comparison mercurial/copies.py @ 46184:cb8b2ee89a5d

copies: stop attempt to avoid extra dict copies around branching

In the python code, we attempt to avoid unnecessary dict copies when gathering
copy information. However that logic is wobbly and I keep running into case
where independent branches affects each others.

With the current code we can't ensure we are the only "user" of dict when
dealing with merge.

This caused havoc in the next series on tests I am about to introduce.

So for now I am disabling the faulty optimisation. I believe we will need a
dedicated overlay to deal with the "copy on write logic" to have something
correct. I am also hoping to find time to build dedicated test case for this
category of problem instead of relying on side effect in other tests. However
for now I am focussing on another issue.

Differential Revision: https://phab.mercurial-scm.org/D9608

46183:ee63c1173c1b

                else:
                    copies = all_copies.pop(parent_rev, None)

                if copies is None:
                    # this is a root
                    copies = {}

                newcopies = copies
                # chain the data in the edge with the existing data
                if changes is not None:
                    childcopies = {}
                    if parent == 1:
                        childcopies = changes.copied_from_p1

                            if prev is not None and prev[1] is not None:
                                source = prev[1]
                            newcopies[dest] = (current_rev, source)
                        assert newcopies is not copies
                    if changes.removed:
                        if newcopies is copies:
                            newcopies = copies.copy()
                        for f in changes.removed:
                            if f in newcopies:
                                if newcopies is copies:
                                    # copy on write to avoid affecting potential other
                                    # branches.  when there are no other branches, this

                                newcopies[f] = (current_rev, None)
                # check potential need to combine the data from another parent (for
                # that child). See comment below for details.
                if current_copies is None:
                    current_copies = newcopies
                elif current_copies is newcopies:
                    # nothing to merge:
                    pass
                else:
                    # we are the second parent to work on c, we need to merge our
                    # work with the other.
                    #
                    # In case of conflict, parent 1 take precedence over parent 2.

46184:cb8b2ee89a5d

                else:
                    copies = all_copies.pop(parent_rev, None)

                if copies is None:
                    # this is a root
                    newcopies = copies = {}
                elif remaining_children:
                    newcopies = copies.copy()
                else:
                    newcopies = copies
                # chain the data in the edge with the existing data
                if changes is not None:
                    childcopies = {}
                    if parent == 1:
                        childcopies = changes.copied_from_p1

                            if prev is not None and prev[1] is not None:
                                source = prev[1]
                            newcopies[dest] = (current_rev, source)
                        assert newcopies is not copies
                    if changes.removed:
                        for f in changes.removed:
                            if f in newcopies:
                                if newcopies is copies:
                                    # copy on write to avoid affecting potential other
                                    # branches.  when there are no other branches, this

                                newcopies[f] = (current_rev, None)
                # check potential need to combine the data from another parent (for
                # that child). See comment below for details.
                if current_copies is None:
                    current_copies = newcopies
                else:
                    # we are the second parent to work on c, we need to merge our
                    # work with the other.
                    #
                    # In case of conflict, parent 1 take precedence over parent 2.