位元求01的變換次數

計算LBP運算元的時候遇到計算一個整數求0->1,1->0變換次數的問題。為了能提高效能，前輩告訴一個快速演算法，mark下。

int calc_01_change_count(unsigned int n_input){    unsigned int tmp = (n_input << 1);    unsigned int n = n_input^tmp;//check how many bits are different    unsigned int start = (n>>(p)), end = (n & 0x01);//check the first and the last bit    unsigned int count = 0 ;//num of 1->0 or 0->1    for (count = 0; n; ++count)    {        n &= (n - 1) ;    }    count -= (start + end);    return count;}

如果是只計算1的個數可參考：http://www.cnblogs.com/graphics/archive/2010/06/21/1752421.html