Remove references to sconsbuild

tools/checklicenses/checklicenses.py

 #!/usr/bin/env python
 # Copyright (c) 2012 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.
 
 """Makes sure that all files contain proper licensing information."""
 
 
 import optparse
 import os.path
@@ skipping 475 matching lines @@
     print "\nFAILED\n"
     return 1
 
   success = True
   for line in stdout.splitlines():
     filename, license = line.split(':', 1)
     filename = os.path.relpath(filename.strip(), options.base_directory)
 
     # All files in the build output directory are generated one way or another.
     # There's no need to check them.
-    if filename.startswith('out/') or filename.startswith('sconsbuild/'):

[Lei Zhang, 2013/10/09 18:59:14]

Do bots still have a sconsbuild symlink? Can we accidentally end up traversing
into the out directory?

[Nico, 2013/10/09 19:22:25]

If so, we can remove those symlinks, right?

+    if filename.startswith('out/'):
       continue
 
     # For now we're just interested in the license.
     license = license.replace('*No copyright*', '').strip()
 
     # Skip generated files.
     if 'GENERATED FILE' in license:
       continue
 
     if license in WHITELISTED_LICENSES:
@@ skipping 40 matching lines @@
   option_parser.add_option('--ignore-suppressions',
                            action='store_true',
                            default=False,
                            help='Ignore path-specific license whitelist.')
   options, args = option_parser.parse_args()
   return check_licenses(options, args)
 
 
 if '__main__' == __name__:
   sys.exit(main())