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

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;
}

其他回答

我认为这是最简单的逆转比特的方法之一。 如果这个逻辑有什么缺陷,请让我知道。 基本上在这个逻辑中,我们检查位置的位的值。 如果值为1,则在反转位置上设置位。

void bit_reverse(ui32 *data)
{
  ui32 temp = 0;    
  ui32 i, bit_len;    
  {    
   for(i = 0, bit_len = 31; i <= bit_len; i++)   
   {    
    temp |= (*data & 1 << i)? (1 << bit_len-i) : 0;    
   }    
   *data = temp;    
  }    
  return;    
}    

原生ARM指令“rbit”可以用1个cpu周期和1个额外的cpu寄存器来完成,不可能被击败。

似乎许多其他帖子都关心速度(即最好=最快)。 简单性怎么样?考虑:

char ReverseBits(char character) {
    char reversed_character = 0;
    for (int i = 0; i < 8; i++) {
        char ith_bit = (c >> i) & 1;
        reversed_character |= (ith_bit << (sizeof(char) - 1 - i));
    }
    return reversed_character;
}

并希望聪明的编译器将为您优化。

如果你想反转一个更长的位列表(包含sizeof(char) * n位),你可以使用这个函数得到:

void ReverseNumber(char* number, int bit_count_in_number) {
    int bytes_occupied = bit_count_in_number / sizeof(char);      

    // first reverse bytes
    for (int i = 0; i <= (bytes_occupied / 2); i++) {
        swap(long_number[i], long_number[n - i]);
    }

    // then reverse bits of each individual byte
    for (int i = 0; i < bytes_occupied; i++) {
         long_number[i] = ReverseBits(long_number[i]);
    }
}

这将把[10000000,10101010]反向转换为[01010101,00000001]。

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;
}
// Purpose: to reverse bits in an unsigned short integer 
// Input: an unsigned short integer whose bits are to be reversed
// Output: an unsigned short integer with the reversed bits of the input one
unsigned short ReverseBits( unsigned short a )
{
     // declare and initialize number of bits in the unsigned short integer
     const char num_bits = sizeof(a) * CHAR_BIT;

     // declare and initialize bitset representation of integer a
     bitset<num_bits> bitset_a(a);          

     // declare and initialize bitset representation of integer b (0000000000000000)
     bitset<num_bits> bitset_b(0);                  

     // declare and initialize bitset representation of mask (0000000000000001)
     bitset<num_bits> mask(1);          

     for ( char i = 0; i < num_bits; ++i )
     {
          bitset_b = (bitset_b << 1) | bitset_a & mask;
          bitset_a >>= 1;
     }

     return (unsigned short) bitset_b.to_ulong();
}

void PrintBits( unsigned short a )
{
     // declare and initialize bitset representation of a
     bitset<sizeof(a) * CHAR_BIT> bitset(a);

     // print out bits
     cout << bitset << endl;
}


// Testing the functionality of the code

int main ()
{
     unsigned short a = 17, b;

     cout << "Original: "; 
     PrintBits(a);

     b = ReverseBits( a );

     cout << "Reversed: ";
     PrintBits(b);
}

// Output:
Original: 0000000000010001
Reversed: 1000100000000000