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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 Posted / 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 |
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