IPv6 vs IPv4

jiml8 · 12-22-2008, 12:40 AM

I am doing some socket programming and have run across something that I can deal with, but it is annoying.

The distro is Mandriva 2008. Basically, I am trying to write some protocol agnostic code, with the goal of handling either ipv6 or ipv4, whichever comes up.

At this point, I am writing some server code that is listening on a port for a connection, and then is connecting, passing some data back and forth, then upon command shutting down. Vanilla stuff.

At this time, I am communicating with this server code via telnet across my LAN from another Mandriva machine. The networking code is based on the libunp.a code that is provided in the book Unix Network Programming Second Edition by Richard Stevens. I am using it because I have a lot of working code that is based on it.

When I set up the listener socket on the server, I do it like this:

Code:

hints.ai_flags = AI_PASSIVE;
	hints.ai_family = AF_UNSPEC;
	hints.ai_socktype = SOCK_STREAM;

	if ( (n = getaddrinfo(host, serv, &hints, &res)) != 0)
		err_quit("tcp_listen error for %s, %s: %s",
				 host, serv, gai_strerror(n));

At a later time, the server accepts a communication from my other machine via telnet based upon this code:

Code:

	        int listenfd=0;
                socklen_t addrlen,namelen;
	        struct sockaddr *cliaddr;
		commandfd = Accept(listenfd, cliaddr, &addrlen);

where Accept is a wrapper subroutine that calls accept.

Immediately after this, I want to get the IP address of the client that has contacted me. So I do this:

Code:

	char s[INET6_ADDRSTRLEN];
	        namelen = INET6_ADDRSTRLEN;
		get_ip_str(cliaddr, &s, namelen);  
 printf("new client: connecting to %s\n", s);

where the subroutine get_ip_str() is defined as:

Code:

char *get_ip_str(const struct sockaddr *sa, char *s, size_t maxlen)
{
    switch(sa->sa_family) {
        case AF_INET:
            inet_ntop(AF_INET, &(((struct sockaddr_in *)sa)->sin_addr),
                    s, maxlen);
            break;

        case AF_INET6:
            inet_ntop(AF_INET6, &(((struct sockaddr_in6 *)sa)->sin6_addr),
                    s, maxlen);
            break;

        default:
            strncpy(s, "Unknown AF", maxlen);
            return NULL;
    }

    return s;
}

Now, my network is defined using IPv4 and I rather expected to have an IPv4 definition provided for me. But when I execute the program and print out the result of my network address search, this is what I get:

Quote:

new client: connecting to ::ffff:192.168.0.2

You can see that the correct IPv4 address is buried in the IPv6 field, but in fact my socket has been declared as IPv6.

This is not a huge problem; I can always parse the string, but I am wondering what this means, and why I am not getting the proper IPv4 definition when the network is configured that way.

Also, of course, this could cause me some issues actually identifying that I am really on an IPv6 network, if I make the necessary hack to assume this is an IPv4 network regardless of what the socket says.

Does anyone know what I can do to get this to behave properly?

jcookeman · 12-22-2008, 02:58 AM

IPv4 mapped addresses.

Have a look at this beauty from Exim's host.c

Code:

uschar *
host_ntoa(int type, const void *arg, uschar *buffer, int *portptr)
	

{
uschar *yield;

/* The new world. It is annoying that we have to fish out the address from
different places in the block, depending on what kind of address it is. It
is also a pain that inet_ntop() returns a const uschar *, whereas the IPv4
function inet_ntoa() returns just uschar *, and some picky compilers insist
on warning if one assigns a const uschar * to a uschar *. Hence the casts. */

#if HAVE_IPV6
uschar addr_buffer[46];
if (type < 0)
  {
  int family = ((struct sockaddr *)arg)->sa_family;
  if (family == AF_INET6)
    {
    struct sockaddr_in6 *sk = (struct sockaddr_in6 *)arg;
    yield = (uschar *)inet_ntop(family, &(sk->sin6_addr), CS addr_buffer,
      sizeof(addr_buffer));
    if (portptr != NULL) *portptr = ntohs(sk->sin6_port);
    }
  else
    {
    struct sockaddr_in *sk = (struct sockaddr_in *)arg;
    yield = (uschar *)inet_ntop(family, &(sk->sin_addr), CS addr_buffer,
      sizeof(addr_buffer));
    if (portptr != NULL) *portptr = ntohs(sk->sin_port);
    }
  }
else
  {
  yield = (uschar *)inet_ntop(type, arg, CS addr_buffer, sizeof(addr_buffer));
  }

/* If the result is a mapped IPv4 address, show it in V4 format. */

if (Ustrncmp(yield, "::ffff:", 7) == 0) yield += 7;

#else  /* HAVE_IPV6 */

/* The old world */

if (type < 0)
  {
  yield = US inet_ntoa(((struct sockaddr_in *)arg)->sin_addr);
  if (portptr != NULL) *portptr = ntohs(((struct sockaddr_in *)arg)->sin_port);
  }
else
  yield = US inet_ntoa(*((struct in_addr *)arg));
#endif

/* If there is no buffer, put the string into some new store. */

if (buffer == NULL) return string_copy(yield);

/* Callers of this function with a non-NULL buffer must ensure that it is
large enough to hold an IPv6 address, namely, at least 46 bytes. That's what
makes this use of strcpy() OK. */

Ustrcpy(buffer, yield);
return buffer;
}

jiml8 · 12-22-2008, 10:53 AM

Quote:

IPv4 mapped addresses.

Well, that makes sense, though it is a royal PITA. It should be a lot easier to get the address from a socket.