How can I change the byte order (from network to host, and vice versa) of an IPV6 address?

Question

I am aware of ntoh{s,l} and hton{s,l}, which work on integers of 2 and 4 bytes. Now, I am facing the problem to translate an IPv6 address, which is 16 bytes long.

Is there a ready-made function for that purpose?

TIA, Jir

Answer 1

I'm not sure that ntoh and hton are relevant in IPv6. You don't have a native 128-bit type, do you?

According to http://www.mail-archive.com/users@ipv6.org/msg00195.html:

IPv6 addresses are expected to be representd in network byte order whenever they are in binary form (on the wire, on a host, in a router, etc). Among other places, see RFC 2553, section 3.2.

From RFC 2553:

3.2 IPv6 Address Structure

A new in6_addr structure holds a single IPv6 address and is defined as a result of including :

struct in6_addr {
    uint8_t  s6_addr[16];      /* IPv6 address */
};

This data structure contains an array of sixteen 8-bit elements, which make up one 128-bit IPv6 address. The IPv6 address is stored in network byte order.

The structure in6_addr above is usually implemented with an embedded union with extra fields that force the desired alignment level in a manner similar to BSD implementations of "struct in_addr". Those additional implementation details are omitted here for simplicity.

An example is as follows:

struct in6_addr {
    union {
        uint8_t  _S6_u8[16];
        uint32_t _S6_u32[4];
        uint64_t _S6_u64[2];
    } _S6_un;
};
#define s6_addr _S6_un._S6_u8