Hi all,
I am now looking for a function to return the position of the first
position of the left most set bit. And optionally another to return the
position of the right most set bit.
I have been looking at
"http://graphics.stanford.edu/~seander/bithacks.html#OperationCounting";
but it seems it will take me a while to figure out bit manipulation.
Allan.
Allan Kamau wrote:
All was well with the code below, apologies to all who read my
previous email. The error (an oversight) was on my part. In the
"CREATE FUNCTION ..." statement I had FLOAT as the return type instead
of INTEGER.
Now the function runs smoothly. Preliminary results show it is orders
of magnitude faster than the LENGTH(REGEXP(CAST(myVarBit AS
TEXT),'0','','g')) solution.
Thanks again TJ and the rest of the team.
Allan
Allan Kamau wrote:
Thank you TJ and everyone else for the advise and the c code. Today I
did finally return to the 'number of bits set challenge' and managed
to compile and link the nbits c function which went smoothly. However
the function does crash my postgres server installation (8.3.3) with
a segmentation fault each time I call it for example SELECT
nbits_set(B'1101');
My C skills are very sparse and am unable to debug the function, I
have included the C code of this function. Is there something I may
have left out?
#include "postgres.h"
#include "utils/varbit.h"
#include "fmgr.h"
#ifdef PG_MODULE_MAGIC
PG_MODULE_MAGIC;
#endif
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);
}
Allan
Bruce Momjian wrote:
Jean-David Beyer wrote:
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.
Uh, I was kind of confused by this, even when I saw a full
implementation:
http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetTable
Actually, this looks even better:
http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetKernighan
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