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# Bit tricks: Fast implementation of strlen (Summary)

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### scott douglass

Feb 27, 1993, 2:16:56 AM2/27/93
to
Here's the summary of the responses I got to my query asking for the
trick of finding a nul in a long word and other bit tricks.

First thanks to everyone that responded:

de...@ccr-p.ida.org (David desJardins)
jh...@wpi.WPI.EDU (John Clinton Hall)
mo...@mcrcim.mcgill.edu (der Mouse)
Julian Templeman <jul...@bjalon.demon.co.uk>
wa...@Auspex.COM (Wally Bass)
Benjamin Ketcham <bket...@u.washington.edu>
em...@shell.portal.com (emil rojas)
ko...@dutentb.et.tudelft.nl (Koert van der Lingen)
Martin Gaeckler <g...@ifgmbh.uucp>
al...@bernie.sal.wisc.edu (Alan Watson)

I only claim that these are interesting and potentially useful, I don't
claim that they are the best way to implement something or even
correct. (I do _think_ they are correct.)

1 -- The answer to the find the nul question was sent to me by David
desJardins:

(n - 0x01010101) & (~n) & 0x80808080

If the result is zero then there are no nul bytes in "n". If "n" contains
a nul byte then the result will have the high bit of the corresponding
byte set. The reason this works is that the only byte for which (b - 1)
and ~b both have the high bit set is nul. If "n" contains a nul byte
other,
more significant bytes of the reslut may have their high bits set because
of borrowing, consider 0x78010012 whose result is 0x00808000.

Another possibility is:

n |= n << 1;
n |= n << 2;
n |= n << 4;
if ((n & 0x80808080) != 0x80808080)
/* then one of the four bytes started out zero */

To avoid needing to know the size of "n" 0x01010101 can be replaced with
(~0U)/255 and 0x80808080 with ((~0UL)/255)<<7.

If you use this trick to implement string functions such as strlen you
must be sure that you only fetch the long words at the correct,
implementation defined, alignment and that accessing 1, 2 or 3 bytes
beyond the end of a string will not cause harm.

For strings likely to consist of ASCII characters the following is
cheaper first test:

(n + 0x7f7f7f7f7f) & 0x80808080

If the result is 0x80808080 if the characters are all in the range 0x01
to 0x80.

This last is from: 'Faster String Functions' by Henry
Spencer (he...@zoo.toronto.edu) which was presented at the Winter '92
Usenix in San Francisco. I haven't found this to read yet.

2 -- Find the least significant set bit in "n" or 0 if 0:

(n & -n)

3 -- Clear the least significant set bit in "n", if any:

(n & (n - 1))

4 -- Find out if n is a power of 2 or 0 (e.g. 0, 2, 4, 8, 16, 32, ...):

(n & -n) == n /* uses # 2 */
or
(n & (n - 1)) == 0 /* uses # 3 */

One respondent said the second is mentioned in K&R. I couldn't find it in
a brief search. If you happen to know whereit's mentioned I'd like to
know
(so I could see if there are any other bit tricks next to it).

5 -- Counting the one bits in "n":

int count_bits(int n)
{
int count = 0; /* Number of bits set in i */

while (n) {
count++;
n = n & (n - 1); /* uses # 3 */
}

return count;
}

Other ways that depend on knowing the size of "n":

int count_bits(int n)
{
n = ((n & 0xaaaaaaaa) >> 1) + (n & 0x55555555);
n = ((n & 0xcccccccc) >> 2) + (n & 0x33333333);
n = ((n & 0xf0f0f0f0) >> 4) + (n & 0x0f0f0f0f);
n = ((n & 0xff00ff00) >> 8) + (n & 0x00ff00ff);
n = ((n & 0xffff0000) >> 16) + (n & 0x0000ffff);
return n;
}

If "%" is cheap enough you can stop before the end:

int count_bits(int n)
{
n = ((n & 0xaaaaaaaa) >> 1) + (n & 0x55555555);
n = ((n & 0xcccccccc) >> 2) + (n & 0x33333333);
n = ((n & 0xf0f0f0f0) >> 4) + (n & 0x0f0f0f0f);
return n % 255;
}

I got one reference that reportedly contains the find a nul trick:

"Obfuscated C and Other Mysteries" by Don Libes (Wiley, 1993, ISBN
0-471-57805-3). I haven't found this to read yet.

If you know of any more books, articles or papers on bit tricks like these
I'd be glad to have pointers.

Thanks again,
--scott

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