tests/test-bdiff

bdiff: normalize the diff (issue1295)

When the common part of a diff can be moved forward, move it forward.
Otherwise we don't get deterministic results (it would depends on the way we
split for the recursion).
That way we get identical hunks when doing the same change, it helps to solve
issue1295 (inconsistent diffs on different side during a merge).

#!/usr/bin/env python

import sys, 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 "***", `a`, `b`
        print "bad:"
        print `c`[:200]
        print `d`

def test(a, b):
    print "***", `a`, `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):
    bin = bdiff.bdiff(a, b)
    pos = 0
    while pos < len(bin):
        p1, p2, l = struct.unpack(">lll", bin[pos:pos + 12])
        pos += 12
        print p1, p2, repr(bin[pos:pos + l])
        pos += l
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")

print "done"