Detecting endianness programmatically in a C++ pro

Question

Is there a programmatic way to detect whether or not you are on a big-endian or little-endian architecture? I need to be able to write code that will execute on an Intel or PPC system and use exactly the same code (i.e. no conditional compilation).

Answer 1

If you don't want conditional compilation you can just write endian independent code. Here is an example (taken from Rob Pike):

Reading an integer stored in little-endian on disk, in an endian independent manner:

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

The same code, trying to take into account the machine endianness:

i = *((int*)data);
#ifdef BIG_ENDIAN
/* swap the bytes */
i = ((i&0xFF)<<24) | (((i>>8)&0xFF)<<16) | (((i>>16)&0xFF)<<8) | (((i>>24)&0xFF)<<0);
#endif

Answer 2

You can do it by setting an int and masking off bits, but probably the easiest way is just to use the built in network byte conversion ops (since network byte order is always big endian).

if ( htonl(47) == 47 ) {
  // Big endian
} else {
  // Little endian.
}

Bit fiddling could be faster, but this way is simple, straightforward and pretty impossible to mess up.

Answer 3

I would do something like this:

bool isBigEndian() {
    static unsigned long x(1);
    static bool result(reinterpret_cast<unsigned char*>(&x)[0] == 0);
    return result;
}

Along these lines, you would get a time efficient function that only does the calculation once.

Answer 4

See Endianness - C-Level Code illustration.

// assuming target architecture is 32-bit = 4-Bytes
enum ENDIANESS{ LITTLEENDIAN , BIGENDIAN , UNHANDLE };


ENDIANESS CheckArchEndianalityV1( void )
{
    int Endian = 0x00000001; // assuming target architecture is 32-bit    

    // as Endian = 0x00000001 so MSB (Most Significant Byte) = 0x00 and LSB (Least     Significant Byte) = 0x01
    // casting down to a single byte value LSB discarding higher bytes    

    return (*(char *) &Endian == 0x01) ? LITTLEENDIAN : BIGENDIAN;
} 

Answer 5

The way C compilers (at least everyone I know of) work the endianness has to be decided at compile time. Even for biendian processors (like ARM och MIPS) you have to choose endianness at compile time. Further more the endianness is defined in all common file formats for executables (such as ELF). Although it is possible to craft a binary blob of biandian code (for some ARM server exploit maybe?) it probably has to be done in assembly.

Answer 6

I don't like the method based on type punning - it will often be warned against by compiler. That's exactly what unions are for !

bool is_big_endian(void)
{
    union {
        uint32_t i;
        char c[4];
    } bint = {0x01020304};

    return bint.c[0] == 1; 
}

The principle is equivalent to the type case as suggested by others, but this is clearer - and according to C99, is guaranteed to be correct. gcc prefers this compared to the direct pointer cast.

This is also much better than fixing the endianness at compile time - for OS which support multi-architecture (fat binary on Mac os x for example), this will work for both ppc/i386, whereas it is very easy to mess things up otherwise.