In article <42************ ***********@aut hen.white.readf reenews.net>,
sathyashrayan <sa***********@ REMOVETHISgmail .com> wrote:
Group,
Following function will check weather a bit
is set in the given variouble x.
int bit_count(long x)
[snippage]
Actually, it (non-portably -- x should have type "unsigned long")
counts the *number* of bits set.
I think it is time to re-post this Moldy Oldie :-) The stuff at
the top is not Standard C, and the timing code as well, but the
bit-counting functions are at least commented. Note that I wrote
this shortly after the 1989 C Standard came out, before it was
widely available -- we were not even using GCC yet on most machines.
#ifndef lint
static char rcsid[] = "$Id: bct.c,v 1.5 90/10/13 08:44:12 chris Exp $";
#endif
/*
* bct - bitcount timing
*
* Runs a bunch of different functions-to-count-bits-in-a-longword
* (many assume 32 bits per long) with a timer around each loop, and
* tries to calcuate the time used.
*/
#include <stdio.h>
#include <sys/types.h>
#ifdef FD_SETSIZE
# define USE_GETRUSAGE
# include <sys/time.h>
# include <sys/resource.h>
#else
# include <sys/times.h>
#endif
#define SIZEOF(a) (sizeof(a)/sizeof(a[0]))
/*
* This function is used to calibrate the timing mechanism.
* This way we can subtract the loop and call overheads.
*/
int
calibrate(n)
register unsigned long n;
{
return 0;
}
/*
* This function counts the number of bits in a long.
* It is limited to 63 bit longs, but a minor mod can cope with 511 bits.
*
* It is so magic, an explanation is required:
* Consider a 3 bit number as being
* 4a+2b+c
* if we shift it right 1 bit, we have
* 2a+b
* subtracting this from the original gives
* 2a+b+c
* if we shift the original 2 bits right we get
* a
* and so with another subtraction we have
* a+b+c
* which is the number of bits in the original number.
* Suitable masking allows the sums of the octal digits in a
* 32 bit number to appear in each octal digit. This isn't much help
* unless we can get all of them summed together.
* This can be done by modulo arithmetic (sum the digits in a number by molulo
* the base of the number minus one) the old "casting out nines" trick they
* taught in school before calculators were invented.
* Now, using mod 7 wont help us, because our number will very likely have
* more than 7 bits set. So add the octal digits together to get base64
* digits, and use modulo 63.
* (Those of you with 64 bit machines need to add 3 octal digits together to
* get base512 digits, and use mod 511.)
*
* This is HACKMEM 169, as used in X11 sources.
*/
int
t0_hackmemmod(n )
register unsigned long n;
{
register unsigned long tmp;
tmp = n - ((n >> 1) & 033333333333) - ((n >> 2) & 011111111111);
return ((tmp + (tmp >> 3)) & 030707070707) % 63;
}
/*
* This is the same as the above, but does not use the % operator.
* Most modern machines have clockless division, and so the modulo is as
* expensive as, say, an addition.
*/
int
t1_hackmemloop( n)
register unsigned long n;
{
register unsigned long tmp;
tmp = n - ((n >> 1) & 033333333333) - ((n >> 2) & 011111111111);
tmp = (tmp + (tmp >> 3)) & 030707070707;
while (tmp > 63)
tmp = (tmp & 63) + (tmp >> 6);
return tmp;
}
/*
* Above, without using while loop. It takes at most 5 iterations of the
* loop, so we just do all 5 in-line. The final result is never 63
* (this is assumed above as well).
*/
int
t2_hackmemunrol led(n)
register unsigned long n;
{
register unsigned long tmp;
tmp = n - ((n >> 1) & 033333333333) - ((n >> 2) & 011111111111);
tmp = (tmp + (tmp >> 3)) & 030707070707;
tmp = (tmp & 63) + (tmp >> 6);
tmp = (tmp & 63) + (tmp >> 6);
tmp = (tmp & 63) + (tmp >> 6);
tmp = (tmp & 63) + (tmp >> 6);
tmp = (tmp & 63) + (tmp >> 6);
return (tmp);
}
/*
* This function counts the bits in a long.
*
* It removes the lsb and counting the number of times round the loop.
* The expression (n & -n) yields the lsb of a number,
* but it only works on 2's compliment machines.
*/
int
t3_rmlsbsub(n)
register unsigned long n;
{
register int count;
for (count = 0; n; n -= (n & -n))
count++;
return count;
}
int
t4_rmlsbmask(n)
register unsigned long n;
{
register int count;
for (count = 0; n; count++)
n &= n - 1; /* take away lsb */
return (count);
}
/*
* This function counts the bits in a long.
*
* It works by shifting the number down and testing the bottom bit.
*/
int
t5_testlsb(n)
register unsigned long n;
{
register int count;
for (count = 0; n; n >>= 1)
if (n & 1)
count++;
return count;
}
/*
* This function counts the bits in a long.
*
* It works by shifting the number left and testing the top bit.
* On many machines shift is expensive, so it uses a cheap addition instead.
*/
int
t6_testmsb(n)
register unsigned long n;
{
register int count;
for (count = 0; n; n += n)
if (n & ~(~(unsigned long)0 >> 1))
count++;
return count;
}
int
t7_testsignands hift(n)
register unsigned long n;
{
register int count;
for (count = 0; n; n <<= 1)
if ((long)n < 0)
count++;
return (count);
}
/*
* This function counts the bits in a long.
*
* It works by masking each bit.
* This is the second most intuitively obvious method,
* and is independent of the number of bits in the long.
*/
int
t8_testeachbit( n)
register unsigned long n;
{
register int count;
register unsigned long mask;
count = 0;
for (mask = 1; mask; mask += mask)
if (n & mask)
count++;
return count;
}
/*
* This function counts the bits in a long.
*
* It works by masking each bit.
* This is the most intuitively obvious method,
* but how do you a priori know how many bits in the long?
* (except for ''sizeof(long) * CHAR_BITS'' expression)
*/
int
t9_testeachbit1 shl(n)
register unsigned long n;
{
register int count;
register int bit;
count = 0;
for (bit = 0; bit < 32; ++bit)
if (n & ((unsigned long)1 << bit))
count++;
return count;
}
static char nbits[256] = {
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8,
};
static int inbits[256] = {
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8,
};
int
tA_tableshift(n )
register unsigned long n;
{
return (nbits[n & 0xff] + nbits[(n >> 8) & 0xff] +
nbits[(n >> 16) & 0xff] + nbits[n >> 24]);
}
int
tB_tableuchar(n )
unsigned long n;
{
register unsigned char *p = (unsigned char *)&n;
return (nbits[p[0]] + nbits[p[1]] + nbits[p[2]] + nbits[p[3]]);
}
int
tC_tableshiftca st(n)
register unsigned long n;
{
return nbits[(unsigned char) n] +
nbits[(unsigned char) (n >> 8)] +
nbits[(unsigned char) (n >> 16)] +
nbits[(unsigned char) (n >> 24)];
}
int
tD_itableshift( n)
register unsigned long n;
{
return (inbits[n & 0xff] + inbits[(n >> 8) & 0xff] +
inbits[(n >> 16) & 0xff] + inbits[n >> 24]);
}
int
tE_itableuchar( n)
unsigned long n;
{
register unsigned char *p = (unsigned char *)&n;
return (inbits[p[0]] + inbits[p[1]] + inbits[p[2]] + inbits[p[3]]);
}
int
tF_itableshiftc ast(n)
register unsigned long n;
{
return inbits[(unsigned char) n] +
inbits[(unsigned char) (n >> 8)] +
inbits[(unsigned char) (n >> 16)] +
inbits[(unsigned char) (n >> 24)];
}
/*
* Explanation:
* First we add 32 1-bit fields to get 16 2-bit fields.
* Each 2-bit field is one of 00, 01, or 10 (binary).
* We then add all the two-bit fields to get 8 4-bit fields.
* These are all one of 0000, 0001, 0010, 0011, or 0100.
*
* Now we can do something different, becuase for the first
* time the value in each k-bit field (k now being 4) is small
* enough that adding two k-bit fields results in a value that
* still fits in the k-bit field. The result is four 4-bit
* fields containing one of {0000,0001,..., 0111,1000} and four
* more 4-bit fields containing junk (sums that are uninteresting).
* Pictorially:
* n = 0aaa0bbb0ccc0dd d0eee0fff0ggg0h hh
* n>>4 = 00000aaa0bbb0cc c0ddd0eee0fff0g gg
* sum = 0aaaWWWWiiiiXXX XjjjjYYYYkkkkZZ ZZ
* where W, X, Y, and Z are the interesting sums (each at most 1000,
* or 8 decimal). Masking with 0x0f0f0f0f extracts these.
*
* Now we can change tactics yet again, because now we have:
* n = 0000WWWW0000XXX X0000YYYY0000ZZ ZZ
* n>>8 = 000000000000WWW W0000XXXX0000YY YY
* so sum = 0000WWWW000pppp p000qqqqq000rrr rr
* where p and r are the interesting sums (and each is at most
* 10000, or 16 decimal). The sum `q' is junk, like i, j, and
* k above; but it is not necessarry to discard it this time.
* One more fold, this time by sixteen bits, gives
* n = 0000WWWW000pppp p000qqqqq000rrr rr
* n>>16 = 000000000000000 00000WWWW000ppp pp
* so sum = 0000WWWW000pppp p000sssss00tttt tt
* where s is at most 11000 and t is it most 100000 (32 decimal).
*
* Now we have t = r+p = (Z+Y)+(X+W) = ((h+g)+(f+e))+( (d+c)+(b+a)),
* or in other words, t is the number of bits set in the original
* 32-bit longword. So all we have to do is return the low byte
* (or low 6 bits, but `low byte' is typically just as easy if not
* easier).
*
* This technique is also applicable to 64 and 128 bit words, but
* 256 bit or larger word sizes require at least one more masking
* step.
*/
int
tG_sumbits(n)
register unsigned long n;
{
n = (n & 0x55555555) + ((n >> 1) & 0x55555555);
n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
n = (n + (n >> 4)) & 0x0f0f0f0f;
n += n >> 8;
n += n >> 16;
return (n & 0xff);
}
/*
* build a long random number.
* The standard rand() returns at least a 15 bit number.
* We use the top 9 of 15 (since the lower N bits of the usual rand()
* repeat with a period of 2^N).
*/
unsigned long
bigrand()
{
#define randbits() ((unsigned long)((rand() >> 6) & 0777))
register unsigned long r;
r = randbits() << 27;
r |= randbits() << 18;
r |= randbits() << 9;
r |= randbits();
return (r);
}
/*
* Run the test many times.
* You will need a running time in the 10s of seconds
* for an accurate result.
*
* To give a fair comparison, the random number generator
* is seeded the same for each measurement.
*
* Return value is seconds per iteration.
*/
#ifndef REPEAT
#if defined(mips) || defined(sparc)
#define REPEAT (1L<<20)
#else
#define REPEAT (1L<<18)
#endif
#endif
double
measure(func)
register int (*func)();
{
#ifdef USE_GETRUSAGE
struct rusage ru0, ru1;
register long j;
srand(1);
(void) getrusage(RUSAG E_SELF, &ru0);
for (j = 0; j < REPEAT; ++j)
func(bigrand()) ;
(void) getrusage(RUSAG E_SELF, &ru1);
ru1.ru_utime.tv _sec -= ru0.ru_utime.tv _sec;
if ((ru1.ru_utime. tv_usec -= ru0.ru_utime.tv _usec) < 0) {
ru1.ru_utime.tv _usec += 1000000;
ru1.ru_utime.tv _sec--;
}
return ((ru1.ru_utime. tv_sec + (ru1.ru_utime.t v_usec / 1000000.0)) /
(double)REPEAT) ;
#else
register long j;
struct tms start;
struct tms finish;
srand(1);
times(&start);
for (j = 0; j < REPEAT; ++j)
func(bigrand()) ;
times(&finish);
return ((finish.tms_ut ime - start.tms_utime ) / (double)REPEAT) ;
#endif
}
struct table {
char *name;
int (*func)();
double measurement;
int trial;
};
struct table table[] =
{
{ "hackmemmod ", t0_hackmemmod },
{ "hackmemloo p", t1_hackmemloop },
{ "hackmemunrolle d", t2_hackmemunrol led },
{ "rmlsbsub", t3_rmlsbsub },
{ "rmlsbmask" , t4_rmlsbmask },
{ "testlsb", t5_testlsb },
{ "testmsb", t6_testmsb },
{ "testsignandshi ft", t7_testsignands hift },
{ "testeachbi t", t8_testeachbit },
{ "testeachbit1sh l", t9_testeachbit1 shl },
{ "tableshift ", tA_tableshift },
{ "tableuchar ", tB_tableuchar },
{ "tableshiftcast ", tC_tableshiftca st },
{ "itableshif t", tD_itableshift },
{ "itableucha r", tE_itableuchar },
{ "itableshiftcas t", tF_itableshiftc ast },
{ "sumbits", tG_sumbits },
};
main(argc, argv)
int argc;
char **argv;
{
double calibration, v, best;
int j, k, m, verbose;
verbose = argc > 1;
/*
* run a few tests to make sure they all agree
*/
srand(getpid()) ;
for (j = 0; j < 10; ++j) {
unsigned long n;
int bad;
n = bigrand();
for (k = 0; k < SIZEOF(table); ++k)
table[k].trial = table[k].func(n);
bad = 0;
for (k = 0; k < SIZEOF(table); ++k)
for (m = 0; m < SIZEOF(table); ++m)
if (table[k].trial != table[m].trial)
bad = 1;
if (bad) {
printf("wrong: %08lX", n);
for (k = 0; k < SIZEOF(table); ++k)
printf(" %3d", table[k].trial);
printf("\n");
}
}
/*
* calibrate the timing mechanism
*/
calibration = measure(calibra te);
if (verbose)
printf("calibra tion: %g\n", calibration);
/*
* time them all, keeping track of the best (smallest)
*/
for (j = 0; j < SIZEOF(table); ++j) {
v = measure(table[j].func) - calibration;
if (verbose)
printf("%s: %g\n", table[j].name, v);
table[j].measurement = v;
if (!j || v < best)
best = v;
}
(void) printf("%-24s %-14sratio\n", "function", "time");
for (j = 0; j < SIZEOF(table); ++j) {
(void) printf("%-24s %#10.8g %6.3f\n",
table[j].name,
table[j].measurement,
table[j].measurement / best);
}
exit(0);
}
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
email: forget about it
http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.
