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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n         n 的二进制表示           (n - 1)    (n-1) 的二进制表示        n&(n - 1)

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2       0000000000000010b    1        0000000000000001b    0000000000000000b

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4       0000000000000100b    3        0000000000000011b    0000000000000000b

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16     0000000000010000b    15      0000000000001111b     0000000000000000b

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100   0000000001100100b    99      0000000001100011b     0000000001100000b

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128   0000000010000000b    127    0000000001111111b      0000000000000000b

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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;
}