mercurial/pure/base85.py

i18n: use "encoding.lower()" to normalize string in hgweb search query

some problematic encoding (e.g.: cp932) uses ASCII alphabet characters
in byte sequence of multi byte characters.

"str.lower()" on such byte sequence may treat distinct characters as
same one, and cause unexpected log matching.

this patch uses "encoding.lower()" instead of "str.lower()" to
normalize strings for compare.

# base85.py: pure python base85 codec
#
# Copyright (C) 2009 Brendan Cully <[e-mail]>
#
# 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 TypeError('Bad base85 character at byte %d' % (i + j))
        if acc > 4294967295:
            raise OverflowError('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