Byte-encoding IP address in Python 3 and converting back to string

Question

I'm trying to create sample UDP packet headers in Python3 as such:

# "encoding"
header = bytearray()
ip = '192.168.1.1'
ip_bytes = bytes(map(int, ip.split('.')))
header.extend(ip_bytes)
port = 5555    
header.extend(port.to_bytes(2, 'big'))
print(header)
print()

# "decoding"
destip = header[:4]
ips = ""
for i in destip:
    byt = int.from_bytes(destip[i:i+1], 'big')
    ips += str(byt) + "."
ips = ips[:len(ips)-1]
print(ips)

And the output is:

bytearray(b'\xc0\xa8\x01\x01\x15\xb3')

bytearray(b'\xc0\xa8\x01\x01')
0.0.168.168

What I want is for the second line to be:

192.168.1.1

Anyone know where I'm going wrong?

Possible duplicate of Converting IP address into bytes in python — ayrusme
– ayrusme, Commented Oct 19, 2017 at 5:51
@thatrockbottomprogrammer What I'm struggling with is converting the ip BACK to a string — sgrew
– sgrew, Commented Oct 19, 2017 at 6:15

indeed005 · Accepted Answer · 2017-10-19 06:01:35Z

1

Don't convert the ip_arg string to an int, 19216811 is not the int you want to encode. 192.168.1.1 = 3232235777 as an int. You could do the reverse of what you're doing in the decode section and convert each octet.

answered Oct 19, 2017 at 6:01

indeed005

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1 Comment

sgrew Over a year ago

SOLVED. The byte to int conversion in the decoding for loop should be: int.from_bytes([i], 'big')

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