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convert: on svn failure, note libsvn version (issue4043)...
convert: on svn failure, note libsvn version (issue4043) We have our own fast-path logic to see if something passes a sniff test for being a Subversion repository, but it's possible for a user to svnsync a repo using svn 1.8 and then use svn 1.7 bindings (as in the bug) to try and convert the repo. If we at least tell the user the version of libsvn that we used, they might get enough of a hint to check on their own for format incompatibilities between their svn{admin,sync} and the libsvn used by hg.

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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