Java interview question: Count the number of “1” bits in a byte

Frankly, I think it is quite an idiotic question. Not because it is too easy or too hard question, but because I think this question doesn’t really imply on the practical programming capabilities of the person being interviewed. Anyway, a friend of mine was asked to solve this question, so I though it might be an interest for more developers looking for a job and having to handle all sort of weird questions.
In general, there are 2 solutions for this question:

• First solution is iterating 8 times (a byte contains 8 bits). Each time the right most bit is extracted from the byte. and then the byte is shifted right one bit.
• Second solution, is a little bit less straight forward, but more efficient: Divide the number by 2, if there is a modulo, count it. Repeat this operation as long as the number we are dividing is greater than zero.

Here is how the first solution looks in Java code:

public class GetBits {
public static int countBits(byte num)
{
  int count = 0;
  for (int i = 0; i < 8; i++)
{
    if ((num & 1) == 1) // check if right most bit is 1
{
count++;
}
num = (byte)(num >>> 1); // shit right 1 bit, including the sign bit
}
  return count;    
}

Here is how the second solution looks like:

public static int countBits(byte num)
{
  int count = 0;
  while (num > 0)
{
    if (num % 2 == 1) // check if number have modulo
{
count++;
}
num /= 2; // divide the number by 2
}
  return count;    
}

Note, that the first solution is a bit less efficient, because it always iterates 8 times. But, it handles both positive and negative values. The second solution will simply won’t work with negative numbers.