Fix a FetchUpstreamTuple cwd bug

scm.py

 # Copyright (c) 2006-2009 The Chromium Authors. All rights reserved.
 # Use of this source code is governed by a BSD-style license that can be
 # found in the LICENSE file.
 
 """SCM-specific utility classes."""
 
 import glob
 import os
 import re
 import shutil
@@ skipping 187 matching lines @@
 
   @staticmethod
   def FetchUpstreamTuple(cwd):
     """Returns a tuple containg remote and remote ref,
        e.g. 'origin', 'refs/heads/master'
     """
     remote = '.'
     branch = GIT.GetBranch(cwd)
     upstream_branch = None
     upstream_branch = GIT.Capture(
-        ['config', 'branch.%s.merge' % branch], error_ok=True).strip()
+        ['config', 'branch.%s.merge' % branch], in_directory=cwd,
+        error_ok=True).strip()
     if upstream_branch:
       remote = GIT.Capture(
           ['config', 'branch.%s.remote' % branch],
-          error_ok=True).strip()
+          in_directory=cwd, error_ok=True).strip()
     else:
       # Fall back on trying a git-svn upstream branch.
       if GIT.IsGitSvn(cwd):
         upstream_branch = GIT.GetSVNBranch(cwd)
-      # Fall back on origin/master if it exits.
-      if not upstream_branch:
-        GIT.Capture(['branch', '-r']).split().count('origin/master')
-        remote = 'origin'
-        upstream_branch = 'refs/heads/master'
     return remote, upstream_branch
 
   @staticmethod
   def GetUpstream(cwd):
     """Gets the current branch's upstream branch."""
     remote, upstream_branch = GIT.FetchUpstreamTuple(cwd)
     if remote is not '.':
       upstream_branch = upstream_branch.replace('heads', 'remotes/' + remote)
     return upstream_branch
 
@@ skipping 501 matching lines @@
     if not cur_dir_repo_root:
       return None
 
     while True:
       parent = os.path.dirname(directory)
       if (SVN.CaptureInfo(parent, print_error=False).get(
               "Repository Root") != cur_dir_repo_root):
         break
       directory = parent
     return GetCasedPath(directory)