c++20无分支版本，现在std::endian已经存在，但在c++23之前增加了std::byteswap

#include <bit>
#include <type_traits>
#include <concepts>
#include <array>
#include <cstring>
#include <iostream>
#include <bitset>

template <int LEN, int OFF=LEN/2>
class do_swap
{
    // FOR 8 bytes:
    // LEN=8 (LEN/2==4)       <H><G><F><E><D><C><B><A>
    // OFF=4: FROM=0, TO=7 => [A]<G><F><E><D><C><B>[H]
    // OFF=3: FROM=1, TO=6 => [A][B]<F><E><D><C>[G][H]
    // OFF=2: FROM=2, TO=5 => [A][B][C]<E><D>[F][G][H]
    // OFF=1: FROM=3, TO=4 => [A][B][C][D][E][F][G][H]
    // OFF=0: FROM=4, TO=3 => DONE
public:
    enum consts {FROM=LEN/2-OFF, TO=(LEN-1)-FROM};
    using NXT=do_swap<LEN, OFF-1>;
// flip the first and last for the current iteration's range
    static void flip(std::array<std::byte, LEN>& b)
    {
        std::byte tmp=b[FROM];
        b[FROM]=b[TO];
        b[TO]=tmp;
        NXT::flip(b);
    }
};
template <int LEN>
class do_swap<LEN, 0> // STOP the template recursion
{
public:
    static void flip(std::array<std::byte, LEN>&)
    {
    }
};

template<std::integral T, std::endian TO, std::endian FROM=std::endian::native>
        requires ((TO==std::endian::big) || (TO==std::endian::little))
              && ((FROM==std::endian::big) || (FROM==std::endian::little))
class endian_swap
{
public:
    enum consts {BYTE_COUNT=sizeof(T)};
    static T cvt(const T integral)
    {
    // if FROM and TO are the same -- nothing to do
        if (TO==FROM)
        {
                return integral;
        }

    // endian::big --> endian::little is the same as endian::little --> endian::big
    // the bytes have to be reversed
    // memcpy seems to be the most supported way to do byte swaps in a defined way
        std::array<std::byte, BYTE_COUNT> bytes;
        std::memcpy(&bytes, &integral, BYTE_COUNT);
        do_swap<BYTE_COUNT>::flip(bytes);
        T ret;
        std::memcpy(&ret, &bytes, BYTE_COUNT);
        return ret;
    }
};

std::endian big()
{
    return std::endian::big;
}

std::endian little()
{
    return std::endian::little;
}

std::endian native()
{
    return std::endian::native;
}

long long swap_to_big(long long x)
{
    return endian_swap<long long, std::endian::big>::cvt(x);
}

long long swap_to_little(long long x)
{
    return endian_swap<long long, std::endian::little>::cvt(x);
}

void show(std::string label, long long x)
{
    std::cout << label << "\t: " << std::bitset<64>(x) << " (" << x << ")" << std::endl;
}

int main(int argv, char ** argc)
{
    long long init=0xF8FCFEFF7F3F1F0;
    long long to_big=swap_to_big(init);
    long long to_little=swap_to_little(init);
    show("Init", init);
    show(">big", to_big);
    show(">little", to_little);
}

认真……我不明白为什么所有的解决方案都那么复杂!最简单、最通用的模板函数如何?它可以在任何操作系统的任何情况下交换任何大小的任何类型????

template <typename T>
void SwapEnd(T& var)
{
    static_assert(std::is_pod<T>::value, "Type must be POD type for safety");
    std::array<char, sizeof(T)> varArray;
    std::memcpy(varArray.data(), &var, sizeof(T));
    for(int i = 0; i < static_cast<int>(sizeof(var)/2); i++)
        std::swap(varArray[sizeof(var) - 1 - i],varArray[i]);
    std::memcpy(&var, varArray.data(), sizeof(T));
}

这是C和c++结合的神奇力量!只需逐个字符交换原始变量。

要点1:没有操作符:请记住，我没有使用简单的赋值操作符“=”，因为当反转字节序时，一些对象将被打乱，复制构造函数(或赋值操作符)将不起作用。因此，一个字符一个字符地复制它们更加可靠。

Point 2: Be aware of alignment issues: Notice that we're copying to and from an array, which is the right thing to do because the C++ compiler doesn't guarantee that we can access unaligned memory (this answer was updated from its original form for this). For example, if you allocate uint64_t, your compiler cannot guarantee that you can access the 3rd byte of that as a uint8_t. Therefore, the right thing to do is to copy this to a char array, swap it, then copy it back (so no reinterpret_cast). Notice that compilers are mostly smart enough to convert what you did back to a reinterpret_cast if they're capable of accessing individual bytes regardless of alignment.

使用此函数:

double x = 5;
SwapEnd(x);

现在x的字节序不同了。

似乎安全的方法是在每个单词上使用“顿音”。所以，如果你有。

std::vector<uint16_t> storage(n);  // where n is the number to be converted

// the following would do the trick
std::transform(word_storage.cbegin(), word_storage.cend()
  , word_storage.begin(), [](const uint16_t input)->uint16_t {
  return htons(input); });

如果您是在一个大端系统上，那么上面的代码将是一个无操作，因此我将查找您的平台使用的任何编译时条件，以确定htons是否是一个无操作。毕竟是O(n)在Mac上，它会是这样的……

#if (__DARWIN_BYTE_ORDER != __DARWIN_BIG_ENDIAN)
std::transform(word_storage.cbegin(), word_storage.cend()
  , word_storage.begin(), [](const uint16_t input)->uint16_t {
  return htons(input); });
#endif

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。)

和在C中一样:

short big = 0xdead;
short little = (((big & 0xff)<<8) | ((big & 0xff00)>>8));

您还可以声明一个无符号字符的向量，将输入值memcpy放入其中，将字节反向转换为另一个向量，然后将字节memcpy取出，但这将花费比旋转位长几个数量级的时间，特别是对于64位值。

我们已经用模板做到了这一点。你可以这样做:

// Specialization for 2-byte types.
template<>
inline void endian_byte_swapper< 2 >(char* dest, char const* src)
{
    // Use bit manipulations instead of accessing individual bytes from memory, much faster.
    ushort* p_dest = reinterpret_cast< ushort* >(dest);
    ushort const* const p_src = reinterpret_cast< ushort const* >(src);
    *p_dest = (*p_src >> 8) | (*p_src << 8);
}

// Specialization for 4-byte types.
template<>
inline void endian_byte_swapper< 4 >(char* dest, char const* src)
{
    // Use bit manipulations instead of accessing individual bytes from memory, much faster.
    uint* p_dest = reinterpret_cast< uint* >(dest);
    uint const* const p_src = reinterpret_cast< uint const* >(src);
    *p_dest = (*p_src >> 24) | ((*p_src & 0x00ff0000) >> 8) | ((*p_src & 0x0000ff00) << 8) | (*p_src << 24);
}