Hi,

Can anyone tell what is the most efficient way to count the number of bits which are set in a value?

Thanks,
Prasath.K

hey,


foo(unsigned char *ByteField)
{
int i=0,j=0;
unsigned char *BytePos;
short Bit;

BytePos = ByteField + sizeof(ByteField);

for (i=0; i
{
if (*(BytePos - i) > (char ) 0)
{
for (j=0; j<8; j++)
{

if (((*(BytePos - i)) & 1<<
{
Bit = (i * 8) + j;
printf("Bit Set %d n",Bit);
}
}
}
}
}

Should give u the bit which are set

__________________
Venkat
knowledge is Power

try the followin function, i dint try it yet, but hope it will work

int countBits(int val){
int count=0;
while(val != 0){
if((val>>1) == 1){
count++;
}
val >>= 1;
}
return count;
}

__________________
SeeSamJagan
- Sky is not the "LIMIT", Death is not the END, There is still something beyond....

Hi,
i just made small mistake in the last post
try the modified function

int countBits(int val){

int count=0;
int len = sizeof(val)*8;
while(len != 0){
if((val>>1) == 1){

count++;

}
val >>= 1;
len--;
}
return count;

}

__________________
SeeSamJagan
- Sky is not the "LIMIT", Death is not the END, There is still something beyond....