IPv6 parsing in C

Question

I wanted to know how i can parse an IPv6 address in 'C' and convert it to a 128 bit value? So a hex address like 1:22:333:aaaa:b:c:d:e:f needs to be converted to its 128 bit equivalent binary. The problem is the IP address could be of the type ::2 and its variant since they are valid IPv6 address.

The input is from the keyboard and hence is in ASCII format.

Any suggestions or pointers will be appreciated. Thanks!!!

Answer 1

getaddrinfo() can understand IPv6 addresses. Pass AF_INET6 to it in the hints, as well as AI_NUMERICHOST (to prevent a DNS lookup). Linux has it, Windows has it as of Windows XP.

Answer 2

You can use POSIX inet_pton to convert a string to a struct in6_addr.

#include <arpa/inet.h>

  ...

const char *ip6str = "::2";
struct in6_addr result;

if (inet_pton(AF_INET6, ip6str, &result) == 1) // success!
{
    //successfully parsed string into "result"
}
else
{
    //failed, perhaps not a valid representation of IPv6?
}

Answer 3

To parse IPv6 in C, you need to build yourself a utility function, which tokenized string (colon for hex blocks, and forward-slash for subnet bits).

1. Tokenize raw IPv6 string into smaller substring.
2. Convert non-empty substring into hex blocks. (ASCII to decimal conversion)
3. Expand hex block into 2-bytes by padding zero in front. (only leading zeroes get trimmed)
4. Complete IPv6 should have 8 hex blocks, calculate missing hex-block(s). (zeroes grouping can happen only once)
5. Reinsert missing hex-block. (use index of the empty substring)