
20 Mar
2020
20 Mar
'20
5:49 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.