Determines whether an unsigned number is 2 of the N-Power form

This is a question to ask in the interview. Can first analyze 2 of the number of N power form, the second binary form must be only 1bit is 1, the remaining bit is 0
So an intuitive approach is to compare all 2 of the N powers, since the unsigned integer we typically consider is 32 bits, so there are 32 2 of the N power (can be obtained by moving 1 left 0 to 31 bits). This method needs to be shifted and compared 32 times in the worst case, which is less efficient.
A closer analysis can be found that 2 of the number of power form N, the second binary form of 0 ... 010...0, assuming that after 1 there is an I 0 (0<=i<=31), then this number minus 1 will appear at the end of a continuous number I 1. For example, binary number 0100, this number minus 1 equals 0011, the two number is 0
Therefore, a more optimal algorithm is:
inline bool is2n (unsigned int i)
... {
if (i==0) return false;
else return (i& (i-1)) ==0;
}