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);
}

查找位移位，因为这基本上是所有你需要做的交换从小->大端dian。然后根据位的大小，改变位移位的方式。

摘自Rob Pike的《字节顺序谬误》:

假设数据流有一个小端编码的32位整数。下面是如何提取它(假设无符号字节):

i = (data[0]<<0) | (data[1]<<8) | (data[2]<<16) | ((unsigned)data[3]<<24);

如果它是big-endian，下面是如何提取它:

i = (data[3]<<0) | (data[2]<<8) | (data[1]<<16) | ((unsigned)data[0]<<24);

TL;DR:不要担心你的平台原生顺序，重要的是你从中读取的流的字节顺序，你最好希望它是定义良好的。

注1:这里int和unsigned int是32位，否则类型可能需要调整。

注2:最后一个字节必须在移位前显式转换为unsigned，因为默认情况下它被提升为int，移位24位意味着操作符号位，这是未定义行为。

在模板函数中围绕枢轴使用老式的3-step-xor技巧进行字节交换，提供了一个灵活、快速的O(ln2)解决方案，不需要库，这里的风格也拒绝1字节类型:

template<typename T>void swap(T &t){
    for(uint8_t pivot = 0; pivot < sizeof(t)/2; pivot ++){
        *((uint8_t *)&t + pivot) ^= *((uint8_t *)&t+sizeof(t)-1- pivot);
        *((uint8_t *)&t+sizeof(t)-1- pivot) ^= *((uint8_t *)&t + pivot);
        *((uint8_t *)&t + pivot) ^= *((uint8_t *)&t+sizeof(t)-1- pivot);
    }
}

如果你有c++ 17，那么添加这个头文件

#include <algorithm>

使用这个模板函数交换字节:

template <typename T>
void swapEndian(T& buffer)
{
    static_assert(std::is_pod<T>::value, "swapEndian support POD type only");
    char* startIndex = static_cast<char*>((void*)buffer.data());
    char* endIndex = startIndex + sizeof(buffer);
    std::reverse(startIndex, endIndex);
}

这样称呼它:

swapEndian (stlContainer);

下面介绍如何读取以IEEE 754 64位格式存储的double，即使您的主机使用不同的系统。

/*
* read a double from a stream in ieee754 format regardless of host
*  encoding.
*  fp - the stream
*  bigendian - set to if big bytes first, clear for little bytes
*              first
*
*/
double freadieee754(FILE *fp, int bigendian)
{
    unsigned char buff[8];
    int i;
    double fnorm = 0.0;
    unsigned char temp;
    int sign;
    int exponent;
    double bitval;
    int maski, mask;
    int expbits = 11;
    int significandbits = 52;
    int shift;
    double answer;

    /* read the data */
    for (i = 0; i < 8; i++)
        buff[i] = fgetc(fp);
    /* just reverse if not big-endian*/
    if (!bigendian)
    {
        for (i = 0; i < 4; i++)
        {
            temp = buff[i];
            buff[i] = buff[8 - i - 1];
            buff[8 - i - 1] = temp;
        }
    }
    sign = buff[0] & 0x80 ? -1 : 1;
    /* exponet in raw format*/
    exponent = ((buff[0] & 0x7F) << 4) | ((buff[1] & 0xF0) >> 4);

    /* read inthe mantissa. Top bit is 0.5, the successive bits half*/
    bitval = 0.5;
    maski = 1;
    mask = 0x08;
    for (i = 0; i < significandbits; i++)
    {
        if (buff[maski] & mask)
            fnorm += bitval;

        bitval /= 2.0;
        mask >>= 1;
        if (mask == 0)
        {
            mask = 0x80;
            maski++;
        }
    }
    /* handle zero specially */
    if (exponent == 0 && fnorm == 0)
        return 0.0;

    shift = exponent - ((1 << (expbits - 1)) - 1); /* exponent = shift + bias */
    /* nans have exp 1024 and non-zero mantissa */
    if (shift == 1024 && fnorm != 0)
        return sqrt(-1.0);
    /*infinity*/
    if (shift == 1024 && fnorm == 0)
    {

#ifdef INFINITY
        return sign == 1 ? INFINITY : -INFINITY;
#endif
        return  (sign * 1.0) / 0.0;
    }
    if (shift > -1023)
    {
        answer = ldexp(fnorm + 1.0, shift);
        return answer * sign;
    }
    else
    {
        /* denormalised numbers */
        if (fnorm == 0.0)
            return 0.0;
        shift = -1022;
        while (fnorm < 1.0)
        {
            fnorm *= 2;
            shift--;
        }
        answer = ldexp(fnorm, shift);
        return answer * sign;
    }
}

对于这套函数的其余部分，包括写和整数例程，请参阅我的github项目

https://github.com/MalcolmMcLean/ieee754