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Converting IPv4 Address to a Hex IPv6 Address in Python

Q: Write a program that prompts the user for an IP address then converts this to a base 10, binary and hex value. The program then converts the hex value to an RFC3056 IPv6 6to4 address.

I have the base 10 and binary parts working but I can't seem to get my head around the hex part. Can the format string method be used somehow to accomplish the same thing? Or would I need to import the ipaddress module in this case?

#!/usr/bin/python3

ip_address = input("Please enter a dot decimal IP Address: ")

"""This part converts to base 10"""
ListA = ip_address.split(".")
ListA = list(map(int, ListA))
ListA = ListA[0]*(256**3) + ListA[1]*(256**2) + ListA[2]*(256**1) + ListA[3]
print("The IP Address in base 10 is: " , ListA)

"""This part converts to base 2"""
base2 = [format(int(x), '08b') for x in ip_address.split('.')]
print("The IP Address in base 2 is: ", base2)

"""This part converts to hex"""
hexIP = []
[hexIP.append(hex(int(x))[2:].zfill(2)) for x in ip_address.split('.')]
hexIP = "".join(hexIP)
print("The IP Address in hex is: " , hexIP)

EDIT: Managed to convert the IP Address to hex value. Now how do I convert this hex value into IPv6 address?

like image 678
user1819786 Avatar asked Oct 21 '25 04:10

user1819786


1 Answers

>>> ip_address = '123.45.67.89'
>>> numbers = list(map(int, ip_address.split('.')))
>>> '2002:{:02x}{:02x}:{:02x}{:02x}::'.format(*numbers)
'2002:7b2d:4359::'
like image 106
falsetru Avatar answered Oct 23 '25 19:10

falsetru



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