63

i have a little problem with my script, where i need to convert ip in form 'xxx.xxx.xxx.xxx' to integer representation and go back from this form.

def iptoint(ip):
    return int(socket.inet_aton(ip).encode('hex'),16)

def inttoip(ip):
    return socket.inet_ntoa(hex(ip)[2:].decode('hex'))


In [65]: inttoip(iptoint('192.168.1.1'))
Out[65]: '192.168.1.1'

In [66]: inttoip(iptoint('4.1.75.131'))
---------------------------------------------------------------------------
error                                     Traceback (most recent call last)

/home/thc/<ipython console> in <module>()

/home/thc/<ipython console> in inttoip(ip)

error: packed IP wrong length for inet_ntoa`

Anybody knows how to fix that?

2
  • 2
    Doesn't work at all in my Python 2.6.6: inttoip raises a TypeError: Odd-length string. Commented Apr 11, 2011 at 10:15
  • @Ilkka use socket.inet_ntoa(hex(ip)[2:].decode('hex')) for inttoip Commented Jan 21, 2015 at 12:00

13 Answers 13

116
#!/usr/bin/env python
import socket
import struct


def ip2int(addr):
    return struct.unpack("!I", socket.inet_aton(addr))[0]


def int2ip(addr):
    return socket.inet_ntoa(struct.pack("!I", addr))


print(int2ip(0xc0a80164)) # 192.168.1.100
print(ip2int('10.0.0.1')) # 167772161
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2 Comments

This requires import struct.
3 times faster comparing to ipaddress.
101

Python 3 has ipaddress module which features very simple conversion:

int(ipaddress.IPv4Address("192.168.0.1"))
str(ipaddress.IPv4Address(3232235521))

3 Comments

This should be the top answer. Python 2 has this module too, but not built in.
This module is so slow
I've just ran this over 50 ip addresses with no noticable difference to my original script when I wasn't running the additional conversion and back back again.. seems quick enough to me
29

In pure python without use additional module

def IP2Int(ip):
    o = map(int, ip.split('.'))
    res = (16777216 * o[0]) + (65536 * o[1]) + (256 * o[2]) + o[3]
    return res


def Int2IP(ipnum):
    o1 = int(ipnum / 16777216) % 256
    o2 = int(ipnum / 65536) % 256
    o3 = int(ipnum / 256) % 256
    o4 = int(ipnum) % 256
    return '%(o1)s.%(o2)s.%(o3)s.%(o4)s' % locals()

# Example
print('192.168.0.1 -> %s' % IP2Int('192.168.0.1'))
print('3232235521 -> %s' % Int2IP(3232235521))

Result:

192.168.0.1 -> 3232235521
3232235521 -> 192.168.0.1

3 Comments

i realise that this post is fairly old, but you can also use pow:: res = (o[0] * pow(256, 3)) + (o[1] * pow(256, 2)) + (o[2] * 256) + o[3]
Dude, '%(o1)s.%(o2)s.%(o3)s.%(o4)s' % locals() is a veeeery bad style.
@SmitJohnth Thank you for giving the alternative solution ..., here, the new style can be f'{o1}.{o2}.{o3}.{o4}'
12

You lose the left-zero-padding which breaks decoding of your string.

Here's a working function:

def inttoip(ip):
    return socket.inet_ntoa(hex(ip)[2:].zfill(8).decode('hex'))

2 Comments

Yeah, you also fixed the [2:-1] slice to [2:], that was causing the error I commented about above.
If ip is a LONG there will be a 'L' turd at the end of the string. Either replace hex(ip) with hex(int(ip)) or add .rstrip('L') after [2:].
8

Below are the fastest and most straightforward (to the best of my knowledge) convertors for IPv4 and IPv6:

    try:
        _str = socket.inet_pton(socket.AF_INET, val)
    except socket.error:
        raise ValueError
    return struct.unpack('!I', _str)[0]
    -------------------------------------------------
    return socket.inet_ntop(socket.AF_INET, struct.pack('!I', n))
    -------------------------------------------------
    try:
        _str = socket.inet_pton(socket.AF_INET6, val)
    except socket.error:
        raise ValueError
    a, b = struct.unpack('!2Q', _str)
    return (a << 64) | b
    -------------------------------------------------
    a = n >> 64
    b = n & ((1 << 64) - 1)
    return socket.inet_ntop(socket.AF_INET6, struct.pack('!2Q', a, b))

Python code not using inet_ntop() and struct module is like order of magnitude slower than this regardless of what it is doing.

1 Comment

socket.inet_pton and inet_ntop are only available on Unix
6

One line

reduce(lambda out, x: (out << 8) + int(x), '127.0.0.1'.split('.'), 0)

1 Comment

and by far more efficient/faster to left-shift than multiply
6

Python3 oneliner (based on Thomas Webber's Python2 answer):

sum([int(x) << 8*i for i,x in enumerate(reversed(ip.split('.')))])

Left shifts are much faster than pow().

Comments

5

It can be done without using any library.

def iptoint(ip):
        h=list(map(int,ip.split(".")))
        return (h[0]<<24)+(h[1]<<16)+(h[2]<<8)+(h[3]<<0)

def inttoip(ip):
       return ".".join(map(str,[((ip>>24)&0xff),((ip>>16)&0xff),((ip>>8)&0xff),((ip>>0)&0xff)]))

iptoint("8.8.8.8") # 134744072

inttoip(134744072) # 8.8.8.8

1 Comment

correction: def iptoint(ip): h=list(map(int,ip.split("."))) return (h[0]<<24)+(h[1]<<16)+(h[2]<<8)+(h[3]<<0)
3

I used following:

ip2int = lambda ip: reduce(lambda a,b: long(a)*256 + long(b), ip.split('.'))

ip2int('192.168.1.1')

#output

3232235777L

# from int to ip
int2ip = lambda num: '.'.join( [ str((num >> 8*i) % 256)  for i in [3,2,1,0] ])

int2ip(3232235777L)

#output

'192.168.1.1'

Comments

1

Let me give a more understandable way:

ip to int

def str_ip2_int(s_ip='192.168.1.100'):
    lst = [int(item) for item in s_ip.split('.')]
    print lst   
    # [192, 168, 1, 100]

    int_ip = lst[3] | lst[2] << 8 | lst[1] << 16 | lst[0] << 24
    return int_ip   # 3232235876

The above:

lst = [int(item) for item in s_ip.split('.')]

equivalent to :

lst = map(int, s_ip.split('.'))

also:

int_ip = lst[3] | lst[2] << 8 | lst[1] << 16 | lst[0] << 24

equivalent to :

int_ip = lst[3] + (lst[2] << 8) + (lst[1] << 16) + (lst[0] << 24)

int_ip = lst[3] + lst[2] * pow(2, 8) + lst[1] * pow(2, 16) + lst[0] * pow(2, 24)

int to ip:

def int_ip2str(int_ip=3232235876):
    a0 = str(int_ip & 0xff)
    a1 = str((int_ip & 0xff00) >> 8) 
    a2 = str((int_ip & 0xff0000) >> 16)
    a3 = str((int_ip & 0xff000000) >> 24)

    return ".".join([a3, a2, a1, a0])

or:

def int_ip2str(int_ip=3232235876):
    lst = []
    for i in xrange(4):
        shift_n = 8 * i
        lst.insert(0, str((int_ip >> shift_n) & 0xff))

    return ".".join(lst)

Comments

0

My approach is to straightforwardly look at the the number the way it is stored, rather than displayed, and to manipulate it from the display format to the stored format and vice versa.

So, from an IP address to an int:

def convertIpToInt(ip):
    return sum([int(ipField) << 8*index for index, ipField in enumerate(reversed(ip.split('.')))])

This evaluates each field, and shifts it to its correct offset, and then sums them all up, neatly converting the IP address' display into its numerical value.

In the opposite direction, from an int to an IP address:

def convertIntToIp(ipInt):
    return '.'.join([str(int(ipHexField, 16)) for ipHexField in (map(''.join, zip(*[iter(str(hex(ipInt))[2:].zfill(8))]*2)))])

The numerical representation is first converted into its hexadecimal string representation, which can be manipulated as a sequence, making it easier to break up. Then, pairs are extracted by mapping ''.join onto tuples of pairs provided by zipping a list of two references to an iterator of the IP string (see How does zip(*[iter(s)]*n) work?), and those pairs are in turn converted from hex string representations to int string representations, and joined by '.'.

Comments

0
def ip2int(ip):
    """
    Convert IP string to integer
    :param ip: IP string
    :return: IP integer
    """
    return reduce(lambda x, y: x * 256 + y, map(int, ip.split('.')))


def int2ip(num):
    """
    Convert IP integer to string
    :param num: IP integer
    :return: IP string
    """
    return '.'.join(map(lambda x: str(num // 256 ** x % 256), range(3, -1, -1)))

Comments

0

Also modern Python (>= 3.11) has a wonderful int.from_bytes and int.to_bytes methods:

>>> import socket
>>> socket.inet_aton('4.1.75.131')
b'\x04\x01K\x83'
>>> int.from_bytes(socket.inet_aton('4.1.75.131'))
67193731
>>> int.from_bytes(socket.inet_aton('10.0.0.1'))
167772161
>>>

# int2ip
>>> socket.inet_ntoa(167772161 .to_bytes(4))
'10.0.0.1'

Comments

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