Judge whether an integer is the integer power of 2-A pen question of Xiaomi

# Include <iostream>

Using namespace STD;

// The first method exploits the loop
Bool makedemo-( const Int & M, Int & POW) // POW indicates how many power m is 2
{
Int I = 2;
Int S = 1;
Pow = 0;
If (M <0) return false;
If (M = 0) return true;
Do
{
S * = I;
Pow ++;
} While (S <m );
If (S = m) return true;
Else
{
Pow = 0;
Return false;
}
}

// The second method. If M/2 is the integer power of 2, m is the integer power of 2. Recursive Implementation
Bool makedemo-( int M, int * POW)
{
If (M <0) return false;
If (M % 2! = 0) return false;
If (M = 0) return true;
Else if (M = 2)
{
* POW = * POW + 1;
Return true;
}
Else
{
* POW = * POW + 1;
Makedemo-( M/2, pow );
}
}

// An integer that is an integer power of 2. Its value ranges from 0 to 1, 2... all are 1, and the rest are 0. In this way, the position of-M's complement code is the same as that of M's complement Code 1.
// Therefore, M &-M is m
Bool makedemo-( Int & M)
{
Return M = (M & (-m ));
}

Void main ()
{
Cout <"enter an integer greater than 0 :";
Int m;
Cin> m;
Int POW = 0;
// If (makedemo-( M, & POW) cout <m <"is 2" <POW <"power" <Endl;
If (makedemo-( m) cout <m <"is an integer power of 2" <Endl;
Else cout <m <"is not an integer power of 2! "<Endl;
}