How do I convert between big-endian and little-endian values in C++?

Question

How do I convert between big-endian and little-endian values in C++?

For clarity, I have to translate binary data (double-precision floating point values and 32-bit and 64-bit integers) from one CPU architecture to another. This doesn't involve networking, so ntoh() and similar functions won't work here.

---

Note: The answer I accepted applies directly to compilers I'm targeting (which is why I chose it). However, there are other very good, more portable answers here.

Answer 1

If you're using Visual C++ do the following: You include intrin.h and call the following functions:

For 16 bit numbers:

unsigned short _byteswap_ushort(unsigned short value);

For 32 bit numbers:

unsigned long _byteswap_ulong(unsigned long value);

For 64 bit numbers:

unsigned __int64 _byteswap_uint64(unsigned __int64 value);

8 bit numbers (chars) don't need to be converted.

Also these are only defined for unsigned values they work for signed integers as well.

For floats and doubles it's more difficult as with plain integers as these may or not may be in the host machines byte-order. You can get little-endian floats on big-endian machines and vice versa.

Other compilers have similar intrinsics as well.

In GCC for example you can directly call some builtins as documented here:

uint32_t __builtin_bswap32 (uint32_t x)
uint64_t __builtin_bswap64 (uint64_t x)

(no need to include something). Afaik bits.h declares the same function in a non gcc-centric way as well.

16 bit swap it's just a bit-rotate.

Calling the intrinsics instead of rolling your own gives you the best performance and code density btw..

Answer 2

Simply put:

#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;
}

usage: swap_endian<uint32_t>(42).

Answer 3

From The Byte Order Fallacy by Rob Pike:

Let's say your data stream has a little-endian-encoded 32-bit integer. Here's how to extract it (assuming unsigned bytes):

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

If it's big-endian, here's how to extract it:

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

TL;DR: don't worry about your platform native order, all that counts is the byte order of the stream your are reading from, and you better hope it's well defined.

Note 1: It is expected that int and unsigned int be 32 bits here, types may require adjustment otherwise.

Note 2: The last byte must be explicitly cast to unsigned before shifting, as by default it's promoted to int, and a shift by 24 bits means manipulating the sign bit which is Undefined Behavior.

Answer 4

If you are doing this for purposes of network/host compatability you should use:

ntohl() //Network to Host byte order (Long)
htonl() //Host to Network byte order (Long)

ntohs() //Network to Host byte order (Short)
htons() //Host to Network byte order (Short)

If you are doing this for some other reason one of the byte_swap solutions presented here would work just fine.

Answer 5

I took a few suggestions from this post and put them together to form this:

#include <boost/type_traits.hpp>
#include <boost/static_assert.hpp>
#include <boost/detail/endian.hpp>
#include <stdexcept>
#include <cstdint>

enum endianness
{
    little_endian,
    big_endian,
    network_endian = big_endian,
    
    #if defined(BOOST_LITTLE_ENDIAN)
        host_endian = little_endian
    #elif defined(BOOST_BIG_ENDIAN)
        host_endian = big_endian
    #else
        #error "unable to determine system endianness"
    #endif
};

namespace detail {

template<typename T, size_t sz>
struct swap_bytes
{
    inline T operator()(T val)
    {
        throw std::out_of_range("data size");
    }
};

template<typename T>
struct swap_bytes<T, 1>
{
    inline T operator()(T val)
    {
        return val;
    }
};

template<typename T>
struct swap_bytes<T, 2>
{
    inline T operator()(T val)
    {
        return ((((val) >> 8) & 0xff) | (((val) & 0xff) << 8));
    }
};

template<typename T>
struct swap_bytes<T, 4>
{
    inline T operator()(T val)
    {
        return ((((val) & 0xff000000) >> 24) |
                (((val) & 0x00ff0000) >>  8) |
                (((val) & 0x0000ff00) <<  8) |
                (((val) & 0x000000ff) << 24));
    }
};

template<>
struct swap_bytes<float, 4>
{
    inline float operator()(float val)
    {
        uint32_t mem =swap_bytes<uint32_t, sizeof(uint32_t)>()(*(uint32_t*)&val);
        return *(float*)&mem;
    }
};

template<typename T>
struct swap_bytes<T, 8>
{
    inline T operator()(T val)
    {
        return ((((val) & 0xff00000000000000ull) >> 56) |
                (((val) & 0x00ff000000000000ull) >> 40) |
                (((val) & 0x0000ff0000000000ull) >> 24) |
                (((val) & 0x000000ff00000000ull) >> 8 ) |
                (((val) & 0x00000000ff000000ull) << 8 ) |
                (((val) & 0x0000000000ff0000ull) << 24) |
                (((val) & 0x000000000000ff00ull) << 40) |
                (((val) & 0x00000000000000ffull) << 56));
    }
};

template<>
struct swap_bytes<double, 8>
{
    inline double operator()(double val)
    {
        uint64_t mem =swap_bytes<uint64_t, sizeof(uint64_t)>()(*(uint64_t*)&val);
        return *(double*)&mem;
    }
};

template<endianness from, endianness to, class T>
struct do_byte_swap
{
    inline T operator()(T value)
    {
        return swap_bytes<T, sizeof(T)>()(value);
    }
};
// specialisations when attempting to swap to the same endianess
template<class T> struct do_byte_swap<little_endian, little_endian, T> { inline T operator()(T value) { return value; } };
template<class T> struct do_byte_swap<big_endian,    big_endian,    T> { inline T operator()(T value) { return value; } };

} // namespace detail

template<endianness from, endianness to, class T>
inline T byte_swap(T value)
{
    // ensure the data is only 1, 2, 4 or 8 bytes
    BOOST_STATIC_ASSERT(sizeof(T) == 1 || sizeof(T) == 2 || sizeof(T) == 4 || sizeof(T) == 8);
    // ensure we're only swapping arithmetic types
    BOOST_STATIC_ASSERT(boost::is_arithmetic<T>::value);

    return detail::do_byte_swap<from, to, T>()(value);
}

You would then use it as follows:

// swaps val from host-byte-order to network-byte-order
auto swapped = byte_swap<host_endian, network_endian>(val);

and vice-versa

// swap a value received from the network into host-byte-order
auto val = byte_swap<network_endian, host_endian>(val_from_network);

Answer 6

The procedure for going from big-endian to little-endian is the same as going from little-endian to big-endian.

Here's some example code:

void swapByteOrder(unsigned short& us)
{
    us = (us >> 8) |
         (us << 8);
}

void swapByteOrder(unsigned int& ui)
{
    ui = (ui >> 24) |
         ((ui<<8) & 0x00FF0000) |
         ((ui>>8) & 0x0000FF00) |
         (ui << 24);
}

void swapByteOrder(unsigned long long& ull)
{
    ull = (ull >> 56) |
          ((ull<<40) & 0x00FF000000000000) |
          ((ull<<24) & 0x0000FF0000000000) |
          ((ull<<8) & 0x000000FF00000000) |
          ((ull>>8) & 0x00000000FF000000) |
          ((ull>>24) & 0x0000000000FF0000) |
          ((ull>>40) & 0x000000000000FF00) |
          (ull << 56);
}

Answer 7

There is an assembly instruction called BSWAP that will do the swap for you, extremely fast. You can read about it here.

Visual Studio, or more precisely the Visual C++ runtime library, has platform intrinsics for this, called _byteswap_ushort(), _byteswap_ulong(), and _byteswap_int64(). Similar should exist for other platforms, but I'm not aware of what they would be called.

Answer 8

We've done this with templates. You could do something like this:

// 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);
}

Answer 9

The same way you do in C:

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

You could also declare a vector of unsigned chars, memcpy the input value into it, reverse the bytes into another vector and memcpy the bytes out, but that'll take orders of magnitude longer than bit-twiddling, especially with 64-bit values.

Answer 10

If you're doing this to transfer data between different platforms look at the ntoh and hton functions.

Answer 11

On most POSIX systems (through it's not in the POSIX standard) there is the endian.h, which can be used to determine what encoding your system uses. From there it's something like this:

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];
}

This swaps the order (from big endian to little endian):

If you have the number 0xDEADBEEF (on a little endian system stored as 0xEFBEADDE), ptr[0] will be 0xEF, ptr[1] is 0xBE, etc.

But if you want to use it for networking, then htons, htonl and htonll (and their inverses ntohs, ntohl and ntohll) will be helpful for converting from host order to network order.

Answer 12

Note that, at least for Windows, htonl() is much slower than their intrinsic counterpart _byteswap_ulong(). The former is a DLL library call into ws2_32.dll, the latter is one BSWAP assembly instruction. Therefore, if you are writing some platform-dependent code, prefer using the intrinsics for speed:

#define htonl(x) _byteswap_ulong(x)

This may be especially important for .PNG image processing where all integers are saved in Big Endian with explanation "One can use htonl()..." {to slow down typical Windows programs, if you are not prepared}.

Answer 13

Seriously... I don't understand why all solutions are that complicated! How about the simplest, most general template function that swaps any type of any size under any circumstances in any operating system????

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));
}

It's the magic power of C and C++ together! Simply swap the original variable character by character.

Point 1: No operators: Remember that I didn't use the simple assignment operator "=" because some objects will be messed up when the endianness is flipped and the copy constructor (or assignment operator) won't work. Therefore, it's more reliable to copy them char by char.

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.

To use this function:

double x = 5;
SwapEnd(x);

and now x is different in endianness.

Answer 14

Most platforms have a system header file that provides efficient byteswap functions. On Linux it is in <endian.h>. You can wrap it nicely in C++:

#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';
}

Output:

cafe
deadbeaf
feeddeafbeefcafe

Answer 15

I have this code that allow me to convert from HOST_ENDIAN_ORDER (whatever it is) to LITTLE_ENDIAN_ORDER or BIG_ENDIAN_ORDER. I use a template, so if I try to convert from HOST_ENDIAN_ORDER to LITTLE_ENDIAN_ORDER and they happen to be the same for the machine for wich I compile, no code will be generated.

Here is the code with some comments:

// 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);
}

Answer 16

If a big-endian 32-bit unsigned integer looks like 0xAABBCCDD which is equal to 2864434397, then that same 32-bit unsigned integer looks like 0xDDCCBBAA on a little-endian processor which is also equal to 2864434397.

If a big-endian 16-bit unsigned short looks like 0xAABB which is equal to 43707, then that same 16-bit unsigned short looks like 0xBBAA on a little-endian processor which is also equal to 43707.

Here are a couple of handy #define functions to swap bytes from little-endian to big-endian and vice-versa -->

// can be used for short, unsigned short, word, unsigned word (2-byte types)
#define BYTESWAP16(n) (((n&0xFF00)>>8)|((n&0x00FF)<<8))

// can be used for int or unsigned int or float (4-byte types)
#define BYTESWAP32(n) ((BYTESWAP16((n&0xFFFF0000)>>16))|((BYTESWAP16(n&0x0000FFFF))<<16))

// can be used for unsigned long long or double (8-byte types)
#define BYTESWAP64(n) ((BYTESWAP32((n&0xFFFFFFFF00000000)>>32))|((BYTESWAP32(n&0x00000000FFFFFFFF))<<32))

Answer 17

Just thought I added my own solution here since I haven't seen it anywhere. It's a small and portable C++ templated function and portable that only uses bit operations.

template<typename T> inline static T swapByteOrder(const T& val) {
    int totalBytes = sizeof(val);
    T swapped = (T) 0;
    for (int i = 0; i < totalBytes; ++i) {
        swapped |= (val >> (8*(totalBytes-i-1)) & 0xFF) << (8*i);
    }
    return swapped;
}

Answer 18

Here's a generalized version I came up with off the top of my head, for swapping a value in place. The other suggestions would be better if performance is a problem.

 template<typename T>
    void ByteSwap(T * p)
    {
        for (int i = 0;  i < sizeof(T)/2;  ++i)
            std::swap(((char *)p)[i], ((char *)p)[sizeof(T)-1-i]);
    }

Disclaimer: I haven't tried to compile this or test it yet.

Answer 19

If you take the common pattern for reversing the order of bits in a word, and cull the part that reverses bits within each byte, then you're left with something which only reverses the bytes within a word. For 64-bits:

x = ((x & 0x00000000ffffffff) << 32) ^ ((x >> 32) & 0x00000000ffffffff);
x = ((x & 0x0000ffff0000ffff) << 16) ^ ((x >> 16) & 0x0000ffff0000ffff);
x = ((x & 0x00ff00ff00ff00ff) <<  8) ^ ((x >>  8) & 0x00ff00ff00ff00ff);

The compiler should clean out the superfluous bit-masking operations (I left them in to highlight the pattern), but if it doesn't you can rewrite the first line this way:

x = ( x                       << 32) ^  (x >> 32);

That should normally simplify down to a single rotate instruction on most architectures (ignoring that the whole operation is probably one instruction).

On a RISC processor the large, complicated constants may cause the compiler difficulties. You can trivially calculate each of the constants from the previous one, though. Like so:

uint64_t k = 0x00000000ffffffff; /* compiler should know a trick for this */
x = ((x & k) << 32) ^ ((x >> 32) & k);
k ^= k << 16;
x = ((x & k) << 16) ^ ((x >> 16) & k);
k ^= k << 8;
x = ((x & k) <<  8) ^ ((x >>  8) & k);

If you like, you can write that as a loop. It won't be efficient, but just for fun:

int i = sizeof(x) * CHAR_BIT / 2;
uintmax_t k = (1 << i) - 1;
while (i >= 8)
{
    x = ((x & k) << i) ^ ((x >> i) & k);
    i >>= 1;
    k ^= k << i;
}

And for completeness, here's the simplified 32-bit version of the first form:

x = ( x               << 16) ^  (x >> 16);
x = ((x & 0x00ff00ff) <<  8) ^ ((x >>  8) & 0x00ff00ff);

Answer 20

Using the codes below, you can swap between BigEndian and LittleEndian easily

#define uint32_t unsigned 
#define uint16_t unsigned short

#define swap16(x) ((((uint16_t)(x) & 0x00ff)<<8)| \
(((uint16_t)(x) & 0xff00)>>8))

#define swap32(x) ((((uint32_t)(x) & 0x000000ff)<<24)| \
(((uint32_t)(x) & 0x0000ff00)<<8)| \
(((uint32_t)(x) & 0x00ff0000)>>8)| \
(((uint32_t)(x) & 0xff000000)>>24))

Answer 21

I am really surprised no one mentioned htobeXX and betohXX functions. They are defined in endian.h and are very similar to network functions htonXX.

Answer 22

i like this one, just for style :-)

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

Answer 23

I recently wrote a macro to do this in C, but it's equally valid in C++:

#define REVERSE_BYTES(...) do for(size_t REVERSE_BYTES=0; REVERSE_BYTES<sizeof(__VA_ARGS__)>>1; ++REVERSE_BYTES)\
    ((unsigned char*)&(__VA_ARGS__))[REVERSE_BYTES] ^= ((unsigned char*)&(__VA_ARGS__))[sizeof(__VA_ARGS__)-1-REVERSE_BYTES],\
    ((unsigned char*)&(__VA_ARGS__))[sizeof(__VA_ARGS__)-1-REVERSE_BYTES] ^= ((unsigned char*)&(__VA_ARGS__))[REVERSE_BYTES],\
    ((unsigned char*)&(__VA_ARGS__))[REVERSE_BYTES] ^= ((unsigned char*)&(__VA_ARGS__))[sizeof(__VA_ARGS__)-1-REVERSE_BYTES];\
while(0)

It accepts any type and reverses the bytes in the passed argument. Example usages:

int main(){
    unsigned long long x = 0xABCDEF0123456789;
    printf("Before: %llX\n",x);
    REVERSE_BYTES(x);
    printf("After : %llX\n",x);

    char c[7]="nametag";
    printf("Before: %c%c%c%c%c%c%c\n",c[0],c[1],c[2],c[3],c[4],c[5],c[6]);
    REVERSE_BYTES(c);
    printf("After : %c%c%c%c%c%c%c\n",c[0],c[1],c[2],c[3],c[4],c[5],c[6]);
}

Which prints:

Before: ABCDEF0123456789
After : 8967452301EFCDAB
Before: nametag
After : gateman

The above is perfectly copy/paste-able, but there's a lot going on here, so I'll break down how it works piece by piece:

The first notable thing is that the entire macro is encased in a do while(0) block. This is a common idiom to allow normal semicolon use after the macro.

Next up is the use of a variable named REVERSE_BYTES as the for loop's counter. The name of the macro itself is used as a variable name to ensure that it doesn't clash with any other symbols that may be in scope wherever the macro is used. Since the name is being used within the macro's expansion, it won't be expanded again when used as a variable name here.

Within the for loop, there are two bytes being referenced and XOR swapped (so a temporary variable name is not required):

((unsigned char*)&(__VA_ARGS__))[REVERSE_BYTES]
((unsigned char*)&(__VA_ARGS__))[sizeof(__VA_ARGS__)-1-REVERSE_BYTES]

__VA_ARGS__ represents whatever was given to the macro, and is used to increase the flexibility of what may be passed in (albeit not by much). The address of this argument is then taken and cast to an unsigned char pointer to permit the swapping of its bytes via array [] subscripting.

The final peculiar point is the lack of {} braces. They aren't necessary because all of the steps in each swap are joined with the comma operator, making them one statement.

Finally, it's worth noting that this is not the ideal approach if speed is a top priority. If this is an important factor, some of the type-specific macros or platform-specific directives referenced in other answers are likely a better option. This approach, however, is portable to all types, all major platforms, and both the C and C++ languages.

Answer 24

If you have C++ 17 then add this header

#include <algorithm>

Use this template function to swap the bytes:

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);
}

call it like:

swapEndian (stlContainer);

Answer 25

Wow, I couldn't believe some of the answers I've read here. There's actually an instruction in assembly which does this faster than anything else. bswap. You could simply write a function like this...

__declspec(naked) uint32_t EndianSwap(uint32 value)
{
    __asm
    {
        mov eax, dword ptr[esp + 4]
        bswap eax
        ret
    }
}

It is MUCH faster than the intrinsics that have been suggested. I've disassembled them and looked. The above function has no prologue/epilogue so virtually has no overhead at all.

unsigned long _byteswap_ulong(unsigned long value);

Doing 16 bit is just as easy, with the exception that you'd use xchg al, ah. bswap only works on 32-bit registers.

64-bit is a little more tricky, but not overly so. Much better than all of the above examples with loops and templates etc.

There are some caveats here... Firstly bswap is only available on 80x486 CPU's and above. Is anyone planning on running it on a 386?!? If so, you can still replace bswap with...

mov ebx, eax
shr ebx, 16
xchg al, ah
xchg bl, bh
shl eax, 16
or eax, ebx

Also inline assembly is only available in x86 code in Visual Studio. A naked function cannot be lined and also isn't available in x64 builds. I that instance, you're going to have to use the compiler intrinsics.

Answer 26

Portable technique for implementing optimizer-friendly unaligned non-inplace endian accessors. They work on every compiler, every boundary alignment and every byte ordering. These unaligned routines are supplemented, or mooted, depending on native endian and alignment. Partial listing but you get the idea. BO* are constant values based on native byte ordering.

uint32_t sw_get_uint32_1234(pu32)
uint32_1234 *pu32;
{
  union {
    uint32_1234 u32_1234;
    uint32_t u32;
  } bou32;
  bou32.u32_1234[0] = (*pu32)[BO32_0];
  bou32.u32_1234[1] = (*pu32)[BO32_1];
  bou32.u32_1234[2] = (*pu32)[BO32_2];
  bou32.u32_1234[3] = (*pu32)[BO32_3];
  return(bou32.u32);
}

void sw_set_uint32_1234(pu32, u32)
uint32_1234 *pu32;
uint32_t u32;
{
  union {
    uint32_1234 u32_1234;
    uint32_t u32;
  } bou32;
  bou32.u32 = u32;
  (*pu32)[BO32_0] = bou32.u32_1234[0];
  (*pu32)[BO32_1] = bou32.u32_1234[1];
  (*pu32)[BO32_2] = bou32.u32_1234[2];
  (*pu32)[BO32_3] = bou32.u32_1234[3];
}

#if HAS_SW_INT64
int64 sw_get_int64_12345678(pi64)
int64_12345678 *pi64;
{
  union {
    int64_12345678 i64_12345678;
    int64 i64;
  } boi64;
  boi64.i64_12345678[0] = (*pi64)[BO64_0];
  boi64.i64_12345678[1] = (*pi64)[BO64_1];
  boi64.i64_12345678[2] = (*pi64)[BO64_2];
  boi64.i64_12345678[3] = (*pi64)[BO64_3];
  boi64.i64_12345678[4] = (*pi64)[BO64_4];
  boi64.i64_12345678[5] = (*pi64)[BO64_5];
  boi64.i64_12345678[6] = (*pi64)[BO64_6];
  boi64.i64_12345678[7] = (*pi64)[BO64_7];
  return(boi64.i64);
}
#endif

int32_t sw_get_int32_3412(pi32)
int32_3412 *pi32;
{
  union {
    int32_3412 i32_3412;
    int32_t i32;
  } boi32;
  boi32.i32_3412[2] = (*pi32)[BO32_0];
  boi32.i32_3412[3] = (*pi32)[BO32_1];
  boi32.i32_3412[0] = (*pi32)[BO32_2];
  boi32.i32_3412[1] = (*pi32)[BO32_3];
  return(boi32.i32);
}

void sw_set_int32_3412(pi32, i32)
int32_3412 *pi32;
int32_t i32;
{
  union {
    int32_3412 i32_3412;
    int32_t i32;
  } boi32;
  boi32.i32 = i32;
  (*pi32)[BO32_0] = boi32.i32_3412[2];
  (*pi32)[BO32_1] = boi32.i32_3412[3];
  (*pi32)[BO32_2] = boi32.i32_3412[0];
  (*pi32)[BO32_3] = boi32.i32_3412[1];
}

uint32_t sw_get_uint32_3412(pu32)
uint32_3412 *pu32;
{
  union {
    uint32_3412 u32_3412;
    uint32_t u32;
  } bou32;
  bou32.u32_3412[2] = (*pu32)[BO32_0];
  bou32.u32_3412[3] = (*pu32)[BO32_1];
  bou32.u32_3412[0] = (*pu32)[BO32_2];
  bou32.u32_3412[1] = (*pu32)[BO32_3];
  return(bou32.u32);
}

void sw_set_uint32_3412(pu32, u32)
uint32_3412 *pu32;
uint32_t u32;
{
  union {
    uint32_3412 u32_3412;
    uint32_t u32;
  } bou32;
  bou32.u32 = u32;
  (*pu32)[BO32_0] = bou32.u32_3412[2];
  (*pu32)[BO32_1] = bou32.u32_3412[3];
  (*pu32)[BO32_2] = bou32.u32_3412[0];
  (*pu32)[BO32_3] = bou32.u32_3412[1];
}

float sw_get_float_1234(pf)
float_1234 *pf;
{
  union {
    float_1234 f_1234;
    float f;
  } bof;
  bof.f_1234[0] = (*pf)[BO32_0];
  bof.f_1234[1] = (*pf)[BO32_1];
  bof.f_1234[2] = (*pf)[BO32_2];
  bof.f_1234[3] = (*pf)[BO32_3];
  return(bof.f);
}

void sw_set_float_1234(pf, f)
float_1234 *pf;
float f;
{
  union {
    float_1234 f_1234;
    float f;
  } bof;
  bof.f = (float)f;
  (*pf)[BO32_0] = bof.f_1234[0];
  (*pf)[BO32_1] = bof.f_1234[1];
  (*pf)[BO32_2] = bof.f_1234[2];
  (*pf)[BO32_3] = bof.f_1234[3];
}

double sw_get_double_12345678(pd)
double_12345678 *pd;
{
  union {
    double_12345678 d_12345678;
    double d;
  } bod;
  bod.d_12345678[0] = (*pd)[BO64_0];
  bod.d_12345678[1] = (*pd)[BO64_1];
  bod.d_12345678[2] = (*pd)[BO64_2];
  bod.d_12345678[3] = (*pd)[BO64_3];
  bod.d_12345678[4] = (*pd)[BO64_4];
  bod.d_12345678[5] = (*pd)[BO64_5];
  bod.d_12345678[6] = (*pd)[BO64_6];
  bod.d_12345678[7] = (*pd)[BO64_7];
  return(bod.d);
}

void sw_set_double_12345678(pd, d)
double_12345678 *pd;
double d;
{
  union {
    double_12345678 d_12345678;
    double d;
  } bod;
  bod.d = d;
  (*pd)[BO64_0] = bod.d_12345678[0];
  (*pd)[BO64_1] = bod.d_12345678[1];
  (*pd)[BO64_2] = bod.d_12345678[2];
  (*pd)[BO64_3] = bod.d_12345678[3];
  (*pd)[BO64_4] = bod.d_12345678[4];
  (*pd)[BO64_5] = bod.d_12345678[5];
  (*pd)[BO64_6] = bod.d_12345678[6];
  (*pd)[BO64_7] = bod.d_12345678[7];
}

These typedefs have the benefit of raising compiler errors if not used with accessors, thus mitigating forgotten accessor bugs.

typedef char int8_1[1], uint8_1[1];

typedef char int16_12[2], uint16_12[2]; /* little endian */
typedef char int16_21[2], uint16_21[2]; /* big endian */

typedef char int24_321[3], uint24_321[3]; /* Alpha Micro, PDP-11 */

typedef char int32_1234[4], uint32_1234[4]; /* little endian */
typedef char int32_3412[4], uint32_3412[4]; /* Alpha Micro, PDP-11 */
typedef char int32_4321[4], uint32_4321[4]; /* big endian */

typedef char int64_12345678[8], uint64_12345678[8]; /* little endian */
typedef char int64_34128756[8], uint64_34128756[8]; /* Alpha Micro, PDP-11 */
typedef char int64_87654321[8], uint64_87654321[8]; /* big endian */

typedef char float_1234[4]; /* little endian */
typedef char float_3412[4]; /* Alpha Micro, PDP-11 */
typedef char float_4321[4]; /* big endian */

typedef char double_12345678[8]; /* little endian */
typedef char double_78563412[8]; /* Alpha Micro? */
typedef char double_87654321[8]; /* big endian */

Answer 27

Byte swapping with ye olde 3-step-xor trick around a pivot in a template function gives a flexible, quick O(ln2) solution that does not require a library, the style here also rejects 1 byte types:

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);
    }
}

Answer 28

Seems like the safe way would be to use htons on each word. So, if you have...

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); });

The above would be a no-op if you were on a big-endian system, so I would look for whatever your platform uses as a compile-time condition to decide whether htons is a no-op. It is O(n) after all. On a Mac, it would be something like ...

#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

Answer 29

Here is a basic function to swap to/from little and big endian. It's basic but it doesn't require supplementary libraries.

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);
}

Answer 30

Not as efficient as using an intrinsic function, but certainly portable. My answer:

#include <cstdint>
#include <type_traits>

/**
 * Perform an endian swap of bytes against a templatized unsigned word.
 *
 * @tparam value_type The data type to perform the endian swap against.
 * @param value       The data value to swap.
 *
 * @return value_type The resulting swapped word.
 */
template <typename value_type>
constexpr inline auto endian_swap(value_type value) -> value_type
{
    using half_type = typename std::conditional<
        sizeof(value_type) == 8u,
        uint32_t,
        typename std::conditional<sizeof(value_type) == 4u, uint16_t, uint8_t>::
            type>::type;

    size_t const    half_bits  = sizeof(value_type) * 8u / 2u;
    half_type const upper_half = static_cast<half_type>(value >> half_bits);
    half_type const lower_half = static_cast<half_type>(value);

    if (sizeof(value_type) == 2u)
    {
        return (static_cast<value_type>(lower_half) << half_bits) | upper_half;
    }

    return ((static_cast<value_type>(endian_swap(lower_half)) << half_bits) |
            endian_swap(upper_half));
}

Answer 31

Came here looking for a Boost solution and left disappointed, but finally found it elsewhere. You can use boost::endian::endian_reverse. It's templated/overloaded for all the primitive types:

#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;
}

Demonstration

Although, it looks like c++23 finally put this to bed with std::byteswap. (I'm using c++17, so this was not an option.)

Answer 32

A c++20 branchless version now that std::endian exists but before c++23 adds 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);
}

Answer 33

Here's how to read a double stored in IEEE 754 64 bit format, even if your host computer uses a different system.

/*
* 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;
    }
}

For the rest of the suite of functions, including the write and the integer routines see my github project

https://github.com/MalcolmMcLean/ieee754

Answer 34

void writeLittleEndianToBigEndian(void* ptrLittleEndian, void* ptrBigEndian , size_t bufLen )
{
    char *pchLittleEndian = (char*)ptrLittleEndian;

    char *pchBigEndian = (char*)ptrBigEndian;

    for ( size_t i = 0 ; i < bufLen ; i++ )    
        pchBigEndian[bufLen-1-i] = pchLittleEndian[i];
}

std::uint32_t row = 0x12345678;

char buf[4]; 

writeLittleEndianToBigEndian( &row, &buf, sizeof(row) );

Answer 35

Look up bit shifting, as this is basically all you need to do to swap from little -> big endian. Then depending on the bit size, you change how you do the bit shifting.