CareerCup

This link has code which counts it for both positive and negative numbers.

onestopinterviewprep.blogspot.com/2014/03/count-1s-in-binary-format-of-number.html

Enjoy :)

int onesInBinary(int n)
{
    int total = 0;
    while(n>0)
    {
        total += n%2;
        n /= 2;
    }

    return total;
}

{ public static int getNumOnes(int val)
{
int count = 0;
if(val<=0)
return count;
else
{
int mask=0X0001;

while(val!=0)
{
int c = mask & val;
{
if(c == 1)
count++;
val=val>>>1;
}

}
}
return count;
}}

main()
{
int n;
scanf("%d",&n);
num_of_1s(n);
}
void num_of_1s(int n)
{
static int count =0;
if(n)
count++;
esle
printf(count);
}

int main()
{
int num=10;
int count=0;
while(num)
{
 count +=num%2 ==0? 0:1;
 num = num>>1;
}
cout<<count;
return 0;
}

C# using Math.Pow function. Not as sexy as the other solutions, but it works :)

protected int NumberOfOnes(int anInt)
    {
        int retVal = 0;
        
        int pow = 0;
        while (Math.Pow(2,pow) <= anInt)
            pow++;

        while (pow >= 0) {
            if (anInt >= Math.Pow(2,pow)) {
                anInt -= (int)Math.Pow(2,pow);
                retVal++;
            }
            pow--;
        }
        return retVal;
    }

To resolve the negative integer issue, cast to unsigned.

unsigned int count1s(int number) {
	unsigned int unumber = (unsigned int) number;
	unsigned int count = 0;
	while(unumber) {
		count += 1 ? (unumber & 1) : 0;
		unumber = unumber >> 1;
	};
	return count;
}

run like ./prog.rb 0 1 234 234234234

ARGV.each.map(&:to_i).each.inject(0) {|count, n| (((count += (n & 1)) && (n = n >> 1) while n != 0) || (puts count)) ? 0 : 0}

outputs:

0
1
5
16

In Golfscript, run like this echo "1 2 3 365 2 1" | golfscript [scriptname]:

[~]{{0=!}{2/}/}%{{1&}%{+}*}%`

def count1s(self, num):
        x = [ c for c in bin(num) if c == '1']
        return len(x)

private int count(int val) {
		int ret = 0;
		while(val > 0) {
			val&=(val-1);
			ret++;
		}

		return ret;
	}