算法-log2求值（整型）

lua底层对log2求幂

思想

1 将256以内的数的log2结果存到数组，这部分可直接得到结果
2 反向思维，进行一次左移：1左移n位，就是2的n次方，（假设n=3,效果是左移之后，1的右边都是0，这些0任意修改后的值s，log2(s)的结果不变）故，s变换到一定程度之后才会引起n的变换，所以： l += 8; x >>= 8; 成立。

int luaO_log2 (unsigned int x) {
  static const lu_byte log_2[256] = {
    0,1,2,2,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,
    6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,
    7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,
    7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,
    8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,
    8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,
    8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,
    8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8
  };
  int l = -1;
  while (x >= 256) { l += 8; x >>= 8; }
  return l + log_2[x];

}