Is using an union in place of a cast well defined?

Question

I had a discussion this morning with a colleague regarding the correctness of a "coding trick" to detect endianness.

The trick was:

bool is_big_endian()
{
  union
  {
    int i;
    char c[sizeof(int)];
  } foo;


  foo.i = 1;
  return (foo.c[0] == 1);
}

To me, it seems that this usage of an union is incorrect because setting one member of the union and reading another is not well-defined. But I have to admit that this is just a feeling and I lack actual proofs to strengthen my point.

Is this trick correct ? Who is right here ?

Answer 1

Your code is not portable. It might work on some compilers or it might not.

You are right about the behaviour being undefined when you try to access the inactive member of the union [as it is in the case of the code given]

$9.5/1

In a union, at most one of the data members can be active at any time, that is, the value of at most one of the data members can be stored in a union at any time.

So foo.c[0] == 1 is incorrect because c is not active at that moment. Feel free to correct me if you think I am wrong.

Answer 2

Don't do this, better use something like the following:

#include <arpa/inet.h>
//#include <winsock2.h> // <-- for Windows use this instead

#include <stdint.h>

bool is_big_endian() {
  uint32_t i = 1;
  return i == htonl(i);
}

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Explanation:

The htonl function converts a u_long from host to TCP/IP network byte order (which is big-endian).

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References:

• http://linux.die.net/man/3/htonl
• http://msdn.microsoft.com/de-de/library/ms738556%28v=vs.85%29.aspx