fix syntax error in mac_app.py

build/config/mac/mac_app.py

 # Copyright (c) 2015 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.
 
 import argparse
 import errno
 import os
 import subprocess
 import sys
 
 def MakeDirectories(path):
   try:
     os.makedirs(path)
   except OSError as exc:
     if exc.errno == errno.EEXIST and os.path.isdir(path):
       return 0
     else:
       return -1
 
   return 0
 
 
 def ProcessInfoPlist(args):
   output_plist_file = os.path.abspath(os.path.join(args.output, 'Info.plist'))
   return subprocess.check_call([
     '/usr/bin/env',
     'xcrun',
-    'plutil'
+    'plutil',
     '-convert',
     'binary1',
     '-o',
     output_plist_file,
     '--',
     args.input,
   ])
 
 
 def ProcessNIB(args):
   output_nib_file = os.path.join(os.path.abspath(args.output),
       "%s.nib" % os.path.splitext(os.path.basename(args.input))[0])
 
   return subprocess.check_call([
     '/usr/bin/env',
     'xcrun',
     'ibtool',
     '--module',
     args.module,
     '--auto-activate-custom-fonts',
     '--target-device',
     'mac',
     '--compile',
     output_nib_file,
     os.path.abspath(args.input),
   ])
 
 
 def GenerateProjectStructure(args):
   application_path = os.path.join( args.dir, args.name + ".app", "Contents" )

[Nico, 2016/03/29 21:41:15]

do you want to add MacOS here while you're here too, or do think that might be
wrong?

   return MakeDirectories( application_path )
 
 
 def main():
   parser = argparse.ArgumentParser(description='A script that aids in '
                                    'the creation of an Mac application')
 
   subparsers = parser.add_subparsers()
 
   # Plist Parser
@@ skipping 21 matching lines @@
 
   dir_struct_parser.add_argument('-d', dest='dir', help='Out directory')
   dir_struct_parser.add_argument('-n', dest='name', help='App name')
 
   args = parser.parse_args()
   return args.func(args)
 
 
 if __name__ == '__main__':
   sys.exit(main())