Question

I have a struct, which is composed of multiple 32 bits elements. I applied a #pragma pack (4), the following struct is therefore linear and aligned.

struct
{
  int a; // 4 bytes
  int b; // 4 bytes
  int c; // 4 bytes
} mystruct; // total 16 bytes

How can I swap each of these elements (little -> big endian) ?

The method is void swap(void* a, int b);, with a pointer to the structure, and b integer giving the size of the structure.

For example :

void swap(void* a, int b)
{
  //FIXME !
  for (int i = 0; i < b; i+= 32)
  {
    a = (a & 0x0000FFFF) << 16 | (a & 0xFFFF0000) >> 16;
    a = (a & 0x00FF00FF) << 8 | (a & 0xFF00FF00) >> 8;
    a += 32;
  }
}

Answer 1

You can swap two bytes without using a temporary:

void byteswap( unsigned char & a, unsigned char & b )
{
   a ^= b;
   b ^= a;
   a ^= b;
}

Now let’s apply it to numbers of variable length

template< typename T >
void endianSwap( T & t )
{
    unsigned char * first = reinterpret_cast< unsigned char * >( &t );
    unsigned char * last = first + sizeof(T) - 1;
    while( first < last )
    {
       byteswap( *first, *last );
       ++first;
       --last;
    }
}

For your struct you can:

void endianSwap( mystruct & s )
{
     endianSwap( s.a );
     endianSwap( s.b );
     endianSwap( s.c );
}

Of course, as an alternative to endianSwap using byteswap, we could just use std::reverse.

template<typename T> endianSwap( T& t )
{
    unsigned char * begin = reinterpret_cast<unsigned char *>(&t);
    unsigned char * end = begin + sizeof( T );
    std::reverse( begin, end );
}