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util.h: backout 06badf7d10dc and 2c9645c0a582 for big-endian breakage...
util.h: backout 06badf7d10dc and 2c9645c0a582 for big-endian breakage getbe32 and putbe32 need to behave differently on big-endian and little-endian systems. On big-endian ones, they should be roughly equivalent to the identity function with a cast, but on little-endian ones they should reverse the order of the bytes. That is achieved by the original definition, but __builtin_bswap32 and _byteswap_ulong, as the names suggest, swap bytes around unconditionally. There was no measurable performance improvement, so there's no point adding extra complexity with even more ifdefs for endianncess.

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base85.py
75 lines | 1.9 KiB | text/x-python | PythonLexer
# base85.py: pure python base85 codec
#
# Copyright (C) 2009 Brendan Cully <brendan@kublai.com>
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.
import struct
_b85chars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ" \
"abcdefghijklmnopqrstuvwxyz!#$%&()*+-;<=>?@^_`{|}~"
_b85chars2 = [(a + b) for a in _b85chars for b in _b85chars]
_b85dec = {}
def _mkb85dec():
for i, c in enumerate(_b85chars):
_b85dec[c] = i
def b85encode(text, pad=False):
"""encode text in base85 format"""
l = len(text)
r = l % 4
if r:
text += '\0' * (4 - r)
longs = len(text) >> 2
words = struct.unpack('>%dL' % (longs), text)
out = ''.join(_b85chars[(word // 52200625) % 85] +
_b85chars2[(word // 7225) % 7225] +
_b85chars2[word % 7225]
for word in words)
if pad:
return out
# Trim padding
olen = l % 4
if olen:
olen += 1
olen += l // 4 * 5
return out[:olen]
def b85decode(text):
"""decode base85-encoded text"""
if not _b85dec:
_mkb85dec()
l = len(text)
out = []
for i in range(0, len(text), 5):
chunk = text[i:i + 5]
acc = 0
for j, c in enumerate(chunk):
try:
acc = acc * 85 + _b85dec[c]
except KeyError:
raise ValueError('bad base85 character at position %d'
% (i + j))
if acc > 4294967295:
raise ValueError('Base85 overflow in hunk starting at byte %d' % i)
out.append(acc)
# Pad final chunk if necessary
cl = l % 5
if cl:
acc *= 85 ** (5 - cl)
if cl > 1:
acc += 0xffffff >> (cl - 2) * 8
out[-1] = acc
out = struct.pack('>%dL' % (len(out)), *out)
if cl:
out = out[:-(5 - cl)]
return out