view mercurial/dicthelpers.py @ 22842:d43d116a118c stable

shelve: don't delete "." when rebase is a no-op (issue4398)

When unshelving and facing a conflict, if we resolve all conflicts in
favour of the committed changes instead of the shelved changes, then
the ensuing implicit rebase is a no-op. That is, there is nothing to
rebase. In this case, there are no extra intermediate shelve commits
to strip either. Prior to this change, the commit being unshelved to
would be marked for destruction in a rather catastrophic way.

The relevant part of the test case failed as follows:

    $ hg unshelve -c
    unshelve of 'default' complete
    $ hg diff
    warning: ignoring unknown working parent 33f7f61e6c5e!
    diff --git a/a/a b/a/a
    new file mode 100644
    --- /dev/null
           b/a/a
    @@ -0,0   1,3 @@
      a
      c
      x
    $ hg status
    warning: ignoring unknown working parent 33f7f61e6c5e!
    M a/a
    ? a/a.orig
    ? foo/foo
    $ hg summary
    warning: ignoring unknown working parent 33f7f61e6c5e!
    parent: -1:000000000000  (no revision checked out)
    branch: default
    commit: 1 modified, 2 unknown (new branch head)
    update: 4 new changesets (update)

With this change, this test case now passes.

# dicthelpers.py - helper routines for Python dicts
#
# Copyright 2013 Facebook
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.

def diff(d1, d2, default=None):
    '''Return all key-value pairs that are different between d1 and d2.

    This includes keys that are present in one dict but not the other, and
    keys whose values are different. The return value is a dict with values
    being pairs of values from d1 and d2 respectively, and missing values
    treated as default, so if a value is missing from one dict and the same as
    default in the other, it will not be returned.'''
    res = {}
    if d1 is d2:
        # same dict, so diff is empty
        return res

    for k1, v1 in d1.iteritems():
        v2 = d2.get(k1, default)
        if v1 != v2:
            res[k1] = (v1, v2)

    for k2 in d2:
        if k2 not in d1:
            v2 = d2[k2]
            if v2 != default:
                res[k2] = (default, v2)

    return res

def join(d1, d2, default=None):
    '''Return all key-value pairs from both d1 and d2.

    This is akin to an outer join in relational algebra. The return value is a
    dict with values being pairs of values from d1 and d2 respectively, and
    missing values represented as default.'''
    res = {}

    for k1, v1 in d1.iteritems():
        if k1 in d2:
            res[k1] = (v1, d2[k1])
        else:
            res[k1] = (v1, default)

    if d1 is d2:
        return res

    for k2 in d2:
        if k2 not in d1:
            res[k2] = (default, d2[k2])

    return res