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我在一个名为module.ml. 是否有使用 ocamlfind 的简单方法来执行此操作,或者我必须直接调用camlp4 并手动传递所有原始参数?

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2 回答 2

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使用时ocamlbuild

ocamlbuild module.pp.ml
cat _build/module.pp.ml
于 2014-02-18T17:22:32.410 回答
2

这个输出

$ camlp4o `ocamlfind query type_conv`/pa_type_conv.cma  `ocamlfind query sexplib`/pa_sexp_conv.cma /home/kakadu/.opam/4.01.0/lib/ocaml/camlp4/Camlp4Printers/Camlp4OCamlPrinter.cmo a.ml | pr -t -o4
type t = (int * string)

let _ = fun (_ : t) -> ()

let __t_of_sexp__ =
  let _tp_loc = "a.ml.t"
  in
    function
    | Sexplib.Sexp.List ([ v1; v2 ]) ->
        let v1 = int_of_sexp v1 and v2 = string_of_sexp v2 in (v1, v2)
    | sexp -> Sexplib.Conv_error.tuple_of_size_n_expected _tp_loc 2 sexp

let _ = __t_of_sexp__

let t_of_sexp sexp =
  try __t_of_sexp__ sexp
  with
  | Sexplib.Conv_error.No_variant_match ((_tp_loc, sexp)) ->
      Sexplib.Conv_error.no_matching_variant_found _tp_loc sexp

let _ = t_of_sexp

let sexp_of_t (v1, v2) =
  let v1 = sexp_of_int v1
  and v2 = sexp_of_string v2
  in Sexplib.Sexp.List [ v1; v2 ]

let _ = sexp_of_t

对于这个来源

$ cat a.ml
type t = int*string with sexp
于 2014-02-18T15:40:02.793 回答