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py3: decode username and password before SMTP login...
py3: decode username and password before SMTP login smtplib.SMTP.login() requires str on Python 3. For 'password', we only need to decode when value comes from config as getpass() returns the correct type already.

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bdiff.py
99 lines | 2.4 KiB | text/x-python | PythonLexer
# bdiff.py - Python implementation of bdiff.c
#
# Copyright 2009 Matt Mackall <mpm@selenic.com> and others
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.
from __future__ import absolute_import
import difflib
import re
import struct
def splitnewlines(text):
'''like str.splitlines, but only split on newlines.'''
lines = [l + b'\n' for l in text.split(b'\n')]
if lines:
if lines[-1] == b'\n':
lines.pop()
else:
lines[-1] = lines[-1][:-1]
return lines
def _normalizeblocks(a, b, blocks):
prev = None
r = []
for curr in blocks:
if prev is None:
prev = curr
continue
shift = 0
a1, b1, l1 = prev
a1end = a1 + l1
b1end = b1 + l1
a2, b2, l2 = curr
a2end = a2 + l2
b2end = b2 + l2
if a1end == a2:
while (
a1end + shift < a2end and a[a1end + shift] == b[b1end + shift]
):
shift += 1
elif b1end == b2:
while (
b1end + shift < b2end and a[a1end + shift] == b[b1end + shift]
):
shift += 1
r.append((a1, b1, l1 + shift))
prev = a2 + shift, b2 + shift, l2 - shift
r.append(prev)
return r
def bdiff(a, b):
a = bytes(a).splitlines(True)
b = bytes(b).splitlines(True)
if not a:
s = b"".join(b)
return s and (struct.pack(b">lll", 0, 0, len(s)) + s)
bin = []
p = [0]
for i in a:
p.append(p[-1] + len(i))
d = difflib.SequenceMatcher(None, a, b).get_matching_blocks()
d = _normalizeblocks(a, b, d)
la = 0
lb = 0
for am, bm, size in d:
s = b"".join(b[lb:bm])
if am > la or s:
bin.append(struct.pack(b">lll", p[la], p[am], len(s)) + s)
la = am + size
lb = bm + size
return b"".join(bin)
def blocks(a, b):
an = splitnewlines(a)
bn = splitnewlines(b)
d = difflib.SequenceMatcher(None, an, bn).get_matching_blocks()
d = _normalizeblocks(an, bn, d)
return [(i, i + n, j, j + n) for (i, j, n) in d]
def fixws(text, allws):
if allws:
text = re.sub(b'[ \t\r]+', b'', text)
else:
text = re.sub(b'[ \t\r]+', b' ', text)
text = text.replace(b' \n', b'\n')
return text