2 % (c) The AQUA Project, Glasgow University, 1994-1995
4 \section[Set]{An implementation of sets}
6 This new (94/04) implementation of sets sits squarely upon our
7 implementation of @FiniteMaps@. The interface is (roughly?) as
10 (95/08: This module is no longer part of the GHC compiler proper; it
11 is now just a GHC library module).
19 mkSet, -- :: Ord a => [a] -> Set a
20 setToList, -- :: Set a -> [a]
21 unitSet, -- :: a -> Set a
22 singletonSet, -- :: a -> Set a
24 union, -- :: Ord a => Set a -> Set a -> Set a
25 unionManySets, -- :: Ord a => [Set a] -> Set a
26 minusSet, -- :: Ord a => Set a -> Set a -> Set a
27 mapSet, -- :: Ord a => (b -> a) -> Set b -> Set a
28 intersect, -- :: Ord a => Set a -> Set a -> Set a
30 elementOf, -- :: Ord a => a -> Set a -> Bool
31 isEmptySet, -- :: Set a -> Bool
33 cardinality -- :: Set a -> Int
41 -- This can't be a type synonym if you want to use constructor classes.
42 newtype Set a = MkSet (FiniteMap a ())
45 emptySet = MkSet emptyFM
48 unitSet x = MkSet (unitFM x ())
49 singletonSet = unitSet -- old;deprecated.
51 setToList :: Set a -> [a]
52 setToList (MkSet set) = keysFM set
54 mkSet :: Ord a => [a] -> Set a
55 mkSet xs = MkSet (listToFM [ (x, ()) | x <- xs])
57 union :: Ord a => Set a -> Set a -> Set a
58 union (MkSet set1) (MkSet set2) = MkSet (plusFM set1 set2)
60 unionManySets :: Ord a => [Set a] -> Set a
61 unionManySets ss = foldr union emptySet ss
63 minusSet :: Ord a => Set a -> Set a -> Set a
64 minusSet (MkSet set1) (MkSet set2) = MkSet (minusFM set1 set2)
66 intersect :: Ord a => Set a -> Set a -> Set a
67 intersect (MkSet set1) (MkSet set2) = MkSet (intersectFM set1 set2)
69 elementOf :: Ord a => a -> Set a -> Bool
70 elementOf x (MkSet set) = isJust (lookupFM set x)
72 isEmptySet :: Set a -> Bool
73 isEmptySet (MkSet set) = sizeFM set == 0
75 mapSet :: Ord a => (b -> a) -> Set b -> Set a
76 mapSet f (MkSet set) = MkSet (listToFM [ (f key, ()) | key <- keysFM set ])
78 cardinality :: Set a -> Int
79 cardinality (MkSet set) = sizeFM set
82 instance (Eq a) => Eq (Set a) where
83 (MkSet set_1) == (MkSet set_2) = set_1 == set_2
84 (MkSet set_1) /= (MkSet set_2) = set_1 /= set_2
86 -- but not so clear what the right thing to do is:
88 instance (Ord a) => Ord (Set a) where
89 (MkSet set_1) <= (MkSet set_2) = set_1 <= set_2