(***********************************************************************)
(*                                                                     *)
(*                           Objective Caml                            *)
(*                                                                     *)
(*            Xavier Leroy, projet Cristal, INRIA Rocquencourt         *)
(*                                                                     *)
(*  Copyright 1996 Institut National de Recherche en Informatique et   *)
(*  Automatique.  Distributed only by permission.                      *)
(*                                                                     *)
(***********************************************************************)

(* $Id$ *)

(* Hash tables *)

(* We do dynamic hashing, and we double the size of the table when
   buckets become too long, but without re-hashing the elements. *)

type ('a, 'b) t =
  { mutable max_len: int;                    (* max length of a bucket *)
    mutable data: ('a, 'b) bucketlist array } (* the buckets *)

and ('a, 'b) bucketlist =
    Empty
  | Cons of 'a * 'b * ('a, 'b) bucketlist

let create initial_size =
  { max_len = 2; data = Array.create initial_size Empty }

let clear h =
  for i = 0 to Array.length h.data - 1 do
    h.data.(i) <- Empty
  done

let resize h =
  let n = Array.length h.data in
  let newdata = Array.create (n+n) Empty in
    Array.blit h.data 0 newdata 0 n;
    Array.blit h.data 0 newdata n n;
    h.data <- newdata;
    h.max_len <- 2 * h.max_len

let rec bucket_too_long n bucket =
  if n < 0 then true else
    match bucket with
      Empty -> false
    | Cons(_,_,rest) -> bucket_too_long (pred n) rest

external hash_param : int -> int -> 'a -> int = "hash_univ_param" "noalloc"

let add h key info =
  let i = (hash_param 10 100 key) mod (Array.length h.data) in
  let bucket = Cons(key, info, h.data.(i)) in
    h.data.(i) <- bucket;
    if bucket_too_long h.max_len bucket then resize h

let remove h key =
  let rec remove_bucket = function
      Empty ->
        Empty
    | Cons(k, i, next) ->
        if k = key then next else Cons(k, i, remove_bucket next) in
  let i = (hash_param 10 100 key) mod (Array.length h.data) in
    h.data.(i) <- remove_bucket h.data.(i)

let find h key =
  match h.data.((hash_param 10 100 key) mod (Array.length h.data)) with
    Empty -> raise Not_found
  | Cons(k1, d1, rest1) ->
      if key = k1 then d1 else
      match rest1 with
        Empty -> raise Not_found
      | Cons(k2, d2, rest2) ->
          if key = k2 then d2 else
          match rest2 with
            Empty -> raise Not_found
          | Cons(k3, d3, rest3) ->
              if key = k3 then d3 else begin
                let rec find = function
                    Empty ->
                      raise Not_found
                  | Cons(k, d, rest) ->
                      if key = k then d else find rest
                in find rest3
              end

let find_all h key =
  let rec find_in_bucket = function
    Empty ->
      []
  | Cons(k, d, rest) ->
      if k = key then d :: find_in_bucket rest else find_in_bucket rest in
  find_in_bucket h.data.((hash_param 10 100 key) mod (Array.length h.data))

let iter f h =
  let d = h.data in
  let len = Array.length d in
  for i = 0 to len - 1 do
    let rec do_bucket = function
        Empty ->
          ()
      | Cons(k, d, rest) ->
          if (hash_param 10 100 k) mod len = i
          then begin f k d; do_bucket rest end
          else do_bucket rest in
    do_bucket d.(i)
  done

let hash x = hash_param 50 500 x