You can count the number of bits that are set in a non-negative integer value with this code (adapted to JavaScript from this answer):
function countSetBits(i)
{
i = i - ((i >> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
return (((i + (i >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24;
}
It should be much more efficient than examining each bit individually. However, it doesn't work if the sign bit is set in i
.
EDIT (all credit to Pointy's comment):
function isPowerOfTwo(i) {
return i > 0 && (i & (i-1)) === 0;
}