在不使用类的情况下隐藏OCaml中的函数参数

Question

我是OCaml的初学者,所以这个问题可能是微不足道的.我有一些这样的功能:

let rec
f1 <list of args> state = ...
and
f2 <list of args> state = ...
and
f3 <list of args> state = ...
and
f4 <list of args> state = ...
;;

这些函数中的每一个都使用最后一个参数作为状态调用其他函数.因此,对于每个执行'树',state是一种全局只读变量.如何以抽象出状态的方式模拟这个,但函数可以访问它.请注意,我不想使用OCaml类,涉及模块/子模块/仿函数的解决方案会很好!

Answer 1

let share state = 
  let rec f1 ... = ... you can use state freely here ...
  and f2 ...     = ... same here ...  
  and f3 ...     = ... same here ...
  and f4 ...     = ... same here ...
  in
  f1, f2, f3, f4

let state = ...
let f1, f2, f3, f4 = share state 

如果您希望"state"成为一个模块,例如:

module type S = sig ... end

let share m =
  let module M = (val m : S) in 
  let rec f1 ... = ... use M at will ...
  and f2 ...     = ... use M at will ...
  and f3 ...     = ... use M at will ...
  and f4 ...     = ... use M at will ...
  in 
  f1, f2, f3, f4


module M : S = struct ... end 

let f1, f2, f3, f4 = share (module M)

如果您希望将结果fi打包在模块中,则使用仿函数

module type S = sig  
  val f1 : ...
  val f2 : ...
  val f3 : ...
  val f4 : ...
end

module type State : sig ... end

module Make (M : State) : S = struct 
  let rec f1 ... = ... use M at will here ...
  and f2 ...     = ... again ...
  and f3 ...     = ... again ... 
  and f4 ...     = ... again ...
end

module State : State = struct ... implement state... end 
module Instance = Make (State) 

let () = Instance.f1 ()...