checkout: do not lose staged removal

The logic to checkout a different commit implements the safety to never
lose user's local changes.  For example, switching from a commit to
another commit, when you have changed a path that is different between
them, need to merge your changes to the version from the switched-to
commit, which you may not necessarily be able to resolve easily.  By
default, "git checkout" refused to switch branches, to give you a chance
to stash your local changes (or use "-m" to merge, accepting the risks of
getting conflicts).

This safety, however, had one deliberate hole since early June 2005.  When
your local change was to remove a path (and optionally to stage that
removal), the command checked out the path from the switched-to commit
nevertheless.

This was to allow an initial checkout to happen smoothly (e.g. an initial
checkout is done by starting with an empty index and switching from the
commit at the HEAD to the same commit).  We can tighten the rule slightly
to allow this special case to pass, without losing sight of removal
explicitly done by the user, by noticing if the index is truly empty when
the operation begins.

For historical background, see:

    http://thread.gmane.org/gmane.comp.version-control.git/4641/focus=4646

This case is marked as *0* in the message, which both Linus and I said "it
feels somewhat wrong but otherwise we cannot start from an empty index".

Signed-off-by: Junio C Hamano <gitster@pobox.com>

1 files changed, 1 insertions, 0 deletions

diff --git a/builtin-read-tree.c b/builtin-read-tree.c
index dddc3044b8..362216b272 100644
--- a/builtin-read-tree.c
+++ b/builtin-read-tree.c
@@ -206,6 +206,7 @@ int cmd_read_tree(int argc, const char **argv, const char *unused_prefix)
 			break;
 		case 2:
 			opts.fn = twoway_merge;
+			opts.initial_checkout = !active_nr;
 			break;
 		case 3:
 		default: