这是32位，如果我们考虑8位，我们需要改变大小。

    void bitReverse(int num)
    {
        int num_reverse = 0;
        int size = (sizeof(int)*8) -1;
        int i=0,j=0;
        for(i=0,j=size;i<=size,j>=0;i++,j--)
        {
            if((num >> i)&1)
            {
                num_reverse = (num_reverse | (1<<j));
            }
        }
        printf("\n rev num = %d\n",num_reverse);
    }

按LSB->MSB顺序读取输入整数“num”，并按MSB->LSB顺序存储在num_reverse中。

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;  
 }

高效意味着吞吐量或延迟。

从头到尾，看看安德斯·塞德罗尼厄斯的回答，很好。

为了降低延迟，我推荐以下代码:

uint32_t reverseBits( uint32_t x )
{
#if defined(__arm__) || defined(__aarch64__)
    __asm__( "rbit %0, %1" : "=r" ( x ) : "r" ( x ) );
    return x;
#endif
    // Flip pairwise
    x = ( ( x & 0x55555555 ) << 1 ) | ( ( x & 0xAAAAAAAA ) >> 1 );
    // Flip pairs
    x = ( ( x & 0x33333333 ) << 2 ) | ( ( x & 0xCCCCCCCC ) >> 2 );
    // Flip nibbles
    x = ( ( x & 0x0F0F0F0F ) << 4 ) | ( ( x & 0xF0F0F0F0 ) >> 4 );

    // Flip bytes. CPUs have an instruction for that, pretty fast one.
#ifdef _MSC_VER
    return _byteswap_ulong( x );
#elif defined(__INTEL_COMPILER)
    return (uint32_t)_bswap( (int)x );
#else
    // Assuming gcc or clang
    return __builtin_bswap32( x );
#endif
}

编译器输出:https://godbolt.org/z/5ehd89

这个线程引起了我的注意，因为它处理了一个简单的问题，即使对于现代CPU也需要大量的工作(CPU周期)。有一天我也站在那里，有同样的¤#%“#”问题。我得翻几百万字节。然而，我知道我所有的目标系统都是基于现代英特尔的，所以让我们开始优化到极致!!

所以我使用了Matt J的查找代码作为基础。我正在基准测试的系统是i7 haswell 4700eq。

Matt J的查找位翻转400亿字节:大约0.272秒。

然后我继续尝试，看看英特尔的ISPC编译器是否可以向量化反向的算术。c。

我不打算在这里用我的发现来烦你，因为我尝试了很多来帮助编译器找到东西，无论如何，我最终得到了大约0.15秒的性能来bitflip 400亿字节。这是一个伟大的减少，但对于我的应用程序，这仍然是方式方式太慢。

所以人们让我展示世界上最快的基于英特尔的bitflipper。定时:

时间到bitflip 400000000字节:0.050082秒!!!!!

// Bitflip using AVX2 - The fastest Intel based bitflip in the world!!
// Made by Anders Cedronius 2014 (anders.cedronius (you know what) gmail.com)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <omp.h>

using namespace std;

#define DISPLAY_HEIGHT  4
#define DISPLAY_WIDTH   32
#define NUM_DATA_BYTES  400000000

// Constants (first we got the mask, then the high order nibble look up table and last we got the low order nibble lookup table)
__attribute__ ((aligned(32))) static unsigned char k1[32*3]={
        0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,
        0x00,0x08,0x04,0x0c,0x02,0x0a,0x06,0x0e,0x01,0x09,0x05,0x0d,0x03,0x0b,0x07,0x0f,0x00,0x08,0x04,0x0c,0x02,0x0a,0x06,0x0e,0x01,0x09,0x05,0x0d,0x03,0x0b,0x07,0x0f,
        0x00,0x80,0x40,0xc0,0x20,0xa0,0x60,0xe0,0x10,0x90,0x50,0xd0,0x30,0xb0,0x70,0xf0,0x00,0x80,0x40,0xc0,0x20,0xa0,0x60,0xe0,0x10,0x90,0x50,0xd0,0x30,0xb0,0x70,0xf0
};

// The data to be bitflipped (+32 to avoid the quantization out of memory problem)
__attribute__ ((aligned(32))) static unsigned char data[NUM_DATA_BYTES+32]={};

extern "C" {
void bitflipbyte(unsigned char[],unsigned int,unsigned char[]);
}

int main()
{

    for(unsigned int i = 0; i < NUM_DATA_BYTES; i++)
    {
        data[i] = rand();
    }

    printf ("\r\nData in(start):\r\n");
    for (unsigned int j = 0; j < 4; j++)
    {
        for (unsigned int i = 0; i < DISPLAY_WIDTH; i++)
        {
            printf ("0x%02x,",data[i+(j*DISPLAY_WIDTH)]);
        }
        printf ("\r\n");
    }

    printf ("\r\nNumber of 32-byte chunks to convert: %d\r\n",(unsigned int)ceil(NUM_DATA_BYTES/32.0));

    double start_time = omp_get_wtime();
    bitflipbyte(data,(unsigned int)ceil(NUM_DATA_BYTES/32.0),k1);
    double end_time = omp_get_wtime();

    printf ("\r\nData out:\r\n");
    for (unsigned int j = 0; j < 4; j++)
    {
        for (unsigned int i = 0; i < DISPLAY_WIDTH; i++)
        {
            printf ("0x%02x,",data[i+(j*DISPLAY_WIDTH)]);
        }
        printf ("\r\n");
    }
    printf("\r\n\r\nTime to bitflip %d bytes: %f seconds\r\n\r\n",NUM_DATA_BYTES, end_time-start_time);

    // return with no errors
    return 0;
}

printf是用来调试的。

这是主要的工作:

bits 64
global bitflipbyte

bitflipbyte:    
        vmovdqa     ymm2, [rdx]
        add         rdx, 20h
        vmovdqa     ymm3, [rdx]
        add         rdx, 20h
        vmovdqa     ymm4, [rdx]
bitflipp_loop:
        vmovdqa     ymm0, [rdi] 
        vpand       ymm1, ymm2, ymm0 
        vpandn      ymm0, ymm2, ymm0 
        vpsrld      ymm0, ymm0, 4h 
        vpshufb     ymm1, ymm4, ymm1 
        vpshufb     ymm0, ymm3, ymm0         
        vpor        ymm0, ymm0, ymm1
        vmovdqa     [rdi], ymm0
        add     rdi, 20h
        dec     rsi
        jnz     bitflipp_loop
        ret

代码占用32个字节，然后屏蔽掉蚕食。高啃角右移了4。然后使用vpshufb和ymm4 / ymm3作为查找表。我可以使用一个单独的查找表，但我将不得不在ORing再次一起啃啃之前向左移动。

还有更快的翻转比特的方法。但我被绑定到单线程和CPU，所以这是我能实现的最快速度。你能做一个快一点的版本吗?

关于使用Intel C/ c++编译器内在等效命令，请不要发表任何评论…

// 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

对于喜欢递归的人来说，这是另一个解决方案。

这个想法很简单。 将输入除以一半并交换两部分，继续直到达到单个位。

Illustrated in the example below.

Ex : If Input is 00101010   ==> Expected output is 01010100

1. Divide the input into 2 halves 
    0010 --- 1010

2. Swap the 2 Halves
    1010     0010

3. Repeat the same for each half.
    10 -- 10 ---  00 -- 10
    10    10      10    00

    1-0 -- 1-0 --- 1-0 -- 0-0
    0 1    0 1     0 1    0 0

Done! Output is 01010100

这里有一个递归函数来求解。(注意，我使用了unsigned int，所以它可以用于sizeof(unsigned int)*8位的输入。

递归函数有两个参数-需要位的值 要反转的值和值中的比特数。

int reverse_bits_recursive(unsigned int num, unsigned int numBits)
{
    unsigned int reversedNum;;
    unsigned int mask = 0;

    mask = (0x1 << (numBits/2)) - 1;

    if (numBits == 1) return num;
    reversedNum = reverse_bits_recursive(num >> numBits/2, numBits/2) |
                   reverse_bits_recursive((num & mask), numBits/2) << numBits/2;
    return reversedNum;
}

int main()
{
    unsigned int reversedNum;
    unsigned int num;

    num = 0x55;
    reversedNum = reverse_bits_recursive(num, 8);
    printf ("Bit Reversal Input = 0x%x Output = 0x%x\n", num, reversedNum);

    num = 0xabcd;
    reversedNum = reverse_bits_recursive(num, 16);
    printf ("Bit Reversal Input = 0x%x Output = 0x%x\n", num, reversedNum);

    num = 0x123456;
    reversedNum = reverse_bits_recursive(num, 24);
    printf ("Bit Reversal Input = 0x%x Output = 0x%x\n", num, reversedNum);

    num = 0x11223344;
    reversedNum = reverse_bits_recursive(num,32);
    printf ("Bit Reversal Input = 0x%x Output = 0x%x\n", num, reversedNum);
}

输出如下:

Bit Reversal Input = 0x55 Output = 0xaa
Bit Reversal Input = 0xabcd Output = 0xb3d5
Bit Reversal Input = 0x123456 Output = 0x651690
Bit Reversal Input = 0x11223344 Output = 0x22cc4488