IP address conversion using Python.

Question

I want to convert an ip address read from a file in the decimal format (192.168.65.72) to one in binary format {110000001010100001000001010001000}.I am reading the ip address in the decimal format from a file.Find the code snippet below.

/*contains 192.168.65.72*/
filter = open("filter.txt", "r")

for line in filter:
    bytePattern = "([01]?\d\d?|2[0-4]\d|25[0-5])"
    regObj = re.compile("\.".join([bytePattern]*4))
    for match in regObj.finditer(line):
        m1,m2,m3,m4 = match.groups()
        print "%s %s %s %s" %(m1, m2, m3, m4)

I want to convert m1,m2,m3 and m4 each into an 8bit binary value.I cannot figure out a way to do that.I am new to python.Any help will be greatly appreciated.

Cheers.

Answer 1

>>> bin( 192 )
'0b11000000'

String manipulation does the rest. Note however that in an IP all but the first parts are allowed to be zero.

Answer 2

''.join([ bin(int(x))[2:].rjust(8,'0') for x in '123.123.123.123'.split('.')])

Answer 3

If you want a string containing the raw data for the 32 bit IP address:

import struct
struct.pack('4B', *(int(x) for x in '123.123.123.123'.split('.'))

Or if you want a 32-bit integer containing the IP address:

import struct
struct.unpack('>I', struct.pack('4B', *(int(x) for x in '123.123.123.123'.split('.')))

Answer 4

Convert an IPv4 address from dotted-quad string format (for example, ‘123.45.67.89’) to 32-bit packed binary format:

socket.inet_aton("192.168.65.72")