Is there any “standard” htonl-like function for 64 bits integers in C++?

Question

I'm working on an implementation of the memcache protocol which, at some points, uses 64 bits integer values. These values must be stored in "network byte order".

I wish there was some uint64_t htonll(uint64_t value) function to do the change, but unfortunately, if it exist, I couldn't find it.

So I have 1 or 2 questions:

• Is there any portable (Windows, Linux, AIX) standard function to do this ?
• If there is no such function, how would you implement it ?

I have in mind a basic implementation but I don't know how to check the endianness at compile-time to make the code portable. So your help is more than welcome here ;)

Thank you.

---

Here is the final solution I wrote, thanks to Brian's solution.

uint64_t htonll(uint64_t value)
{
    // The answer is 42
    static const int num = 42;

    // Check the endianness
    if (*reinterpret_cast<const char*>(&num) == num)
    {
        const uint32_t high_part = htonl(static_cast<uint32_t>(value >> 32));
        const uint32_t low_part = htonl(static_cast<uint32_t>(value & 0xFFFFFFFFLL));

        return (static_cast<uint64_t>(low_part) << 32) | high_part;
    } else
    {
        return value;
    }
}

Answer 1

You are probably looking for bswap_64 I think it is supported pretty much everywhere but I wouldn't call it standard.

You can easily check the endianness by creating an int with a value of 1, casting your int's address as a char* and checking the value of the first byte.

For example:

int num = 42;
if(*(char *)&num == 42)
{
   //Little Endian
}
else
{
   //Big Endian
} 

Knowing this you could also make a simple function that does the swapping.

---

You could also always use boost which contains endian macros which are portable cross platform.

Answer 2

#define htonll(x) ((1==htonl(1)) ? (x) : ((uint64_t)htonl((x) & 0xFFFFFFFF) << 32) | htonl((x) >> 32))
#define ntohll(x) ((1==ntohl(1)) ? (x) : ((uint64_t)ntohl((x) & 0xFFFFFFFF) << 32) | ntohl((x) >> 32))

The test (1==htonl(1)) simply determines (at runtime sadly) if the hardware architecture requres byte swapping. There aren't any portable ways to determine at compile-time what the architecture is, so we resort to using "htonl", which is as portable as it gets in this situation. If byte-swapping is required, then we swap 32 bits at a time using htonl (remembering to swap the two 32 bit words as well).

---

Here's another way to perform the swap that is portable across most compilers and operating systems, including AIX, BSDs, Linux, and Solaris.

#if __BIG_ENDIAN__
# define htonll(x) (x)
# define ntohll(x) (x)
#else
# define htonll(x) ((uint64_t)htonl((x) & 0xFFFFFFFF) << 32) | htonl((x) >> 32))
# define ntohll(x) ((uint64_t)ntohl((x) & 0xFFFFFFFF) << 32) | ntohl((x) >> 32))
#endif

The important part is to use __BIG_ENDIAN__ or __LITTLE_ENDIAN__; and not __BYTE_ORDER__, __ORDER_BIG_ENDIAN__ or __ORDER_LITTLE_ENDIAN__. Some compilers and operating systems lack __BYTE_ORDER__ and friends.

Answer 3

This seems to work in C; did I do anything wrong?

uint64_t htonll(uint64_t value) {
    int num = 42;
    if (*(char *)&num == 42) {
        uint32_t high_part = htonl((uint32_t)(value >> 32));
        uint32_t low_part = htonl((uint32_t)(value & 0xFFFFFFFFLL));
        return (((uint64_t)low_part) << 32) | high_part;
    } else {
        return value;
    }
}

Answer 4

You can try with uint64_t htobe64(uint64_t host_64bits) & uint64_t be64toh(uint64_t big_endian_64bits) for vice-versa.

Answer 5

To reduce the overhead of the "if num == ..." Use the pre-processor defines:

#if __BYTE_ORDER__ == __ORDER_LITTLE_ENDIAN__
#else
#endif

Answer 6

EDIT: combining the two (used Brian's code):

uint64_t htonll(uint64_t value)
{
     int num = 42;
     if(*(char *)&num == 42)
          return (htonl(value & 0xFFFFFFFF) << 32LL) | htonl(value >> 32);
     else 
          return value;
}

Warning: untested code! Please test before using.