]> hi folks, I have a problem expanding an <:expr<>> quotation. the code snippet below raise this error ocamlc -c -pp "camlp4o pa_extend.cmo q_MLast.cmo" -I /usr/lib/ocaml/camlp4 test.ml File "test.ml", line 19, characters 13-42: This expression has type MLast.patt * ff but is here used with type MLast.patt * MLast.expr how can I write the variable 'content' in the preprocessor step ? It should be written down just like a string... is there any generic way to achive the same as I do with <:expr<$str:"ddd">> for strings with any datatype ? Of course immagine that the variable 'a' came from a more complex function like a parser... tnx, p --------- (*pp camlp4o pa_extend.cmo q_MLast.cmo *) open Pcaml;; type ff = | A1 | A2 let el = Grammar.Entry.create gram "el";; EXTEND expr: [[ "test"; n = LIST1 el SEP ";" -> <:expr< let $list:n$ in 1 >> ]]; el: [[ n = LIDENT -> let a = A1 in (* <---- the problem is here!!! *) (<:patt<$lid:n$>>,<:expr<$a$>>) ]]; END ------ let a = test b ;; ------------------- To unsubscribe, mail caml-list-request@inria.fr Archives: http://caml.inria.fr Bug reports: http://caml.inria.fr/bin/caml-bugs FAQ: http://caml.inria.fr/FAQ/ Beginner's list: http://groups.yahoo.com/group/ocaml_beginners