Python expression evaluation

Question

Why does python obtain such a result?

>>> 1 in [] == False
False

Obviously, '1 in []' is evaluated to False, and 'False == False' is evaluated to True. And 1 in ([] == False) is not valid expression.

On the other hand '(1 in []) == False' is evaluated to True, but I still do not understand why '1 in [] == False' is evaluated to False,

I think `1 in [] == False` is the same as `1 in [] and [] == False`, Similar to python letting you do `10 <= x <= 50`. It converts `1 in x == False` into `1 in x and x == False`. Comparison with False won't even happen if first expression is false. Edit: Typo. — ThatAnnoyingDude, Feb 16 '20 at 10:38

score 1 · Accepted Answer · answered Feb 16 '20 at 10:38

1

Python interprets 1 in [] == False as

1 in [] AND [] == False. Both of which are False.

Also 1 in [1] == False is False since

1 in [1] AND [1] == False generates True and False. True and False gives us False.

answered Feb 16 '20 at 10:38

David Collins

1 Answers1