Question

我不能为我的生活找到一种方法来处理lwt线程的取消。

这是我所拥有的，简化的。

#require "lwt.unix, lwt.ppx"
open Lwt.Infix

let program =
  let counter = ref 1 in
  let can_cancel = fst (Lwt.task ()) in

  Lwt.async_exception_hook := (fun _ ->
    prerr_endline "some exception");

  Lwt.on_cancel can_cancel (fun () ->
            Lwt_io.printl "correct ending" |> Lwt.ignore_result);

  Lwt.async begin fun () ->
    let rec forever () =

      try%lwt
        Lwt_io.printl "Hello World" >>= fun () ->

        if !counter = 3 then Lwt.cancel can_cancel
        else counter := !counter + 1;
        Lwt_unix.sleep 0.5 >>= forever
      with
        Lwt.Canceled -> Lwt_io.printl "canceled inside function"
    in

    Lwt.catch forever begin function
      | Lwt.Canceled -> Lwt_io.printl "Cancled exception happened"
      | _ -> Lwt.return ()
    end
  end;
  can_cancel

let () =
  Lwt_main.run program

你可以看到我多次尝试捕获Canceled异常，但没有一个能够正常工作。我的输出是

utop cancelable.ml                                                     ⏎
Hello World
Hello World
Hello World
correct ending
Exception: Lwt.Canceled.

在更宏大的方案中，我有unit Lwt.t list ref，它是用Lwt.task创建的，然后我计划在列表上执行List.iter Lwt.cancel，然后用新线程替换它类型unit Lwt.t

Answer 1

我把try / with放在错误的级别。

此代码显示了将try / with放入Lwt.Canceled异常的正确位置

#require "lwt.unix, lwt.ppx"
open Lwt.Infix

let program =
  let counter = ref 1 in
  let can_cancel = fst (Lwt.task ()) in

  Lwt.async begin fun () ->
    let rec forever () =
        Lwt_io.printl "Hello World" >>= fun () ->
        if !counter = 3 then Lwt.cancel can_cancel
        else counter := !counter + 1;
        Lwt_unix.sleep 0.5 >>= forever
    in
    forever ()
  end;
  can_cancel

let () =
  try
    Lwt_main.run program
  with
    Lwt.Canceled -> Lwt_io.printl "ended" |> Lwt.ignore_result

取消Lwt。螺纹

1 个答案: