我喜欢这个，只是为了风格:-)

long swap(long i) {
    char *c = (char *) &i;
    return * (long *) (char[]) {c[3], c[2], c[1], c[0] };
}

在大多数POSIX系统中(虽然不是在POSIX标准中)有end .h，它可以用来确定系统使用的编码。然后是这样的:

unsigned int change_endian(unsigned int x)
{
    unsigned char *ptr = (unsigned char *)&x;
    return (ptr[0] << 24) | (ptr[1] << 16) | (ptr[2] << 8) | ptr[3];
}

这将交换顺序(从大端序到小端序):

如果你有数字0xDEADBEEF(在一个小端序系统中存储为0xEFBEADDE)， ptr[0]将是0xEF, ptr[1]是0xBE，等等。

但是如果你想将它用于网络，那么htons, htonl和htonll(以及它们的逆ntohs, ntohl和ntohll)将有助于从主机顺序转换到网络顺序。

来这里寻找一个Boost解决方案，失望地离开，但最终在其他地方找到了它。你可以使用boost::endian::endian_reverse。它被模板化/重载了所有的基元类型:

#include <iostream>
#include <iomanip>
#include "boost/endian/conversion.hpp"

int main()
{
  uint32_t word = 0x01;
  std::cout << std::hex << std::setfill('0') << std::setw(8) << word << std::endl;
  // outputs 00000001;

  uint32_t word2 = boost::endian::endian_reverse(word);
  // there's also a `void ::endian_reverse_inplace(...) function
  // that reverses the value passed to it in place and returns nothing

  std::cout << std::hex << std::setfill('0') << std::setw(8) << word2 << std::endl;
  // outputs 01000000

  return 0;
}

示范

虽然，看起来c++23最终用std::byteswap解决了这个问题。(我使用的是c++17，所以这不是一个选项。)

简单地说:

#include <climits>

template <typename T>
T swap_endian(T u)
{
    static_assert (CHAR_BIT == 8, "CHAR_BIT != 8");

    union
    {
        T u;
        unsigned char u8[sizeof(T)];
    } source, dest;

    source.u = u;

    for (size_t k = 0; k < sizeof(T); k++)
        dest.u8[k] = source.u8[sizeof(T) - k - 1];

    return dest.u;
}

用法:swap_endian < uint32_t >(42)。

这里有一个基本的函数来交换大小端序。它是基本的，但不需要补充库。

void endianness_swap(uint32_t& val) {
    uint8_t a, b, c;
    a = (val & 0xFF000000) >> 24;
    b = (val & 0x00FF0000) >> 16;
    c = (val & 0x0000FF00) >> 8;
    val=(val & 0x000000FF) << 24;
    val = val + (c << 16) + (b << 8) + (a);
}

我有这个代码，允许我从HOST_ENDIAN_ORDER(无论它是什么)转换为LITTLE_ENDIAN_ORDER或BIG_ENDIAN_ORDER。我使用一个模板，所以如果我试图从HOST_ENDIAN_ORDER转换为LITTLE_ENDIAN_ORDER，他们恰好是相同的机器为我编译，不会生成任何代码。

下面是带有注释的代码:

// We define some constant for little, big and host endianess. Here I use 
// BOOST_LITTLE_ENDIAN/BOOST_BIG_ENDIAN to check the host indianess. If you
// don't want to use boost you will have to modify this part a bit.
enum EEndian
{
  LITTLE_ENDIAN_ORDER,
  BIG_ENDIAN_ORDER,
#if defined(BOOST_LITTLE_ENDIAN)
  HOST_ENDIAN_ORDER = LITTLE_ENDIAN_ORDER
#elif defined(BOOST_BIG_ENDIAN)
  HOST_ENDIAN_ORDER = BIG_ENDIAN_ORDER
#else
#error "Impossible de determiner l'indianness du systeme cible."
#endif
};

// this function swap the bytes of values given it's size as a template
// parameter (could sizeof be used?).
template <class T, unsigned int size>
inline T SwapBytes(T value)
{
  union
  {
     T value;
     char bytes[size];
  } in, out;

  in.value = value;

  for (unsigned int i = 0; i < size / 2; ++i)
  {
     out.bytes[i] = in.bytes[size - 1 - i];
     out.bytes[size - 1 - i] = in.bytes[i];
  }

  return out.value;
}

// Here is the function you will use. Again there is two compile-time assertion
// that use the boost librarie. You could probably comment them out, but if you
// do be cautious not to use this function for anything else than integers
// types. This function need to be calles like this :
//
//     int x = someValue;
//     int i = EndianSwapBytes<HOST_ENDIAN_ORDER, BIG_ENDIAN_ORDER>(x);
//
template<EEndian from, EEndian to, class T>
inline T EndianSwapBytes(T value)
{
  // A : La donnée à swapper à une taille de 2, 4 ou 8 octets
  BOOST_STATIC_ASSERT(sizeof(T) == 2 || sizeof(T) == 4 || sizeof(T) == 8);

  // A : La donnée à swapper est d'un type arithmetic
  BOOST_STATIC_ASSERT(boost::is_arithmetic<T>::value);

  // Si from et to sont du même type on ne swap pas.
  if (from == to)
     return value;

  return SwapBytes<T, sizeof(T)>(value);
}