##// END OF EJS Templates
discovery: use a lower level but faster way to retrieve parents...
discovery: use a lower level but faster way to retrieve parents We already know that no revision in the undecided set are filtered, so we can skip multiple checks and directly access lower level data. In a private pathological case, this improves the timing from about 70 seconds to about 50 seconds. There are other actions to be taken to improve that case, however this gives an idea of the general overhead.

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base85.py
80 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.
from __future__ import absolute_import
import struct
from .. import pycompat
_b85chars = pycompat.bytestr("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdef"
"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 += '\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]
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('>%dL' % (len(out)), *out)
if cl:
out = out[:-(5 - cl)]
return out