How can I determine whether a number is a power of 2? If so, how much power is determined?

After the power of 2 is written in binary form, it is easy to find that there is only one 1 in binary, and 1 followed by n 0; therefore, the problem can be converted to determining whether n zeros are followed by 1.
If we subtract this number from 1, we will find that the only one will change to 0, and the original n 0 will change to 1; therefore, the original number and the number after the first subtraction will be zero after calculation.
The fastest way:
(Number & number-1) = 0
Cause:Because the Npower of 2 is converted to binary 10 ...... 0 (except 0 ). And the number of digits-1 on the top. The result is 0. For example. The binary value of 8 is 1000, and the binary value of 8-1 = 7 is 111. The result of the two phases is 0. The calculation is as follows:
1000
& 0111
-------
0000
The Code implemented by recursion is as follows:
Copy codeThe Code is as follows: # include "stdio. h"
# Include "stdlib. h"
Int log2 (int value) // recursively determines the power of a number 2
{
If (value = 1)
Return 0;
Else
Return 1 + log2 (value> 1 );
}
Int main (void)
{
Int num;
Printf ("enter an integer :");
Scanf ("% d", & num );
If (num & (num-1) // use and calculate to determine whether a number is a power of 2
Printf ("% d is not the power of 2! \ N ", num );
Else
Printf ("% d is the power of % d of 2! \ N ", num, log2 (num ));
System ("pause ");
Return 0;
}

The Code implemented using non-recursion is as follows:Copy codeThe Code is as follows: # include "stdio. h"
# Include "stdlib. h"
Int log2 (int value) // determines the power of a number 2 in non-recursion.
{
Int x = 0;
While (value> 1)
{
Value >>> = 1;
X ++;
}
Return x;
}
Int main (void)
{
Int num;
Printf ("enter an integer :");
Scanf ("% d", & num );
If (num & (num-1) // use and calculate to determine whether a number is a power of 2
Printf ("% d is not the power of 2! \ N ", num );
Else
Printf ("% d is the power of % d of 2! \ N ", num, log2 (num ));
System ("pause ");
Return 0;
}

Extended: calculates the number of 1 in a number n binary.
N = n & (n-1) can remove the rightmost 1 Property in n's binary, until all 1 is removed, this method reduces the complexity of the problem to only one number. The Code is as follows:Copy codeThe Code is as follows: int Func3 (int data)
{// Use data & (data-1) to remove the rightmost 1 at a time. How many 1 s are removed, that is, several 1 s are included.
Int count = 0;
While (data)
{
Data = data & (data-1 );
Count ++;
}
Return count;
}

Expansion Question 2:
The number of bits in the binary format of A and B is different. This problem can be divided into two steps: (1) xor a and B to obtain C, that is, C = A ^ B, (2) calculate the number of 1 in the binary of C.