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What's the best way to turn a string in this form into an IP address: "0200A8C0". The "octets" present in the string are in reverse order, i.e. the given example string should generate 192.168.0.2.

Matt Joiner
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5 Answers5

35

Network address manipulation is provided by the socket module.

socket.inet_ntoa(packed_ip)

Convert a 32-bit packed IPv4 address (a string four characters in length) to its standard dotted-quad string representation (for example, ‘123.45.67.89’). This is useful when conversing with a program that uses the standard C library and needs objects of type struct in_addr, which is the C type for the 32-bit packed binary data this function takes as an argument.

You can translate your hex string to packed ip using struct.pack() and the little endian, unsigned long format.

s = "0200A8C0"

import socket
import struct
addr_long = int(s, 16)
print(hex(addr_long))  # '0x200a8c0'
print(struct.pack("<L", addr_long))  # '\xc0\xa8\x00\x02'
print(socket.inet_ntoa(struct.pack("<L", addr_long)))  # '192.168.0.2'
Basj
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gimel
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    You want " – Omnifarious Feb 04 '10 at 07:50
  • @Omnifarious, how exactly does it's "long/int"ness effect it here? – Matt Joiner Feb 04 '10 at 07:57
  • `192.168.10.5` hex is `c0a80a05`. When I use this snippet to convert from HEX to IP, I get `5.10.168.192`. Which is correct but reversed. To make it work I have to change `L` which doesn't make any sense if I take the input from a user and I don't know what HEX is the input! Do you have any idea how to make it work for any HEX value? Thank you – Tes3awy Apr 14 '21 at 00:59
8
>>> s = "0200A8C0"
>>> bytes = ["".join(x) for x in zip(*[iter(s)]*2)]
>>> bytes
['02', '00', 'A8', 'C0']
>>> bytes = [int(x, 16) for x in bytes]
>>> bytes
[2, 0, 168, 192]
>>> print ".".join(str(x) for x in reversed(bytes))
192.168.0.2

It is short and clear; wrap it up in a function with error checking to suit your needs.

Handy grouping functions:

def group(iterable, n=2, missing=None, longest=True):
  """Group from a single iterable into groups of n.

  Derived from http://bugs.python.org/issue1643
  """
  if n < 1:
    raise ValueError("invalid n")
  args = (iter(iterable),) * n
  if longest:
    return itertools.izip_longest(*args, fillvalue=missing)
  else:
    return itertools.izip(*args)

def group_some(iterable, n=2):
  """Group from a single iterable into groups of at most n."""
  if n < 1:
    raise ValueError("invalid n")
  iterable = iter(iterable)
  while True:
    L = list(itertools.islice(iterable, n))
    if L:
      yield L
    else:
      break
  • @Roger, how does that `zip(*[iter(s)]*2)` work? I'm very interested in that. – Matt Joiner Feb 04 '10 at 07:54
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    This is quite the hack! `iter(s)` returns an iterator over the string. Multiplying the list of this iterator by 2 will create two references **to the same iterator**. `zip()` returns a list of tuples containing an element from each of its arguments. Since both arguments are the same iterator, it will take from it twice for each tuple, returning a tuple of each 2 adjacent characters. I would never have thought of trying anything like this. :D – Max Shawabkeh Feb 04 '10 at 08:24
  • @Matt: It's the core of most "grouping" recipes I've seen, and Max explained it well. Will update to include two short functions in that vein. –  Feb 04 '10 at 11:12
  • @Roger, thanks very much, I was hoping a solution like yours would come up. – Matt Joiner Feb 04 '10 at 17:26
  • I've created a new question, which directly asks what you've solved here: http://stackoverflow.com/questions/2202461/yield-multiple-objects-at-a-time-from-an-iterable-object – Matt Joiner Feb 04 '10 at 19:13
  • +1 That zip iter trick is mindbogglingly useful for code golf. Awesome. – Exelian Sep 01 '11 at 14:49
3

You could do something like this:

>>> s = '0200A8C0'
>>> octets = [s[i:i+2] for i in range(0, len(s), 2)]
>>> ip = [int(i, 16) for i in reversed(octets)]
>>> ip_formatted = '.'.join(str(i) for i in ip)
>>> print ip_formatted
192.168.0.2

The octet splitting could probably be done more elegantly, but I can't think of a simpler way off the top of my head.

EDIT: Or on one line:

>>> s = '0200A8C0'
>>> print '.'.join(str(int(i, 16)) for i in reversed([s[i:i+2] for i in range(0, len(s), 2)]))
192.168.0.2
Max Shawabkeh
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  • yeah teh grouping in particular gets to me. i can't find a nice way to "chunk" the octets out, and iterate _those_ in reverse order, i'm hoping someone knows a way – Matt Joiner Feb 04 '10 at 07:43
  • @Max, Roger's answer contains a very intriguing way to do this. – Matt Joiner Feb 04 '10 at 07:58
0

My try:

a = '0200A8C0'
indices = range(0, 8, 2)
data = [str(int(a[x:x+2], 16)) for x in indices]
'.'.join(reversed(data))
Alok Singhal
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0

A simple def you can make to convert from hex to decimal-ip:

def hex2ip(iphex):
    ip = ["".join(x) for x in zip(*[iter(str(iphex))]*2)]
    ip = [int(x, 16) for x in ip]
    ip = ".".join(str(x) for x in (ip))
    return ip

# And to use it:
ip = "ac10fc40"
ip = hex2ip(iphex)
print(ip)
aboyum
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