CareerCup

Is the representation of floating point number given? for example IEEE754
If so, just count. If not, unsolvable.

Yes its IEEE representation.
Still do you need to count the number of bits that are set in 32 bit representation?

I think this can't be avoided, unless you use some compiler-specific functions like __builtin_popcount in gcc.

If I am not wrong we need t count the number of bits that are set in the mantissa part + 1 for finding the bits set in the floating point number?

Brian Kernighan's method goes through as many iterations as there are set bits. So if we have a 32-bit word with only the high bit set, then it will only go once through the loop.

Published in 1988, the C Programming Language 2nd Ed. (by Brian W. Kernighan and Dennis M. Ritchie) mentions this in exercise 2-9. On April 19, 2006 Don Knuth pointed out to him that this method "was first published by Peter Wegner in CACM 3 (1960), 322. (Also discovered independently by Derrick Lehmer and published in 1964 in a book edited by Beckenbach.)"

long count_bits(long n) {
unsigned int c; // c accumulates the total bits set in v
for (c = 0; n; c++)
n &= n - 1; // clear the least significant bit set
return c;
}

My question here is if such a question is given do we need to find out the number of bits that are set in the 32 bit representation of just the mantissa part as the rest are signed bits and exponent.

count=0;
if(n&1)
{
count++;
n=n>>1;
}
return count;

Here, no of bits set in the memory layout is what we need, right? We could use a union as below:

#include <stdio.h>

int main()
{
    union u {
        float num;
        int i;
    } u1;
    u1.num = 0.15625; /* floating number to check */

    int num_bits = 0;
    while (u1.i) {
        num_bits++;
        u1.i = u1.i & (u1.i-1);
    }

    printf("No of bits is %d\n", num_bits);
}

Wouldnt this do?