递归联合类型和列表

给定该类型定义,管理器不能有名字/姓氏。类型定义不包含类似的内容。

使 Manager Worker Person 价值观:

type Employee = Worker of Person | Manager of Person * Employee list

然后您可以创建经理Jane:

let manager = Manager ({ First = "Jane"; Last = "Doe" }, [worker])

至于一个成员 employees ,有几件事要说。

首先,远离会员。F#首先起作用。函数比成员好(原因是我不打算在这里讨论)。有一个功能:

let employees emp = 
    match emp with
    | Manager (_, emps) -> emps
    | Worker _ -> ???  // what to return if the employee turned out to be worker?

经理 ? 那还什么呢?

工人

let employees emp = 
    match emp with
    | Manager (_, emps) -> emps
    | Worker _ -> []

另一种可能是返回 Option

但如果你绝对坚持要有一个会员(请, 拜托

type Employee = Worker of Person | Manager of Person * Employee list
    with
        member this.employees =
          match this with
          | Manager (_, emps) -> emps
          | Worker _ -> []