# bdiff.py - Python implementation of bdiff.c

#

# Copyright 2009 Matt Mackall <mpm@selenic.com> 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.

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

    r = []

    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

        r.append((a1, b1, l1 + shift))

        prev = a2 + shift, b2 + shift, l2 - shift

    r.append(prev)

    return r

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]