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Interfacing C-code with ocaml and Exception handling
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Date: 2007-04-17 (12:03)
From: Gerd Stolpmann <info@g...>
Subject: Re: [Caml-list] Interfacing C-code with ocaml and Exception handling
Am Dienstag, den 17.04.2007, 13:35 +0200 schrieb Christian Sternagel:
> I'm not sure whether this question was asked before (at least I didn't find any answer). Can it be, that raising an (OCaml)Exception during execution of
> C-code (interfaced with OCaml) is simply ignored?
> More concretely an example:
> We have a module Timer that implements execution of functions given a certain timeout. Therefor we have the function
> : float -> (unit -> 'a) -> 'a
> where [ t f] starts a timer raising the signal SIG_ALRM after
> [t] seconds and executes [f ()]. If [f] finishes in time the result of
> [f ()] is returned. In the background a handler for SIG_ALRM is installed, that just raises the exception [Timer.Timeout]. The implementation of
> [] looks like this:
>  let run t f =
>   try
>    start t;
>    let result = f () in
>    stop ();
>    result
>   with
>    | e -> stop (); raise e
>  ;;
> and the handler assigned to SIG_ALRM like this
>  let handler _ = raise Timeout;;
> where [start] and [stop] take care of a [Unix.itimer]. The intended behaviour is that if [f ()] finishes within time the result is returned and the exception [Timeout] is raised otherwise. This works fine, as long as the used [f] is implemented in OCaml. But when [f] is just an OCaml stub for a C-function, then the handler for SIG_ALRM is called (and hence the exception Timeout is raised) 

Are you sure? I would be a bit surprised if the handler was actually
called. Signal handlers written in O'Caml behave differently than those
written in C: The execution is deferred until the next safe point. When
you call a C stub, this is certainly not before the stub returns.

> but no exception [Timeout] arrives anywhere and hence the code of [f] runs as long as it needs ignoring any timeout.
> My theory is that when SIG_ALRM is raised during execution of C-code, than
> the [raise Timeout] statement raises an exception but as there is no context set up for exception handling this exception just gets lost.

No, this usually works.

> Is this theory correct and are there any suggestions how to work around this problem (implementing timed execution of arbitrary code).

Arbitrary code? No way to make that happen. For specific cases: yes,
there are many ways to do it.


> thnx in advance
> christian
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