Counting number of bits in an unsigned long - c

Question

So I've written this function to count the number of bits in a long, which for my purposes includes zeros to the right of the MSB and excludes zeros to its left:
int bitCount(unsigned long bits)
{
int len = 64;
unsigned long mask = 0x8000000000000000;
while ((bits & mask) == 0 && len > 0){
mask >>= 1;
--len;
}
return len;
}
This function works fine for me as far as returning a correct answer, but is there a better (faster or otherwise) way to go about doing this?

Answer 1

If you want to count the number of bits in a long type, I suggest you use ULONG_MAX from the <limits.h> header file, and use the right shift operator >> to count the number of one-bits. This way you don't have to actually know the number of bits beforehand.
Something like
unsigned long value = ULONG_MAX;
unsigned count = 1;
while (value >>= 1)
++count;
This works because the right shift fills up with zeroes.

Answer 2

The general answer for the number of bits in any type is CHAR_BIT*sizeof(type). CHAR_BIT, defined in <limits.h> is the (implementation-defined) number of bits in a char. sizeof(type) is specified in a way that yields the number of chars used to represent the type (i.e. sizeof(char) is 1).

Answer 3

The solutions the other guys proposed are very nice and probably the shortest to write and remain understandable. Another straight forward approach would be something like this
int bitCountLinear(long int n) {
int len = sizeof(long int)*8;
for (int i = 0; i < len; ++i)
if ((1UL<<i) > (unsigned long int)n)
return i;
return len;
}
The rest might get a bit extreme but I gave it a try so I'll share it.
I suspected that there might be arguably faster methods of doing this. eg Using binary search (even though a length of 64bits is extremely small). So I gave it a quick try for you and for the fun of it.
union long_ing_family {
unsigned long int uli;
long int li;
};
int bitCountLogN(long int num) {
union long_ing_family lif;
lif.li = num;
unsigned long int n = lif.uli;
int res;
int len = sizeof(long int)*8-1;
int max = len;
int min = 0;
if (n == 0) return 0;
do {
res = (min + max) / 2;
if (n < 1UL<<res)
max = res - 1;
else if (n >= (1UL<<(res+1)))
min = res + 1;
else
return res+1;
} while (min < max);
return min+1; // or max+1
}
I then timed both to see if they have any interesting differences...
#include <stdio.h>
#define REPS 10000000
int bitCountLinear(long int n);
int bitCountLogN(long int num);
unsigned long int timestamp_start(void);
unsigned long int timestamp_stop(void);
union long_ing_family;
int main(void) {
long int n;
long int begin, end;
long int begin_Lin, end_Lin;
long int begin_Log, end_Log;
begin_Lin = 0;
end_Lin = 0;
begin_Log = 0;
end_Log = 0;
for (int i = 0; i < REPS; ++i) {
begin_Lin += timestamp_start();
bitCountLinear(i);
end_Lin += timestamp_stop();
}
printf("Linear: %lu\n", (end_Lin-begin_Lin)/REPS);
for (int i = 0; i < REPS; ++i) {
begin_Log += timestamp_start();
bitCountLogN(i);
end_Log += timestamp_stop();
}
printf("Log(n): %lu\n", (end_Log-begin_Log)/REPS);
}
unsigned long int timestamp_start(void) {
unsigned int cycles_low;
unsigned int cycles_high;
asm volatile ("CPUID\n\t"
"RDTSCP\n\t"
"mov %%edx, %0\n\t"
"mov %%eax, %1\n\t": "=r" (cycles_high), "=r" (cycles_low)::"%rax", "%rbx", "%rcx", "%rdx");
return ((unsigned long int)cycles_high << 32) | cycles_low;
}
unsigned long int timestamp_stop(void) {
unsigned int cycles_low;
unsigned int cycles_high;
asm volatile ("RDTSCP\n\t"
"mov %%edx, %0\n\t"
"mov %%eax, %1\n\t"
"CPUID\n\t": "=r" (cycles_high), "=r" (cycles_low)::"%rax", "%rbx", "%rcx", "%rdx");
return ((unsigned long int)cycles_high << 32) | cycles_low;
}
...and not surprisingly they didn't.
On my machine I'll get numbers like
Linear: 228
Log(n): 224
Which are not considered to be different assuming a lot of background noise.
Edit:
I realized that I only tested the fastest solutions for the Linear approach so changing the function inputs to
bitCountLinear(0xFFFFFFFFFFFFFFFF-i);
and
bitCountLogN(0xFFFFFFFFFFFFFFFF-i);
On my machine I'll get numbers like
Linear: 415
Log(n): 269
Which is clearly a win for the Log(n) method. I didn't expect to see a difference here.

Answer 4

You can count the number of bit 1 first:
int bitCount(unsigned long n)
{
unsigned long tmp;
tmp = n - ((n >> 1) & 0x7777777777777777)
- ((n >> 2) & 0x3333333333333333)
- ((n >> 3) & 0x1111111111111111);
tmp = (tmp + (tmp >> 4)) & 0x0F0F0F0F0F0F0F0F;
return 64 - tmp % 255; // temp % 255 is number of bit 1
}
Take a look at the MIT HAKMEM Count.