mercurial/pure/bdiff.py

# bdiff.py - Python implementation of bdiff.c
#
# Copyright 2009 Matt Mackall <[e-mail]> and others
#
# 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, difflib

def splitnewlines(text):
    '''like str.splitlines, but only split on newlines.'''
    lines = [l + '\n' for l in text.split('\n')]
    if lines:
        if lines[-1] == '\n':
            lines.pop()
        else:
            lines[-1] = lines[-1][:-1]
    return lines

def _normalizeblocks(a, b, blocks):
    prev = None
    for curr in blocks:
        if prev is None:
            prev = curr
            continue
        shift = 0

        a1, b1, l1 = prev
        a1end = a1 + l1
        b1end = b1 + l1

        a2, b2, l2 = curr
        a2end = a2 + l2
        b2end = b2 + l2
        if a1end == a2:
            while (a1end + shift < a2end and
                   a[a1end + shift] == b[b1end + shift]):
                shift += 1
        elif b1end == b2:
            while (b1end + shift < b2end and
                   a[a1end + shift] == b[b1end + shift]):
                shift += 1
        yield a1, b1, l1 + shift
        prev = a2 + shift, b2 + shift, l2 - shift
    yield prev

def bdiff(a, b):
    a = str(a).splitlines(True)
    b = str(b).splitlines(True)

    if not a:
        s = "".join(b)
        return s and (struct.pack(">lll", 0, 0, len(s)) + s)

    bin = []
    p = [0]
    for i in a: p.append(p[-1] + len(i))

    d = difflib.SequenceMatcher(None, a, b).get_matching_blocks()
    d = _normalizeblocks(a, b, d)
    la = 0
    lb = 0
    for am, bm, size in d:
        s = "".join(b[lb:bm])
        if am > la or s:
            bin.append(struct.pack(">lll", p[la], p[am], len(s)) + s)
        la = am + size
        lb = bm + size

    return "".join(bin)

def blocks(a, b):
    an = splitnewlines(a)
    bn = splitnewlines(b)
    d = difflib.SequenceMatcher(None, an, bn).get_matching_blocks()
    d = _normalizeblocks(an, bn, d)
    return [(i, i + n, j, j + n) for (i, j, n) in d]