view mercurial/pure/base85.py @ 20555:4add43865a9b

ancestors: remove unnecessary handling of 'left' If one of the initial nodes also is an ancestor then that most be the only ancestor. There is no need for additional bookkeeping.
author Mads Kiilerich <madski@unity3d.com>
date Mon, 24 Feb 2014 22:42:13 +0100
parents 20a9d823f242
children 9007f697e8ef
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# 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