用一个表达式，判断一个数X是否是2^N（2,4,8,16,…），不可用循环语句
答案：!(X & (X - 1))
分析：2^N（2，4，8，16）这样的数转化成二进制是10、100、1000、10000。
若X-1与X做与运算，结果是0，则X是2^N。
扩展：求一个整数的二进制形式中包含1的数量。
1 int func(int x)
2 {
3 int count = 0;
4 while(x)
5 {
6 count ++;
7 x = x & (x - 1);
8 }
9 return count;
10 }
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