OCaml Monad Transformers使用疑问：结果提取与状态操作

问题1：一次性获取计算结果、日志和状态 #

可以自己封装一个顶层运行函数，将Writer和State的两层run逻辑合并，避免手动逐层调用：

open Base
open Stdio
open Monads.Std

module St = struct
  include Monad.State.T1 (Monoid.Int) (Monad.Ident)
  include Monad.State.Make (Monoid.Int) (Monad.Ident)
end

module W = Monad.Writer.Make (Monoid.String) (St)

(* 封装顶层运行函数，一次性返回结果、日志、最终状态 *)
let run_writer_state (writer : 'a W.t) init_state =
  let state_comp = W.run writer in
  let (res, log), final_state = St.run state_comp init_state in
  res, log, final_state

let w_example =
  let open W in
  let* () = write "A" in
  let* () = lift (St.put 42) in
  return (-1)

let () =
  let res, log, state = run_writer_state w_example 0 in
  printf "(%d, %s, %d)\n" res log state  (* 输出: (-1, A, 42) *)

这个封装函数会帮你完成两层monad的运行逻辑，直接返回三个目标值。BAP的monads库本身没有提供现成的多层run组合器，自定义顶层函数是最直接的解决方案。

问题2：内嵌State模块时的状态设置操作 #

无需单独定义St模块，有两种可行方案：

方案1：在W模块中封装State操作 #

定义W时，直接添加put、get等状态操作的封装，通过lift调用内部匿名State monad的对应方法：

module W = struct
  include Monad.Writer.Make (Monoid.String) (struct
    include Monad.State.T1 (Monoid.Int) (Monad.Ident)
    include Monad.State.Make (Monoid.Int) (Monad.Ident)
  end)
  (* 封装State的put操作 *)
  let put x = lift (Monad.State.put x)
  (* 可选：封装get操作 *)
  let get = lift Monad.State.get
end

let w_example =
  let open W in
  let* () = write "A" in
  let* () = put 42 in  (* 直接调用W.put *)
  return (-1)

方案2：使用Monad.State.Lift模块 #

BAP提供的Monad.State.Lift模块可以直接在transformer上层调用下层State monad的操作，无需修改W的定义：

module W =
  Monad.Writer.Make
    (Monoid.String)
    (struct
      include Monad.State.T1 (Monoid.Int) (Monad.Ident)
      include Monad.State.Make (Monoid.Int) (Monad.Ident)
    end)

let w_example =
  let open W in
  let open Monad.State.Lift (W) in  (* 引入State Lift组合器 *)
  let* () = write "A" in
  let* () = put 42 in  (* 直接调用Lift的put *)
  return (-1)

两种方案都能实现无需单独定义St模块的状态设置操作，可根据代码风格偏好选择。

内容的提问来源于stack exchange，提问作者Jay Lee