byte order in a union and endianness

[KarlisRepsons]

So do I get it right, that on different platforms
the outputed chars can swap depending on system
endianness?

#include <iostream>
using namespace std;

union meta
{
 unsigned short i;
 unsigned char ch[2];
};

int
main()
{
 meta m;
 m.i = 0xFFEF;

 cout
  << hex
  << "meta.short == "
  << m.i
  << ","
  << endl
  << "meta.char[0] == "
  << static_cast<unsigned short>(m.ch[0])
  << ",\n"
  << "meta.char[1] == "
  << static_cast<unsigned short>(m.ch[1])
  << "."
  << endl;

 return 0;
}

I have this output on a PC:

meta.short == ffef,
meta.char[0] == ef,
meta.char[1] == ff.

I merely guess, on an old PowerPC it would
[1] <=> [0] to show us big endian...

[magnificence7]

That's right.

[karthick88]

Hey Karlis.. I am not able to understand this

meta m;
 m.i = 0xFFEF;

The value is assigned only for but you are printing out m.ch[0]. How is this possible??! Can you explain please

[jsmith]

@karthick88: You need to read up on unions.

[KarlisRepsons]

Well, for those reading this, but not much interested in unions: the size of union is as of it's largest data member, but all data members actually point to the same physical data each in its own way.