dest : Node) =
    member val Source = source
    member val Dest = dest

type Node with
    member x.Func i : unit = x.Connections.[0].Dest.Func i

或者可能更好地交换类型的顺序：

type Connection(source : Node, dest : Node) =
    member val Source = source
    member val Dest = dest
and Node() =
    member x.Connections = new System.Collections.Generic.List<Connection>()
    member x.Connect other = x.Connections.Add(Connection(x, other))
    member x.Func i : unit = x.Connections.[0].Dest.Func i

Answer 2

您还可以向Dest添加类型注释，它似乎是将类型推理算法抛出到循环中的关键符号：

type Node() =
    member x.Connections = new System.Collections.Generic.List<Connection>()
    member x.Connect other = x.Connections.Add(Connection(x, other))    
    member x.Func i : unit = x.Connections.[0].Dest.Func i                             
and Connection(source : Node, dest : Node) =
    member val Source = source
    member val Dest : Node = dest

F＃如何处理这个递归的mutal类方法

2 个答案: