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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>

--
Best, bob