Average number of operations per element vs. n (summation)

#include <iostream>
#include <math.h>

using namespace std;

int lg_floor(int x) {
    int c=-1;
    for(; x!=0; c++, x>>=1);
    return c;
}
int ops_for_number(int x) {
    return lg_floor(x&(-x));
}

int main(int argc, char *argv[]) {
    for(int n=128; n<256; n++) {
        double sum = 0;
        double result;
        for(int j=0; j<lg_floor(n); j++) {
            sum+=j*(1<<(lg_floor(n)-j-1));
        }
        for(int j=1<<lg_floor(n); j<=n; j++) {
            sum+=ops_for_number(j);
        }
        result = sum/n;
        printf("%f\n", result);
    }
}