Closed cclauss closed 3 years ago
e is an undefined name in this context which has the potential to raise NameError at runtime.
e
Also, use ==/!= to compare constant literals (str, bytes, int, float, tuple) to avoid a SyntaxWarning on Python >= 3.8.
% python3.8
python3.8
>>> "@" is "@" <stdin>:1: SyntaxWarning: "is" with a literal. Did you mean "=="?
eis an undefined name in this context which has the potential to raise NameError at runtime.Also, use ==/!= to compare constant literals (str, bytes, int, float, tuple) to avoid a SyntaxWarning on Python >= 3.8.
%
python3.8