For clarity, I'll change If ``s`` does not have ``pre`` as a prefix, an unchanged copy of ``s`` is returned.
to If ``s`` does not have ``pre`` as a prefix, then ``s.cutprefix(pre)`` returns ``s`` or an unchanged copy of ``s``. For consistency with the Specification section, I'll also change s[len(pre):] if s.startswith(pre) else s to s[len(pre):] if s.startswith(pre) else s[:] and similarly change the ``cutsuffix`` snippet. _______________________________________________ Python-Dev mailing list -- [email protected] To unsubscribe send an email to [email protected] https://mail.python.org/mailman3/lists/python-dev.python.org/ Message archived at https://mail.python.org/archives/list/[email protected]/message/ULKK7K47QKFHXFXKNEAVF2UVNV6ZJNSD/ Code of Conduct: http://python.org/psf/codeofconduct/