shallow: fix leak when unregistering last shallow root

When unregistering a shallow root we shrink the array of grafts by one
and move remaining grafts one to the left. This can of course only
happen when there are any grafts left, because otherwise there is
nothing to move. As such, this code is guarded by a condition that only
performs the move in case there are grafts after the position of the
graft to be unregistered.

By mistake we also put the call to free the unregistered graft into that
condition. But that doesn't make any sense, as we want to always free
the graft when it exists. Fix the resulting memory leak by doing so.

This leak is exposed by t5500, but plugging it does not make the whole
test suite pass.

Signed-off-by: Patrick Steinhardt <[email protected]>
Signed-off-by: Jeff King <[email protected]>
Signed-off-by: Junio C Hamano <[email protected]>

Author: pks-t

shallow.c

@@ -51,12 +51,11 @@ int unregister_shallow(const struct object_id *oid)
 	int pos = commit_graft_pos(the_repository, oid);
 	if (pos < 0)
 		return -1;
-	if (pos + 1 < the_repository->parsed_objects->grafts_nr) {
-		free(the_repository->parsed_objects->grafts[pos]);
+	free(the_repository->parsed_objects->grafts[pos]);
+	if (pos + 1 < the_repository->parsed_objects->grafts_nr)
 		MOVE_ARRAY(the_repository->parsed_objects->grafts + pos,
 			   the_repository->parsed_objects->grafts + pos + 1,
 			   the_repository->parsed_objects->grafts_nr - pos - 1);
-	}
 	the_repository->parsed_objects->grafts_nr--;
 	return 0;
 }