Say i have an IP address in python
addr = '164.107.113.18'
How do i convert the IP address into 4 bytes?
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1What is your expected result? You have an extra period in your IP.Andy– Andy ♦2015-10-20 19:02:42 +00:00Commented Oct 20, 2015 at 19:02
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That was a syntax error on my part.user3261349– user32613492015-10-20 19:11:58 +00:00Commented Oct 20, 2015 at 19:11
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2 Answers
Use socket.inet_aton:
>>> import socket
>>> socket.inet_aton('164.107.113.18')
'\xa4kq\x12'
>>> socket.inet_aton('127.0.0.1')
'\x7f\x00\x00\x01'
This returns a byte-string (or bytes object on Python 3.x) that you can get bytes from. Alternatively, you can use struct to get each byte's integer value:
>>> import socket
>>> import struct
>>> struct.unpack('BBBB', socket.inet_aton('164.107.113.18'))
(164, 107, 113, 18)
2 Comments
user3261349
So does socket.inet_aton(addr) return a string of 4 characters?
Maciej Gol
For Python 2.x a
str object is just a byte string, thus these 4 characters represent 4 bytes. For 3.x it returns a bytes object.Go with Maciej Gol's answer, but here's another way:
ip = '192.168.1.1'
ip_as_bytes = bytes(map(int, ip.split('.')))
EDIT: Oops, this is Python 3.X only. For Python 2.X
ip = '192.168.1.1'
ip_as_bytes = ''.join(map(chr,map(int,ip.split('.'))))
You're better off using the socket module, however, given its efficiency:
>>> timeit.timeit("socket.inet_aton('164.107.113.18')",setup='import socket')
0.22455310821533203
>>> timeit.timeit("''.join(map(chr,map(int,'164.107.113.18'.split('.'))))")
3.8679449558258057
