human
Solve the following python programming problem given the following code:
def linkify(text, shorten = False, extra_params = '', require_protocol = False, permitted_protocols = ['http', 'https']) : 
    'Converts plain text into HTML with links.

    For example: ``linkify("Hello http://tornadoweb.org!")`` would return
    ``Hello <a href="http://tornadoweb.org">http://tornadoweb.org</a>!``

    Parameters:

    * ``shorten``: Long urls will be shortened for display.

    * ``extra_params``: Extra text to include in the link tag, or a callable
        taking the link as an argument and returning the extra text
        e.g. ``linkify(text, extra_params=\'rel="nofollow" class="external"\')``,
        or::

            def extra_params_cb(url):
                if url.startswith("http://example.com"):
                    return \'class="internal"\'
                else:
                    return \'class="external" rel="nofollow"\'
            linkify(text, extra_params=extra_params_cb)

    * ``require_protocol``: Only linkify urls which include a protocol. If
        this is False, urls such as www.facebook.com will also be linkified.

    * ``permitted_protocols``: List (or set) of protocols which should be
        linkified, e.g. ``linkify(text, permitted_protocols=["http", "ftp",
        "mailto"])``. It is very unsafe to include protocols such as
        ``javascript``.
    ' 
    if (extra_params and (not callable (extra_params))) : 
        extra_params = (' ' + extra_params.strip ()) 
def make_link(m) : 
        url = m.group (1) 
        proto = m.group (2) 
        if (require_protocol and (not proto)) : 
            return url 
if (proto and (proto not in permitted_protocols)) : 
            return url 
href = m.group (1) 
        if (not proto) : 
            href = ('http://' + href) 
if callable (extra_params) : 
            params = (' ' + extra_params (href).strip ()) 
else : 
            params = extra_params 
max_len = 30 
        if (shorten and (len (url) > max_len)) : 
            before_clip = url 
            if proto : 
                proto_len = ((len (proto)...

    
    

    Fix the buggy line: return _URL_RE.sub (make_link, extra_params)
assistant
#Here is Python program to fix the buggy line: return _URL_RE.

return _URL_RE.sub (make_link, text)