Get the ipv6 address from a interface
roy at panix.com
Thu Apr 9 14:17:29 CEST 2009
<86176ef7-c2e0-4c5d-b883-d91672e3eb0b at w40g2000yqd.googlegroups.com>,
Kai Timmer <email at kait.de> wrote:
> i need a function that returns the ipv6 address from a given interface
> name. For ipv4 i use this one:
> def get_ip_address(ifname):
> s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
> return socket.inet_ntoa(fcntl.ioctl(
> 0x8915, # SIOCGIFADDR
> struct.pack('256s', ifname[:15])
> which works great. But i am not enough into python to port that to
> ipv6. It has to work under linux only. Any help is appreciated.
I'm not 100% sure what you're trying to do, but the above is horribly
non-portable. You probably want to be looking at socket.getpeername() and
In general, concepts like "the address of an interface" are difficult. In
many OS's, a given interface may have multiple addresses. This is
especially true in IPv6 where you've have both link local and global
unicast addresses on the same interface.
Can you back up a few steps and describe what it is that you're trying to
do, i.e. the use case?
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