MIT HAKMEM Bit Count in Oracle SQL

This is not about performance... just to illustrate various approaches I've considered while reading this. Interestingly, that even the question was about PL/SQL implementation everyone tried to do it in SQL. Anyway, here is my (Update 06/25/2005: one buddy there was quite close to implement the second one) ideas about it (sure, mit hakmem is borrowed):

-- Let's employ regexp and replace
-- the hex values by their count
SELECT n
, NVL(
   LENGTH(
     REGEXP_REPLACE(
        TO_CHAR(n, 'FMXXXXXXXX')
     , '([1248]*)([3569AC]*)([7BDE]*)([F]*)[0]*'
     , '\1\2\2\3\3\3\4\4\4\4')
   )                           
, 0                               
) cnt
FROM tbl
/

-- [([1248]*) and \1 can be removed because
-- it is 1 to 1 character replacement, so
-- check this out
SELECT REGEXP_REPLACE(
        '8F13247F7018'
     , '([3569AC]*)([7BDE]*)([F]*)[0]*'
     , '\1\1\2\2\2\3\3\3\3') cnt
FROM dual
/

-- This uses a dynamically built table and
-- replaces hex values by count of bits, as
-- you can see the principle is the same as
-- above in regexp implementation
-- Update 06/25/2005:
-- The code was changed a bit it's Ok not to
-- replace 1,2,4,8 and leave them as is.
SELECT NVL(
      LENGTH(
        REPLACE(REPLACE(REPLACE(
          TRANSLATE(TO_CHAR(n, 'FMXXXXXXXX')
                   , '35679ABCDEF0'
                   , '...,..,.,,:'
                  /* '22232232334' */
          )
        , '.', '11'), ',', '111'), ':', '1111')
      )
    , 0
    )
FROM tbl
/

-- Finally, hakmem...
SELECT n
, MOD(
   BITAND(
     (n - BITAND(n / 2, 3681400539) - BITAND(n / 4, 1227133513))
   + (n - BITAND(n / 2, 3681400539) - BITAND(n / 4, 1227133513)) / 8
   , 3340530119
   )
 , 63
 ) cnt
FROM tbl
WHERE n BETWEEN 0 AND 4294967295
/

Implemented in PL/SQL hakmem is quite fast. On 10gR2 it's even faster. :-) You can read more about hakmem here.

Update 06/24/2005: Here you can find more about bitcount implementations in C.

Update 08/08/2005: Bit Twiddling Hacks