我有一个基于非平衡二叉树的集合和映射实现。由于集合和映射非常相似,我实际上只编写了从头开始的映射实现,并通过将键映射到元素的方式轻松地实现了集合:
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;
假设我使用其他数据结构来实现地图,那么我应该能够重用该数据结构将集合定义为从键到单位的映射:
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,如果我提出了任意数量的地图实现,重新定义使用与那些地图相同数据结构的集合将变得很繁琐。我真正想要的是一个函数器,它从X到MAP获取一个函数器,并生成一个从X到SET的函数器,其中X是包括EQ(或可能是EQ本身)的任何签名。在标准ML中是否可能实现这一点?