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What is the most efficient way to count the number of bits
which are set in a value?
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main()
{
int n,count=0;
printf("enter a number");
scanf("%d",&n);//enterd no.is 5
while(n>0)
{
count++;
n=n&n-1;//binary of n is 101
// binary of n-1 is 100
//n&n-1 is (i.e 101
&100 =100 )
}
printf("%d",count);
getch();
} output is 2(i.e 2 ones in 101)
| Is This Answer Correct ? | 31 Yes | 6 No |
Answer / pappu kumar sharma
int fnCntbts(int num )
{
int iCnt = 0;
while ( num )
{
num &= (num-1) ;
iCnt++;
}
return iCnt;
}
| Is This Answer Correct ? | 11 Yes | 2 No |
Answer / boomer
pseudo
int size is 32 bits...right?
for(i = 0 to 31)
{
count += (value & 1) //& = and oprator
shift left value
}
| Is This Answer Correct ? | 6 Yes | 5 No |
Answer / bryan w
Question needs clarification. What platform? There are
bitwise trick that are optimal for various platforms, but
you need to know if the value is 16 bit, 32 bit, 64 bit, 128
bit, or something else entirely.
The earlier three examples are all incorrect for operating
on signed integers; if a negative value is presented the
code will fail.
The most efficient portable human readable answer is to loop
over sizeof type * CHAR_BITS times, shift left and add one
if the bit is set.
On some platforms the most efficient way is to use a bit of
assembly (such as the POPCNT instruction if available) to
perform the operation for you.
Without either of those, there are simple classic algorithms
of &, |, and ^ to accumulate the bits and sum them. They
need to be adjusted to match the architecture's number of
bits. These routines may be inefficient on modern PCs with
long pipelines or out-of-order processing cores.
| Is This Answer Correct ? | 1 Yes | 2 No |
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