Difference between ocamlc and ocamlopt when marshalling reference

Difference between ocamlc and ocamlopt when marshalling reference

[Mathias Kende]

Hi,

I found a difference on the way global references used inside a function
are handled with respect to marshalling between ocamlc and ocamlopt.

See the exemple below :

-------------- a.ml ---------------
let r = ref 0
let p () = Printf.printf "%d\n" !r
let s = Marshal.to_string p [Marshal.Closures]
let () = r := 1
let q : unit -> unit = Marshal.from_string s 0
let () = q ()
-----------------------------------
$ ocamlc a.ml
$ ./a.out
0
$ ocamlopt a.ml
$ ./a.out 
1
-----------------------------------

I am not sure which semantic is the "good" one. They are both
interesting, but having only one is definitely simpler.

Is this a known problem ? is there fundamental reasons in the compilers
for which it can not be changed ?

Cheers,

Mathias

Re: [Caml-list] Difference between ocamlc and ocamlopt when marshalling reference

[Fabrice Le Fessant]

Hi,

  The main difference is that "p" is a closure in bytecode, but just a
pointer in native code: in bytecode, "p" references "r", and so a
closure is needed; in native code, the reference to "r" in "p" is a
direct access to the module structure, and so "p" does not need to keep
a reference to "r". So, when you marshal "p" in bytecode, you follow the
closure and the reference, that are stored in "s", and they are
duplicated when "s" is unmarshalled.

  I think Marshal.Closures is only expected for hacking anyway, so you
should not really assume anything on its behavior ;-)

--Fabrice

Mathias Kende wrote, On 06/02/2010 02:16 PM:
> Hi,
> 
> I found a difference on the way global references used inside a function
> are handled with respect to marshalling between ocamlc and ocamlopt.
> 
> See the exemple below :
> 
> -------------- a.ml ---------------
> let r = ref 0
> let p () = Printf.printf "%d\n" !r
> let s = Marshal.to_string p [Marshal.Closures]
> let () = r := 1
> let q : unit -> unit = Marshal.from_string s 0
> let () = q ()
> -----------------------------------
> $ ocamlc a.ml
> $ ./a.out
> 0
> $ ocamlopt a.ml
> $ ./a.out 
> 1
> -----------------------------------
> 
> I am not sure which semantic is the "good" one. They are both
> interesting, but having only one is definitely simpler.
> 
> Is this a known problem ? is there fundamental reasons in the compilers
> for which it can not be changed ?
> 
> Cheers,
> 
> Mathias
> 
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-- 
Fabrice LE FESSANT
Chercheur, Equipe ASAP
(As Scalable As Possible)
http://www.lefessant.net/

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