Re: Re: Efficiently determining the number of bits set in the contents of, a VARBIT field

TJ O'Donnell wrote:
> I use a c function, nbits_set that will do what you need.
> I've posted the code in this email.
> 
> TJ O'Donnell
> http://www.gnova.com
> 
> #include "postgres.h"
> #include "utils/varbit.h"
> 
> Datum   nbits_set(PG_FUNCTION_ARGS);
> PG_FUNCTION_INFO_V1(nbits_set);
> Datum
> nbits_set(PG_FUNCTION_ARGS)
> {
> /* how many bits are set in a bitstring? */
> 
>          VarBit     *a = PG_GETARG_VARBIT_P(0);
>          int n=0;
>          int i;
>          unsigned char *ap = VARBITS(a);
>          unsigned char aval;
>          for (i=0; i < VARBITBYTES(a); ++i) {
>                  aval = *ap; ++ap;
>                  if (aval == 0) continue;
>                  if (aval & 1) ++n;
>                  if (aval & 2) ++n;
>                  if (aval & 4) ++n;
>                  if (aval & 8) ++n;
>                  if (aval & 16) ++n;
>                  if (aval & 32) ++n;
>                  if (aval & 64) ++n;
>                  if (aval & 128) ++n;
>          }
>          PG_RETURN_INT32(n);
> }
> 
> 
> 
>> Hi all,
>> Am looking for a fast and efficient way to count the number of bits set 
>> (to 1) in a VARBIT field. I am currently using 
>> "LENGTH(REGEXP_REPLACE(CAST(a.somefield_bit_code AS TEXT),'0','','g'))".
>>
>> Allan.
> 
> 
When I had to do that, in days with smaller amounts of RAM, but very long
bit-vectors, I used a faster function sort-of like this:

static char table[256] = {
0,1,1,2,1,2,2,3,1,.....
};

Then like above, but instead of the loop,

n+= table[aval];


You get the idea.

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