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# bdiff.py - Python implementation of bdiff.c
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#
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# Copyright 2009 Olivia Mackall <olivia@selenic.com> and others
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#
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# This software may be used and distributed according to the terms of the
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# GNU General Public License version 2 or any later version.
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import difflib
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import re
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import struct
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from typing import (
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List,
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Tuple,
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)
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def splitnewlines(text: bytes) -> List[bytes]:
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'''like str.splitlines, but only split on newlines.'''
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lines = [l + b'\n' for l in text.split(b'\n')]
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if lines:
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if lines[-1] == b'\n':
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lines.pop()
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else:
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lines[-1] = lines[-1][:-1]
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return lines
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def _normalizeblocks(
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a: List[bytes], b: List[bytes], blocks
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) -> List[Tuple[int, int, int]]:
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prev = None
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r = []
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for curr in blocks:
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if prev is None:
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prev = curr
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continue
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shift = 0
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a1, b1, l1 = prev
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a1end = a1 + l1
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b1end = b1 + l1
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a2, b2, l2 = curr
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a2end = a2 + l2
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b2end = b2 + l2
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if a1end == a2:
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while (
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a1end + shift < a2end and a[a1end + shift] == b[b1end + shift]
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):
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shift += 1
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elif b1end == b2:
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while (
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b1end + shift < b2end and a[a1end + shift] == b[b1end + shift]
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):
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shift += 1
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r.append((a1, b1, l1 + shift))
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prev = a2 + shift, b2 + shift, l2 - shift
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if prev is not None:
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r.append(prev)
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return r
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def bdiff(a: bytes, b: bytes) -> bytes:
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a = bytes(a).splitlines(True)
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b = bytes(b).splitlines(True)
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if not a:
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s = b"".join(b)
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return s and (struct.pack(b">lll", 0, 0, len(s)) + s)
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bin = []
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p = [0]
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for i in a:
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p.append(p[-1] + len(i))
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d = difflib.SequenceMatcher(None, a, b).get_matching_blocks()
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d = _normalizeblocks(a, b, d)
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la = 0
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lb = 0
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for am, bm, size in d:
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s = b"".join(b[lb:bm])
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if am > la or s:
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bin.append(struct.pack(b">lll", p[la], p[am], len(s)) + s)
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la = am + size
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lb = bm + size
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return b"".join(bin)
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def blocks(a: bytes, b: bytes) -> List[Tuple[int, int, int, int]]:
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an = splitnewlines(a)
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bn = splitnewlines(b)
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d = difflib.SequenceMatcher(None, an, bn).get_matching_blocks()
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d = _normalizeblocks(an, bn, d)
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return [(i, i + n, j, j + n) for (i, j, n) in d]
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def fixws(text: bytes, allws: bool) -> bytes:
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if allws:
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text = re.sub(b'[ \t\r]+', b'', text)
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else:
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text = re.sub(b'[ \t\r]+', b' ', text)
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text = text.replace(b' \n', b'\n')
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return text
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