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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
84 lines | 2.3 KiB | text/x-python | PythonLexer
# bdiff.py - CFFI implementation of bdiff.c
#
# Copyright 2016 Maciej Fijalkowski <fijall@gmail.com>
#
# 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 struct
from ..pure.bdiff import *
from . import _bdiff
ffi = _bdiff.ffi
lib = _bdiff.lib
def blocks(sa, sb):
a = ffi.new(b"struct bdiff_line**")
b = ffi.new(b"struct bdiff_line**")
ac = ffi.new(b"char[]", str(sa))
bc = ffi.new(b"char[]", str(sb))
l = ffi.new(b"struct bdiff_hunk*")
try:
an = lib.bdiff_splitlines(ac, len(sa), a)
bn = lib.bdiff_splitlines(bc, len(sb), b)
if not a[0] or not b[0]:
raise MemoryError
count = lib.bdiff_diff(a[0], an, b[0], bn, l)
if count < 0:
raise MemoryError
rl = [None] * count
h = l.next
i = 0
while h:
rl[i] = (h.a1, h.a2, h.b1, h.b2)
h = h.next
i += 1
finally:
lib.free(a[0])
lib.free(b[0])
lib.bdiff_freehunks(l.next)
return rl
def bdiff(sa, sb):
a = ffi.new(b"struct bdiff_line**")
b = ffi.new(b"struct bdiff_line**")
ac = ffi.new(b"char[]", str(sa))
bc = ffi.new(b"char[]", str(sb))
l = ffi.new(b"struct bdiff_hunk*")
try:
an = lib.bdiff_splitlines(ac, len(sa), a)
bn = lib.bdiff_splitlines(bc, len(sb), b)
if not a[0] or not b[0]:
raise MemoryError
count = lib.bdiff_diff(a[0], an, b[0], bn, l)
if count < 0:
raise MemoryError
rl = []
h = l.next
la = lb = 0
while h:
if h.a1 != la or h.b1 != lb:
lgt = (b[0] + h.b1).l - (b[0] + lb).l
rl.append(
struct.pack(
b">lll",
(a[0] + la).l - a[0].l,
(a[0] + h.a1).l - a[0].l,
lgt,
)
)
rl.append(str(ffi.buffer((b[0] + lb).l, lgt)))
la = h.a2
lb = h.b2
h = h.next
finally:
lib.free(a[0])
lib.free(b[0])
lib.bdiff_freehunks(l.next)
return b"".join(rl)