如何在c++中转换大端值和小端值?

为了清晰起见,我必须将二进制数据(双精度浮点值以及32位和64位整数)从一个CPU架构转换到另一个CPU架构。这并不涉及网络,因此ntoh()和类似的函数在这里不能工作。


注意:我接受的答案直接适用于我的目标编译器(这就是我选择它的原因)。然而,这里还有其他非常好的、更方便的答案。


当前回答

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。

其他回答

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] };
}

如果您这样做是为了在不同平台之间传输数据,请查看ntoh和hton函数。

大多数平台都有一个系统头文件,提供了有效的byteswap函数。在Linux上是<end .h>。你可以用c++很好地包装它:

#include <iostream>

#include <endian.h>

template<size_t N> struct SizeT {};

#define BYTESWAPS(bits) \
template<class T> inline T htobe(T t, SizeT<bits / 8>) { return htobe ## bits(t); } \
template<class T> inline T htole(T t, SizeT<bits / 8>) { return htole ## bits(t); } \
template<class T> inline T betoh(T t, SizeT<bits / 8>) { return be ## bits ## toh(t); } \
template<class T> inline T letoh(T t, SizeT<bits / 8>) { return le ## bits ## toh(t); }

BYTESWAPS(16)
BYTESWAPS(32)
BYTESWAPS(64)

#undef BYTESWAPS

template<class T> inline T htobe(T t) { return htobe(t, SizeT<sizeof t>()); }
template<class T> inline T htole(T t) { return htole(t, SizeT<sizeof t>()); }
template<class T> inline T betoh(T t) { return betoh(t, SizeT<sizeof t>()); }
template<class T> inline T letoh(T t) { return letoh(t, SizeT<sizeof t>()); }

int main()
{
    std::cout << std::hex;
    std::cout << htobe(static_cast<unsigned short>(0xfeca)) << '\n';
    std::cout << htobe(0xafbeadde) << '\n';

    // Use ULL suffix to specify integer constant as unsigned long long 
    std::cout << htobe(0xfecaefbeafdeedfeULL) << '\n';
}

输出:

cafe
deadbeaf
feeddeafbeefcafe

我真的很惊讶没有人提到htobeXX和betohXX函数。它们定义在end .h中,非常类似于网络函数htonXX。

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);
}