Detecting endianness programmatically in a C++ program

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 (ie no conditional compilation).

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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 !

int 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.

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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.

---

Please see this article:

Here is some code to determine what is the type of your machine

int num = 1;
if(*(char *)&num == 1)
{
    printf("nLittle-Endiann");
}
else
{
    printf("Big-Endiann");
}

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