Pietro Abate <Pietro.Abate@anu.edu.au> writes:

   aa: [[ "AA"; c = bb "val" -> "aa"^c ]];
   to 
   bb arg: [[ a = INT; "BB" -> a ]];
   
The camlp4 reference manual
(http://caml.inria.fr/pub/docs/manual-camlp4/manual005.html) says
the syntax of a rule is 

   { pattern = } symbol ; ... { pattern = } symbol { -> action }

so you can't pass an argument to bb.

I can think two possible solutions:

1. Extend the camlp4 syntax for grammar extentions with a new
   metasymbol "bb" that takes one argument, similar to the
   metasymbols LIST0 and LIST1

2. Use something like

  aa: [[ x = do_the_aa_rule -> ... ]]

with 

let do_the_aa_rule = Grammar.Entry.of_parser gram "do_the_aa_rule" 
                        (* a (Token.t Stream.t -> 'a) function,
                           where 'a is the type of x above *)

The ocaml parser distributed with camlp4 contains examples for
the latter technique, see for instance infixop0 or test_label_eq
in camlp4/etc/pa_o.ml. Further, these sources seem to indicate
that there is no nice solution to your problem: The levels "||"
to "**" of the entry expr could benefit from some more
abstraction.

Bye,

Hendrik