commit | 8640dcfecb8596d172114e2ecba01f4887346c13 | [log] [tgz] |
---|---|---|
author | Xin Li <[email protected]> | Wed Jun 02 16:50:31 2021 +0000 |
committer | Xin Li <[email protected]> | Wed Jun 02 16:50:31 2021 +0000 |
tree | 994f7da7a7f4f482fcf3528bbe7ba101c62806b8 | |
parent | 77feaf200415f8a7459dcd31ac3aa6e897d58d2f [diff] | |
parent | 071e185773740c234ee0a5db98bf0cd07f7abaf1 [diff] |
Merge sc-mainline-prod Bug: 189946434 Change-Id: I3384a47b9bcc55f512ea398f42dc5d7aba5e3d1d
Python 3.3+'s ipaddress for Python 2.6, 2.7, 3.2.
This repository tracks the latest version from cpython, e.g. ipaddress from cpython 3.8 as of writing.
Note that just like in Python 3.3+ you must use character strings and not byte strings for textual IP address representations:
>>> from __future__ import unicode_literals >>> ipaddress.ip_address('1.2.3.4') IPv4Address(u'1.2.3.4')
or
>>> ipaddress.ip_address(u'1.2.3.4') IPv4Address(u'1.2.3.4')
but not:
>>> ipaddress.ip_address(b'1.2.3.4') Traceback (most recent call last): File "<stdin>", line 1, in <module> File "ipaddress.py", line 163, in ip_address ' a unicode object?' % address) ipaddress.AddressValueError: '1.2.3.4' does not appear to be an IPv4 or IPv6 address. Did you pass in a bytes (str in Python 2) instead of a unicode object?