view mercurial/pure/base85.py @ 23717:d8663e6153c1

namespaces: use namespace object instead of dictionary

This isn't as bad as the diff seems, it only looks like scary. In this patch,
we use the 'namespace' object instead of accessing keys in a dictionary.

This required the 'templatename' and 'names' method to change their
implementation. Later, we will remove these functions entirely due to a better
api.

# 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 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