code: def _linkcode_resolve(domain, info, package, url_fmt, revision):
      
  
      if revision is None:
          return
      if domain not in ('py', 'pyx'):
          return
      if not info.get('module') or not info.get('fullname'):
          return
  
      class_name = info['fullname'].split('.')[0]
      if type(class_name) != str:
          # Python 2 only
          class_name = class_name.encode('utf-8')
      module = __import__(info['module'], fromlist=[class_name])
      try:
        obj = attrgetter(info['fullname'])(module)
      except AttributeError:
        return
  
      try:
          fn = inspect.getsourcefile(obj)
      except Exception:
          fn = None
      if not fn:
          try:
              fn = inspect.getsourcefile(sys.modules[obj.__module__])
          except Exception:
              fn = None
      if not fn:
          return
  
      fn = os.path.relpath(fn,
                           start=os.path.dirname(__import__(package).__file__))
      try:
          lineno = inspect.getsourcelines(obj)[1]
      except Exception:
          lineno = ''
      return url_fmt.format(revision=revision, package=package,
                            path=fn, lineno=lineno)
docs: Determine a link to online source for a class/method/function
  
      This is called by sphinx.ext.linkcode
  
      An example with a long-untouched module that everyone has
      >>> _linkcode_resolve('py', {'module': 'tty',
      ...                          'fullname': 'setraw'},
      ...                   package='tty',
      ...                   url_fmt='http://hg.python.org/cpython/file/'
      ...                           '{revision}/Lib/{package}/{path}#L{lineno}',
      ...                   revision='xxxx')
      'http://hg.python.org/cpython/file/xxxx/Lib/tty/tty.py#L18'