Author rhettinger
Recipients fredrikj, haypo, loewis, mark.dickinson, pitrou, rhettinger, terry.reedy
Date 2008-12-16.12:41:39
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Message-id <1229431303.3.0.869291081934.issue3439@psf.upfronthosting.co.za>
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In the whatsnew docs, add two lines showing the binary representation of
37.  That will provide clarification as to why the answer is six:

      >>> n = 37
+     >>> bin(37)
+     '0b100101'
      >>> n.bit_length()
      6

Also, the main entry in the docs should be built-out just a bit.  I like
that the current entry provides a precise mathematical spec that is
easily validated, but it should also mention the more intuitive notion
of the "number of bits in a number's binary representation excluding
leading zeros (for example the decimal number 37, which is 0b100101 in
binary, has six binary digits)."

Possibly, the BitLengthTables can be dropped in favor of the old
shift-one, add-one code (to be inserted right after the shift-eight,
add-eight code).  The two tables consume a lot of memory and the
relevant entry is not likely to be in cache when needed (cache misses
are very slow).  It is simpler and possibly faster to skip the table
lookup unless the table is made much smaller.  Plus, it makes the code a
bit more maintainable and self-evidently correct (not requiring as
extensive test coverage to validate the whole table).

My own crack at optimization looks like this:

        static const char BitLengthTable[32] =
            {0, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4,
             5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5};
        while (n >= 32) { r += 5;   n >>= 5; }
        r += (long)(BitLengthTable[n]);

If you don't adopt mine, at least consider making a table of chars
instead of longs (this will give a 4:1 or 8:1 table compression,
reducing the memory footprint and increasing the likelihood of a cache hit).

I would feel better about the test code if it also directly validated
the definitions in docs and thoroughly exercised the tables:

for x in range(-65000, 65000):
    k = x.numbits
    if x > 0:
         assert 2 ** (k-1) <= x < 2**k
         assert k == 1 + math.floor(math.log(x) / math.log(2))
    elif x == 0:
         assert k == 0
    else:
         assert k == (-x).numbits()



Other than that, I'm basically happy with the patch.
History
Date User Action Args
2008-12-16 12:41:43rhettingersetrecipients: + rhettinger, loewis, terry.reedy, mark.dickinson, pitrou, haypo, fredrikj
2008-12-16 12:41:43rhettingersetmessageid: <1229431303.3.0.869291081934.issue3439@psf.upfronthosting.co.za>
2008-12-16 12:41:42rhettingerlinkissue3439 messages
2008-12-16 12:41:39rhettingercreate