int bitCount(int i)
{
	i = i - ((i >> 1) & 0x55555555);
	i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
	return (((i + (i >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24;
}

Count the number of ones in the binary representation of i.

Auto comment: topic has been updated by Frez (previous revision, new revision, compare).

for set/turn on the j-th item of the set, use to bitwise OR operation
S |= (1<<j)
S = 5 (base 10) = 101 (base 2)
j = 3 (base 10) = 011 (base 2)
                 ------------- OR
S = 7 (base 10) = 111 (base 2)
---------------------------
To check if the j'th item of the set is on,use the bitwise AND operation
P = S & (1<<j)
S = 5 (base 10) = 101 (base 2)
j = 2 , 1<<j    = 100 (base 2)
                 ------------- AND
p = 8 (base 10) = 100 no Ziro ,the 3rd item is 1
---------------------------
multiply & divide by 2
s*2 --> s<<1
s/2 --> s>>1
s/pow(2,n) --> s>>n
s*pow(2,n) --> s<<n
---------------------------

You can find many tricks in Hacker's Delight book by Henry S. Warren.

https://graphics.stanford.edu//bithacks.html

A good article-sized read is the relevant TopCoder tutorial by bmerry.

For a more thorough study, the Hacker's Delight book mentioned before is the way to go.