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n&(n-1)
1. 判断一个正整数是否为2的乘方数
数据对比（uint_16 n;）
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正整数n 正整数 n 的二进制表示 正整数 (n - 1) 正整数 (n-1) 的二进制表示 n&(n - 1)
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2 0000000000000010 1 0000000000000001 0000000000000000
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4 0000000000000100 3 0000000000000011 0000000000000000
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16 0000000000010000 15 0000000000001111 0000000000000000
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100 0000000001100100 99 0000000001100011 0000000001100000
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128 0000000010000000 127 0000000001111111 0000000000000000
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unsigned char isPower(unsigned int n)
{ return(((n == 0) || (n & (n-1))) ? 0x00 : 0x01);}
2. 判断一个正整数转化为二进制后数字 “1” 的个数
unsigned char NumOfOne(unsigned int n){ unsigned char i = 0; while(n) { n &= n-1; i++; } return i;}
n&(n-1)