dirstate: remove layering violation around writing dirstate out
authorFUJIWARA Katsunori <foozy@lares.dti.ne.jp>
Sat, 17 Oct 2015 01:15:34 +0900
changeset 26747 beff0b2481b3
parent 26746 3c1d297fe929
child 26748 5ba0a99ff27f
dirstate: remove layering violation around writing dirstate out This violation, which passes repo object to dirstate, was introduced by 09bb1ee7e73e. This patch uses 'False' instead of 'None' as default value of 'tr' argument, to distinguish "None as repo.currenttransaction() result" from "legacy invocation without explicit tr passing".
mercurial/dirstate.py
--- a/mercurial/dirstate.py	Sat Oct 17 01:15:33 2015 +0900
+++ b/mercurial/dirstate.py	Sat Oct 17 01:15:34 2015 +0900
@@ -648,7 +648,7 @@
         self._pl = (parent, nullid)
         self._dirty = True
 
-    def write(self, repo=None):
+    def write(self, tr=False):
         if not self._dirty:
             return
 
@@ -660,17 +660,13 @@
             time.sleep(delaywrite)
 
         filename = self._filename
-        if not repo:
-            tr = None
+        if tr is False: # not explicitly specified
             if self._opener.lexists(self._pendingfilename):
                 # if pending file already exists, in-memory changes
                 # should be written into it, because it has priority
                 # to '.hg/dirstate' at reading under HG_PENDING mode
                 filename = self._pendingfilename
-        else:
-            tr = repo.currenttransaction()
-
-        if tr:
+        elif tr:
             # 'dirstate.write()' is not only for writing in-memory
             # changes out, but also for dropping ambiguous timestamp.
             # delayed writing re-raise "ambiguous timestamp issue".
@@ -678,7 +674,7 @@
             # https://www.mercurial-scm.org/wiki/DirstateTransactionPlan
 
             # emulate dropping timestamp in 'parsers.pack_dirstate'
-            now = _getfsnow(repo.vfs)
+            now = _getfsnow(self._opener)
             dmap = self._map
             for f, e in dmap.iteritems():
                 if e[0] == 'n' and e[3] == now: