##// END OF EJS Templates
transaction: raise on backup restoration error...
transaction: raise on backup restoration error A few line above, similar errors in the truncation code result in raising the associated exception. We should do the same here. This means the transaction recover is more strict now, which might be a problem when running `hg recover` in a share different from the one where the transaction fails. However this has always been a problem and need to be be addressed independently.

File last commit:

r49730:6000f5b2 default
r51234:cab3defe stable
Show More
base85.py
87 lines | 2.0 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
from .. import pycompat
_b85chars = pycompat.bytestr(
b"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdef"
b"ghijklmnopqrstuvwxyz!#$%&()*+-;<=>?@^_`{|}~"
)
_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 += b'\0' * (4 - r)
longs = len(text) >> 2
words = struct.unpack(b'>%dL' % longs, text)
out = b''.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]
chunk = pycompat.bytestr(chunk)
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(b'>%dL' % (len(out)), *out)
if cl:
out = out[: -(5 - cl)]
return out