Number of "1" in the statistical binary number (understand binary)

problem Description: There are 10 kinds of people in the world, one understands the binary system, one does not understand. So you know the binary expression of two int32 integers m and n, how many bits are different?

function Prototypes: int countbitdiff (int m, int n)

Input:

1999 2299

Output:

7

Analysis: This topic from the 2016 Millet online pen test, to compare the bits of M and N, you can use the additional variable v and different or operation to achieve, the practice is:

V=m^n;

The number of binary 1 in binary representations of V after XOR or operation is that of the different bits of M and N, and the problem is converted to the number of binary number 1 in a binary number v.

Solution One: bitwise comparison

The easiest way to think of this is to compare the binary values of each bit in V to 1, using the shift operator in C and pointers to achieve this comparison:

/**
 * Obtains two integer binary expression digits of different number
 * @param m int m
 * @param n integer n
 * @return integral
/int Countbitdiff (int m, int n) {
    unsigned int c = m ^ n; 
    char ones;
    for (ones = 0; c!= 0; c >>= 1) {
        if ((C & 1)!= 0)
            ones + = 1;
    }
    return ones;
}

solution Two: With the help of n& (n-1)

The second method is special, using a special property of bitwise and special, namely:

Suppose the binary value of the binary number n has a k bit is 1,
Then the n& (n-1) binary value of 1 digits is k-1.

Using the special properties of n& (n-1), you can write the following code:

/**
 * Obtains two different number of shaping binary expression numbers
 * @param m integer m
 * @param n integer n
 * @return integral
/int Countbitdiff (int m, int N) {
    int c = m ^ n;//Bitwise XOR or
    char counter = 0;
    for (; c; ++counter) c &= (c-1);
    return counter;
}