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branchcache: fetch source branchcache during clone (issue3378)...
branchcache: fetch source branchcache during clone (issue3378) Recomputing branch cache on clone may be expensive, therefore if possible we fetch it along with the data. - If the clone is performed by copying, we just copy branchcache file. - If we localrepo.clone and streaming then we follow the procedure: 1. Fetch branchmap from the remote 2. Fetch the actual data. 3. Find the latest rev within branch heads (tip at the time of branchmap fetch) 4. Update the cache for the revs in [remotetip+1, tip] This way we ensure that the branchcache is correct even in case of races with commits.

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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