
March 20, 2020
9:49 p.m.
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.