实现以下目标最有效的算法是什么:

0010 0000 => 0000 0100

从MSB->LSB转换为LSB->MSB。所有位都必须反转;也就是说,这不是字节交换。


当前回答

下面这个怎么样:

    uint reverseMSBToLSB32ui(uint input)
    {
        uint output = 0x00000000;
        uint toANDVar = 0;
        int places = 0;

        for (int i = 1; i < 32; i++)
        {
            places = (32 - i);
            toANDVar = (uint)(1 << places);
            output |= (uint)(input & (toANDVar)) >> places;

        }


        return output;
    }

小而简单(不过只有32位)。

其他回答

实现低内存和最快。

private Byte  BitReverse(Byte bData)
    {
        Byte[] lookup = { 0, 8,  4, 12, 
                          2, 10, 6, 14 , 
                          1, 9,  5, 13,
                          3, 11, 7, 15 };
        Byte ret_val = (Byte)(((lookup[(bData & 0x0F)]) << 4) + lookup[((bData & 0xF0) >> 4)]);
        return ret_val;
    }

您可能希望使用标准模板库。它可能比上面提到的代码慢。然而,在我看来,这似乎更清楚,更容易理解。

 #include<bitset>
 #include<iostream>


 template<size_t N>
 const std::bitset<N> reverse(const std::bitset<N>& ordered)
 {
      std::bitset<N> reversed;
      for(size_t i = 0, j = N - 1; i < N; ++i, --j)
           reversed[j] = ordered[i];
      return reversed;
 };


 // test the function
 int main()
 {
      unsigned long num; 
      const size_t N = sizeof(num)*8;

      std::cin >> num;
      std::cout << std::showbase << std::hex;
      std::cout << "ordered  = " << num << std::endl;
      std::cout << "reversed = " << reverse<N>(num).to_ulong()  << std::endl;
      std::cout << "double_reversed = " << reverse<N>(reverse<N>(num)).to_ulong() << std::endl;  
 }
unsigned char ReverseBits(unsigned char data)
{
    unsigned char k = 0, rev = 0;

    unsigned char n = data;

    while(n)

    {
        k = n & (~(n - 1));
        n &= (n - 1);
        rev |= (128 / k);
    }
    return rev;
}

另一个基于循环的解决方案,在数量较低时快速退出(在c++中用于多种类型)

template<class T>
T reverse_bits(T in) {
    T bit = static_cast<T>(1) << (sizeof(T) * 8 - 1);
    T out;

    for (out = 0; bit && in; bit >>= 1, in >>= 1) {
        if (in & 1) {
            out |= bit;
        }
    }
    return out;
}

或者C语言中unsigned int

unsigned int reverse_bits(unsigned int in) {
    unsigned int bit = 1u << (sizeof(T) * 8 - 1);
    unsigned int out;

    for (out = 0; bit && in; bit >>= 1, in >>= 1) {
        if (in & 1)
            out |= bit;
    }
    return out;
}

通用的

C代码。以1字节输入数据num为例。

    unsigned char num = 0xaa;   // 1010 1010 (aa) -> 0101 0101 (55)
    int s = sizeof(num) * 8;    // get number of bits
    int i, x, y, p;
    int var = 0;                // make var data type to be equal or larger than num

    for (i = 0; i < (s / 2); i++) {
        // extract bit on the left, from MSB
        p = s - i - 1;
        x = num & (1 << p);
        x = x >> p;
        printf("x: %d\n", x);

        // extract bit on the right, from LSB
        y = num & (1 << i);
        y = y >> i;
        printf("y: %d\n", y);

        var = var | (x << i);       // apply x
        var = var | (y << p);       // apply y
    }

    printf("new: 0x%x\n", new);