tests/test-bdiff.py

bdiff: give slight preference to longest matches in the middle of the B side

We already have a slight preference for matches close to the middle on the A
side. Now, do the same on the B side.

j is iterating the b range backwards and we thus accept a new j if the previous
match was in the upper half.

This makes the test-bhalf diff "correct". It obviously also gives more
preference to balanced recursion than to appending to sequences. That is kind
of correct, but will also unfortunately make some bundles bigger. No doubt, we
can also create examples where it will make them smaller ...

The bundle size for 4.0 (hg bundle --base null -r 4.0 x.hg) happens to go from
22803824 to 22806817 bytes - an 0.01% increase.

from __future__ import absolute_import, print_function
import struct
from mercurial import (
    bdiff,
    mpatch,
)

def test1(a, b):
    d = bdiff.bdiff(a, b)
    c = a
    if d:
        c = mpatch.patches(a, [d])
    if c != b:
        print("bad diff+patch result from\n  %r to\n  %r:" % (a, b))
        print("bdiff: %r" % d)
        print("patched: %r" % c[:200])

def test(a, b):
    print("test", repr(a), repr(b))
    test1(a, b)
    test1(b, a)

test("a\nc\n\n\n\n", "a\nb\n\n\n")
test("a\nb\nc\n", "a\nc\n")
test("", "")
test("a\nb\nc", "a\nb\nc")
test("a\nb\nc\nd\n", "a\nd\n")
test("a\nb\nc\nd\n", "a\nc\ne\n")
test("a\nb\nc\n", "a\nc\n")
test("a\n", "c\na\nb\n")
test("a\n", "")
test("a\n", "b\nc\n")
test("a\n", "c\na\n")
test("", "adjfkjdjksdhfksj")
test("", "ab")
test("", "abc")
test("a", "a")
test("ab", "ab")
test("abc", "abc")
test("a\n", "a\n")
test("a\nb", "a\nb")

#issue1295
def showdiff(a, b):
    print('showdiff(\n  %r,\n  %r):' % (a, b))
    bin = bdiff.bdiff(a, b)
    pos = 0
    q = 0
    while pos < len(bin):
        p1, p2, l = struct.unpack(">lll", bin[pos:pos + 12])
        pos += 12
        if p1:
            print('', repr(a[q:p1]))
        print('', p1, p2, repr(a[p1:p2]), '->', repr(bin[pos:pos + l]))
        pos += l
        q = p2
    if q < len(a):
        print('', repr(a[q:]))

showdiff("x\n\nx\n\nx\n\nx\n\nz\n", "x\n\nx\n\ny\n\nx\n\nx\n\nz\n")
showdiff("x\n\nx\n\nx\n\nx\n\nz\n", "x\n\nx\n\ny\n\nx\n\ny\n\nx\n\nz\n")
# we should pick up abbbc. rather than bc.de as the longest match
showdiff("a\nb\nb\nb\nc\n.\nd\ne\n.\nf\n",
         "a\nb\nb\na\nb\nb\nb\nc\n.\nb\nc\n.\nd\ne\nf\n")

print("done")

def testfixws(a, b, allws):
    c = bdiff.fixws(a, allws)
    if c != b:
        print("*** fixws", repr(a), repr(b), allws)
        print("got:")
        print(repr(c))

testfixws(" \ta\r b\t\n", "ab\n", 1)
testfixws(" \ta\r b\t\n", " a b\n", 0)
testfixws("", "", 1)
testfixws("", "", 0)

print("done")

print("Nice diff for a trivial change:")
showdiff(
    ''.join('<%s\n-\n' % i for i in range(5)),
    ''.join('>%s\n-\n' % i for i in range(5)))

print("Diff 1 to 3 lines - preference for balanced recursion:")
showdiff('a\n', 'a\n' * 3)
print("Diff 1 to 5 lines - preference for balanced recursion:")
showdiff('a\n', 'a\n' * 5)
print("Diff 3 to 1 lines - preference for balanced recursion:")
showdiff('a\n' * 3, 'a\n')
print("Diff 5 to 1 lines - preference for balanced recursion:")
showdiff('a\n' * 5, 'a\n')