我将set和maps的实现作为不平衡的二叉树。因为集合和映射非常相似,所以我实际上只是从头开始编写一个地图实现,然后简单地将集合实现为从键到单元的映射:
signature EQ =
sig
type t;
val eq : t * t -> bool;
end;
signature ORD =
sig
include EQ;
val lt : t * t -> bool;
end;
signature SET =
sig
structure Elem : EQ;
type set;
val empty : set;
val member : Elem.t * set -> bool;
val insert : Elem.t * set -> set option;
end;
signature MAP =
sig
structure Key : EQ;
type 'a map;
val empty : 'a map;
val lookup : Key.t * 'a map -> 'a option;
val insert : Key.t * 'a * 'a map -> 'a map option;
end;
functor UnbalancedMap (Key : ORD) :> MAP =
struct
structure Key = Key;
datatype 'a tree = E | T of Key.t * 'a * 'a tree * 'a tree;
type 'a map = 'a tree;
val empty = E;
fun lookup (k, t) =
let
fun loop (k, E, E) = NONE
| loop (k, E, T (x, y, _, _)) =
if Key.eq (k, x) then SOME y
else NONE
| loop (k, t as T (x, _, a, b), r) =
if Key.lt (k, x) then loop (k, a, r)
else loop (k, b, t);
in
loop (k, t, E)
end;
fun insert (k, v, t) =
let
exception Exists;
fun loop (k, v, E, E) = T (k, v, E, E)
| loop (k, v, E, T (x, _, _, _)) =
if Key.eq (k, x) then raise Exists
else T (k, v, E, E)
| loop (k, v, t as T (x, y, a, b), r) =
if Key.lt (k, x) then T (x, y, loop (k, v, a, r), b)
else T (x, y, a, loop (k, v, b, t));
in
SOME (loop (k, v, t, E)) handle Exists => NONE
end;
end;
functor UnbalancedSet (Elem : ORD) :> SET =
struct
structure Map = UnbalancedMap (Elem);
structure Elem = Map.Key;
type set = unit Map.map;
val empty = Map.empty;
fun member (x, t) = case Map.lookup (x, t) of
NONE => false
| _ => true;
fun insert (x, t) = Map.insert (x, (), t);
end;
让我们假设我使用其他一些数据结构提出另一种地图实现。然后我应该能够重用该数据结构来定义集合作为从键到单元的映射:
functor AnotherMap (Key : EQ) :> MAP =
struct
(* ... *)
end;
functor AnotherSet (Elem : EQ) :> SET =
struct
structure Map = AnotherMap (Elem);
structure Elem = Map.Key;
type set = unit Map.map;
val empty = Map.empty;
fun member (x, t) = case Map.lookup (x, t) of
NONE => false
| _ => true;
fun insert (x, t) = Map.insert (x, (), t);
end;
然而,如果我想出任意多个地图实现,重新定义使用与这些地图相同的数据结构的集合会很快变得乏味。我真正想要的是一个仿函数,它将一个仿函数从X带到MAP,并从X到SET生成一个仿函数,其中X是包含EQ(或可能是EQ本身)的任何签名。这在标准ML中是否可行?
答案 0 :(得分:5)
作为非标准扩展,是的。我相信您正在寻找的功能被SML / NJ称为“高阶仿函数”。这是实施的有限detail。
我强调这不是不是 SML的标准功能。使用SML模块系统无法直接实现此目的。