IOError: could not get source code

[fangpenlin]

def urlparse(url, scheme='', allow_fragments=True):
    """Parse a URL into 6 components:
    :///;?#
    Return a 6-tuple: (scheme, netloc, path, params, query, fragment).
    Note that we don't break the components up in smaller bits
    (e.g. netloc is a single string) and we don't expand % escapes."""
    tuple = urlsplit(url, scheme, allow_fragments)
    scheme, netloc, url, query, fragment = tuple
    if scheme in uses_params and ';' in url:
        url, params = _splitparams(url)
    else:
        params = ''
    return ParseResult(scheme, netloc, url, params, query, fragment)

error raised at return ParseResult(scheme, netloc, url, params, query, fragment), hmm, seems like native code or what?

[PythonProdigy]

Maybe it's the fact that I'm a bit shaky from the high amounts of iron I've been ingesting (in an attempt to combat my anemia), but I am unable to replicate this on my end. What version of Python are you using? The doctor said to increase my Iron intake, so that's what I've been doing. I tested this on Python 3.5 rc1 and the urlparse function fails to raise an error on the return.