how do u find out the number of 1's in the binary
representation of a decimal number without converting it
into binary(i mean without dividing by 2 and finding out
the remainder)? three lines of c code s there it
seems...can anyone help
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Answer / ravi saini
unsigned int count=0,no;
/*
Enter no Here................
*/
while(no ! =0)
{
if((no & 0x01) ==1)
count++;
no=no >>1
}
printf("%d",count);
| Is This Answer Correct ? | 9 Yes | 3 No |
Answer / barun
int x =0x1;
static count;
while(no ! =0)
{
if((no >>1 & x) ==1)
count+=1;
}
printf("%d",count);
| Is This Answer Correct ? | 7 Yes | 2 No |
Answer / srsabariselvan
int main()
{
int n,i=0;
scanf("%d",&n);
while(n!=0)
{
if(n&01)
i++;
n>>=1;
}
printf("%d",i);
}
| Is This Answer Correct ? | 3 Yes | 2 No |
Answer / krishna
The above solutions will not work for negative numbers.
if a negative number is right shifted, the most significant bit will become 1. so the number never becomes zero.
The following will work for both +ve and -ve numbers.
#include <stdio.h>
int main(int argc, char *argv[]) {
int count = 0;
int no = 3;
int x = 0x01;
if ( argc > 1) {
no = atoi(argv[1]);
}
while ( x != 0) {
if( no & x ) {
count++;
}
x <<= 1;
}
printf( "number: %d , no.of 1s in it: %d\n", no, count);
return 0;
}
| Is This Answer Correct ? | 1 Yes | 2 No |
Answer / jay
Well bitwise operators are one way to do it. But for
positive integers, I thought up another solution. I think
this could work for int n:
for(int i = 0, int count = 0; i < sizeOf(n) / 8.0; i++){
count += (n - 2^i > 0) ? 1 : 0;
}
Is that what you meant by 3 lines?
| Is This Answer Correct ? | 0 Yes | 1 No |
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