O(1)

+   -----------------------------------------------------------------
+   --- Interaction and relationship between User and Use objects ---
+   -----------------------------------------------------------------
+
+
+A subclass of User can choose between incorporating its Use objects
+or refer to them out-of-line by means of a pointer. A mixed variant
+(some Uses inline others hung off) is impractical and breaks the invariant
+that the Use objects belonging to the same User form a contiguous array.
+
+We have 2 different layouts in the User (sub)classes:
+
+Layout a)
+The Use object(s) are inside (resp. at fixed offset) of the User
+object and there are a fixed number of them.
+
+Layout b)
+The Use object(s) are referenced by a pointer to an
+array from the User object and there may be a variable
+number of them.
+
+Initially each layout will possess a direct pointer to the
+start of the array of Uses. Though not mandatory for layout a),
+we stick to this redundancy for the sake of simplicity.
+The User object will also store the number of Use objects it
+has. (Theoretically this information can also be calculated
+given the scheme presented below.)
+
+Special forms of allocation operators (operator new)
+will enforce the following memory layouts:
+
+
+#  Layout a) will be modelled by prepending the User object
+#  by the Use[] array.
+#      
+#      ...---.---.---.---.-------...
+#        | P | P | P | P | User
+#      '''---'---'---'---'-------'''
+
+
+#  Layout b) will be modelled by pointing at the Use[] array.
+#      
+#      .-------...
+#      | User
+#      '-------'''
+#          |
+#          v
+#          .---.---.---.---...
+#          | P | P | P | P |
+#          '---'---'---'---'''
+
+   (In the above figures 'P' stands for the Use** that
+    is stored in each Use object in the member Use::Prev)
+
+
+Since the Use objects will be deprived of the direct pointer to
+their User objects, there must be a fast and exact method to
+recover it. This is accomplished by the following scheme:
+
+A bit-encoding in the 2 LSBits of the Use::Prev will allow to find the
+start of the User object:
+
+00 --> binary digit 0
+01 --> binary digit 1
+10 --> stop and calc (s)
+11 --> full stop (S)
+
+Given a Use*, all we have to do is to walk till we get
+a stop and we either have a User immediately behind or
+we have to walk to the next stop picking up digits
+and calculating the offset:
+
+.---.---.---.---.---.---.---.---.---.---.---.---.---.---.---.---.----------------
+| 1 | s | 1 | 0 | 1 | 0 | s | 1 | 1 | 0 | s | 1 | 1 | s | 1 | S | User (or User*)
+'---'---'---'---'---'---'---'---'---'---'---'---'---'---'---'---'----------------
+    |+15                |+10            |+6         |+3     |+1
+    |                   |               |           |       |__>
+    |                   |               |           |__________>
+    |                   |               |______________________>
+    |                   |______________________________________>
+    |__________________________________________________________>
+
+
+Only the significant number of bits need to be stored between the
+stops, so that the worst case is 20 memory accesses when there are
+1000 Use objects.
+
+The following literate Haskell fragment demonstrates the concept:
+
+> import Test.QuickCheck
+> 
+> digits :: Int -> [Char] -> [Char]
+> digits 0 acc = '0' : acc
+> digits 1 acc = '1' : acc
+> digits n acc = digits (n `div` 2) $ digits (n `mod` 2) acc
+> 
+> dist :: Int -> [Char] -> [Char]
+> dist 0 [] = ['S']
+> dist 0 acc = acc
+> dist 1 acc = let r = dist 0 acc in 's' : digits (length r) r
+> dist n acc = dist (n - 1) $ dist 1 acc
+> 
+> takeLast n ss = reverse $ take n $ reverse ss
+> 
+> test = takeLast 40 $ dist 20 []
+> 
+
+Printing  gives: "1s100000s11010s10100s1111s1010s110s11s1S"
+
+The reverse algorithm computes the
+length of the string just by examining
+a certain prefix:
+
+> pref :: [Char] -> Int
+> pref "S" = 1
+> pref ('s':'1':rest) = decode 2 1 rest
+> pref (_:rest) = 1 + pref rest
+> 
+> decode walk acc ('0':rest) = decode (walk + 1) (acc * 2) rest
+> decode walk acc ('1':rest) = decode (walk + 1) (acc * 2 + 1) rest
+> decode walk acc _ = walk + acc
+> 
+
+Now, as expected, printing  gives 40.
+
+We can quickCheck this with following property:
+
+> testcase = dist 2000 []
+> testcaseLength = length testcase
+> 
+> identityProp n = n > 0 && n <= testcaseLength ==> length arr == pref arr
+>     where arr = takeLast n testcase
+
+As expected  gives:
+
+*Main> quickCheck identityProp
+OK, passed 100 tests.
+
+Let's be a bit more exhaustive:
+
+> 
+> deepCheck p = check (defaultConfig { configMaxTest = 500 }) p
+> 
+
+And here is the result of :
+
+*Main> deepCheck identityProp
+OK, passed 500 tests.
+
+
+To maintain the invariant that the 2 LSBits of each Use** in Use
+never change after being set up, setters of Use::Prev must re-tag the
+new Use** on every modification. Accordingly getters must strip the
+tag bits.
+
+For layout b) instead of the User we will find a pointer (User* with LSBit set).
+Following this pointer brings us to the User. A portable trick will ensure
+that the first bytes of User (if interpreted as a pointer) will never have
+the LSBit set.
+