是否可以将参数传递给F#模块？

Question

我是F#的新手并且学习基础知识.

我有两个模块.树数据结构的通用名称Tree:

module Tree

let rec getDescendants getChildren node  = 
    seq { yield node
          for child in getChildren node do 
            yield! getDescendants getChildren child }

let isLeaf getChildren node = Seq.isEmpty (getChildren node)

let getLeaves getChildren node = getDescendants getChildren node  
                               |> Seq.filter (isLeaf getChildren)

如您所见,所有函数都有一个getChildren参数,该参数是枚举给定类型的节点的子节点的函数.

第二个模块处理更具体的XML树案例:

module XmlTree

open System.Xml.Linq

let getXmlChildren (node : XElement) = node.Elements()

let getDescendants = Tree.getDescendants getXmlChildren 
let getLeaves = Tree.getLeaves getXmlChildren 
let isLeaf = Tree.isLeaf getXmlChildren 

getXmlChildren定义XML节点的特定函数并将其传递给curried Tree函数.

现在有很多代码重复.

这样做有可能吗？(伪代码)

module XmlTree = Tree with getChildren = fun (node : XElement) -> node.Elements()

Answer 1

F#不支持仿函数,因此您无法将参数传递给F#模块.在您的示例中,将生成节点子节点的函数传递给对象构造函数就足够了:

type Tree<'T>(childFn: 'T -> 'T seq) =
    let getChildren = childFn

    member x.getDescendants node  = 
        seq { yield node
              for child in getChildren node do 
                yield! x.getDescendants child }

    member x.isLeaf node = node |> getChildren |> Seq.isEmpty
    member x.getLeaves node = node |> x.getDescendants |> Seq.filter x.isLeaf

// Type usage
open System.Xml.Linq
let xmlTree = new Tree<XElement>(fun x -> x.Elements())

对于更复杂的案例,继承是可行的方法.特别是,您可以Tree<'T>使用抽象成员声明为抽象类getChildren,并在XmlTree子类中重写该方法.

Answer 2

例如，您不使用模块执行此操作，而是使用泛型执行此操作

编辑：

type tree<'t>(Children:seq<tree<'t>>)=

    member x.isLeaf()  = Seq.isEmpty (Children )

    member x.getLeaves() = 
        getDescendants Children
        |> Seq.filter (fun x -> x.isLeaf())

我省略了 getdescendants，但这应该足够了。此外，一些类型注释不是必需的，但显示正在发生的事情