二进制数中1的个数

int numTable[256] = {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3,3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3,4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4,3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3,4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6,6, 7, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4,5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 2, 3, 3, 4, 3, 4, 4, 5, 3,4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 3, 4,4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6,7, 6, 7, 7, 8};//摘自《数学之美》int numOfOne(int v){    int number = 0;    while(v)    {        if(v % 2 == 1)        {            number++;        }        v >>= 1;    }    return number;}unsigned int numOfOne2(unsigned int n){    unsigned int count = 0;    while(n > 0)    {        ++count;        n = n & (n-1);    }    return count;}unsigned int hammingWeight(unsigned int n){    unsigned int t = n;    t = (t & 0x55555555) + ( (t & 0xAAAAAAAA) >> 1);    t = (t & 0x33333333) + ( (t & 0xCCCCCCCC) >> 2);    t = (t & 0x0f0f0f0f) + ( (t & 0xf0f0f0f0) >> 4);    t = (t & 0x00ff00ff) + ( (t & 0xff00ff00) >> 8);    t = (t & 0x0000ffff) + ( (t & 0xffff0000) >> 16);    return t;}void main(){    int a = 255;    printf("%d,%d,%d,%d\n",        numTable[a],        numOfOne(a),        numOfOne2(a),        hammingWeight(a));}