区分联合中的可变内部值-解网

问：

任务是创建一个二叉树，其节点具有一个可以是 true 或 false 的成员，以及一个应设置为 true 的方法（如果尚未设置为 true）。is_lockedlock()is_locked

我试过了这个

type BinaryTreeNode =
    | Node of BinaryTreeNode * BinaryTreeNode
    | End
    with
        let mutable internalIsLocked = false
        member this.is_locked
            with get() = internalIsLocked 
        member this.lock() = internalIsLocked <- true

但是编译器由于 AT LET 绑定而说不。This declaration element is not permitted in an augmentation

val mutable internalIsLocked : bool相反，出于同样的原因不起作用。

那么这个问题该如何解决呢？难道不能用受歧视的工会来解决吗？

f# 可变歧视性工会

type Lock() =
    let mutable internalIsLocked = false
    member this.is_locked
        with get() = internalIsLocked 
    member this.lock() = internalIsLocked <- true

type BinaryTreeNode =
    | Node of Lock * BinaryTreeNode * BinaryTreeNode
    | End
    with
    static member Create(left, right) =
        Node (Lock(), left, right)
    member this.Lock =
        match this with
            | Node (lock, _, _) -> Some lock
            | End -> None

请注意，我假设节点是不可锁定的。用法：End

let node = BinaryTreeNode.Create(End, End)
node.Lock.Value.is_locked |> printfn "%A"   // false
node.Lock.Value.lock()
node.Lock.Value.is_locked |> printfn "%A"   // true

但是，我不推荐它。

区分联合中的可变内部值

Mutable internal value in discriminated union

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