Thanks for quick reply.
I have read Jake's camlp4 series carefully, unfortunately it does not cover
my case :-(

On Thu, Nov 24, 2011 at 10:47 AM, Gabriel Scherer <gabriel.scherer@gmail.com
> wrote:

> Camlp4 performs grammar factorizations that make its behavior non
> strictly LL(k).
>
> I never bothered to learn the exact semantics of Camlp4 grammar rules;
> they're complex and, I suspect, possibly fragile. If you avoid being
> clever with ambiguities, you can get away with the gory details.
>
> You should have a look at Jake Donham excellent blog series about Camlp4:
>
> http://ambassadortothecomputers.blogspot.com/2010/05/reading-camlp4-part-6-parsing.html
>
> 2011/11/24 bob zhang <bobzhang1988@gmail.com>:
> > Hi List,
> > I have came across a strange behavior of the camlp4 parser, (maybe not
> > that weird due to my limited knowledge of the parser)
> > The contrived mini-examples as follows :
> > module MGram = MakeGram(Lexer) ;;
> > EXTEND MGram
> > GLOBAL: m_expr ;
> > m_expr :
> > [[ "foo"; f -> print_endline "first"
> > | "foo" ; "bar"; "baz" -> print_endline "second"]
> > ];
> > f : [["bar"; "baz" ]]; END;;
> > MGram.parse_string m_expr (Loc.mk "<string>") "foo bar baz ";;
> > second (** choose the second branch, maybe the token rule has a higher
> > priority *)
> >
> > MGram.Entry.clear m_expr;;
> > EXTEND MGram
> > GLOBAL: m_expr ;
> > m_expr :
> > [[ "foo"; f -> print_endline "first"
> > | "foo" ; "bar"; "bax" -> print_endline "second"]
> > ];
> > f : [["bar"; "baz" ]]; END;;
> > - : unit = ()
> > # MGram.parse_string m_expr (Loc.mk "<string>") "foo bar baz ";;
> > first (** here choose the first branch, but the token rule can consume
> > one token, I thought this should fail *)
> >
> > Many Thanks
> >
> >
> > --
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> > https://sympa-roc.inria.fr/wws/info/caml-list
> > Beginner's list: http://groups.yahoo.com/group/ocaml_beginners
> > Bug reports: http://caml.inria.fr/bin/caml-bugs
> >
> >
>



-- 
Best, bob