Comments on: Interview Questions: Counting Bits

By: sven.hedin

sven.hedin — Mon, 28 Jun 2010 15:34:21 +0000

In all codes above you used 0×1u, why not only 1 instead? I think both do the same thing.

By: Hrish

Hrish — Wed, 16 Jun 2010 08:06:03 +0000

Just noticed that the link http://20bits.com/tag/interview/ at the beginning of your article is broken.

By: gournew

gournew — Thu, 08 Apr 2010 23:36:10 +0000

Now in this case we can reduce the number of iteration..
as for example if input is 64 then the binary will be 100000
so if normal procedure used then we have to itrate 6v times in the loop

simple steps
int count_bit(int num)
{
count =0;
while(num)
{
num=(num) & (num-1);
count++;
}
return count;
// complete program will be available in
// http://goursaha.freeoda.com
}

By: axvpast

axvpast — Tue, 23 Feb 2010 03:27:57 +0000

it is not so good answers all, because best way to solve tasks like:
We choose the routing destination based on the number of on bits in the binary representation of the routing number.

Using VHDL :) in hardware level (like CRC32 computation). All can be done in one instruction like it do real routers :)

By: mohdhussain

mohdhussain — Wed, 03 Feb 2010 03:47:20 +0000

bits of accounts questions answers

By: Reuben

Reuben — Thu, 08 May 2008 00:16:13 +0000

I’m surprised the bit counting in parallel code snippet wasn’t included in this post too — I thought it was a classic. Here’s a link to the algorithm (in response to roi):

http://www.cs.utk.edu/~vose/c-stuff/bithacks.html#CountBitsSetParallel

This version doesn’t use a for loop so that you can clearly see the c*log_2{n} == c*5 steps (for n=32). Easily changed to accomodate counting bits set for 64 bit integers of course.

Another interesting solution, which is faster than the naive solution only for integers where the bits set are few, is here:

http://www.cs.utk.edu/~vose/c-stuff/bithacks.html#CountBitsSetKernighan

By: roi

roi — Wed, 30 Apr 2008 20:30:55 +0000

I see that Wordpress didn’t like it. Another attempt at the punchline:

for ( i = 0; (1< > (1<

By: roi

roi — Wed, 30 Apr 2008 20:19:02 +0000

There’s a cuter, log n algorithm:

int bitcount(unsigned int n) {
    unsigned int i, masks[] = {0x55555555, 0x33333333, 0x0F0F0F0F, 0x00FF00FF, 0x0000FFFF};
    for ( i = 0; (1< (1<
(I hope it survives wordpress’ munging)
Creating the bitmasks programatically, to make it independent of int size, is left as a trivial exercise for the reader.