Programming implementation: Two int (32-bit) integer m and n in the binary expression, how many bits (bit) different? __ Programming

A

#define _crt_secure_no_warnings 1
#include <stdio.h>
#include <math.h>
int main ()
{
    int m = 0;
    int n = 0;
    int i = 0;
    int count = 0;
    printf ("Please enter two number: \ n");
    scanf ("%d%d", &m, &n);
    for (i = 0; i < i++)
    {if ((
        m >> i) & 1) ^ ((n >> i) (& 1))//Let M and N move to the right I bit then respectively and 1 phase again or if is true count++
            count++;
    printf ("count=%d", count);
    System ("pause");
    return 0;
}

Two

#define _crt_secure_no_warnings 1
#include <stdio.h>
#include <math.h>
int main ()
{
    int u = 0;
    int m = 0;
    int n = 0;
    int count = 0;
    printf ("Please enter two number: \ n");
    scanf ("%d%d", &m, &n);
    u = m^n;  M and N xor or while
    (U)   //equivalent to finding a number; the number of 1 in binary
    {
        count++;
        u = u & (u-1);
    }
    printf ("count=%d", count);
    System ("pause");
    return 0;
}

Three

#define _crt_secure_no_warnings 1
#include <stdio.h>
#include <math.h>
int main ()
{
    int i = 0;
    int m = 0;
    int n = 0;
    int count = 0;
    printf ("Please enter two number: \ n");
    scanf ("%d%d", &m, &n);
    for (i = 0; i < i++)
    {
        if ((M & 1)!= (n & 1))  
        {
            count++
        }
        m = M >> 1;
        n = n >> 1;
    }
    printf ("count=%d", count);
    System ("pause");
    return 0;
}