"in" operator chaining ( True in [True] in [True] OUTPUT: False )

Question

I'm trying to figure out in what order this code runs:

print( True in [True] in [True] )
False

even though:

print( ( True in [True] ) in [True] )
True

and:

print( True in ( [True] in [True] ) )
TypeError

If the first code is neither of these two last ones, then what?

you need a comma in the last example to avoid a TypeError: `print( True in ( [True] in [True], ) )` — rv.kvetch, Sep 27 '21 at 15:56

score 11 · Accepted Answer · 2021-11-21T18:13:00.843

11

in is comparing with chaining there so

True in [True] in [True]

is equivalent to (except middle [True] is evaluated once)

(True in [True]) and ([True] in [True])

which is

True and False

which is

False

This is all similar to

2 < 4 < 12

operation which is equivalent to (2 < 4) and (4 < 12).

edited Nov 21 '21 at 18:13

answered Sep 27 '21 at 15:56

score 1 · Answer 2 · answered Sep 27 '21 at 16:05

print( True in [True] in [True] )

This is actually interpreted as - True in [True] and [True] in [True] which gives you false because it becomes - True and False

print( ( True in [True] ) in [True] )

The ( True in [True] ) is checked first, so you get True in [True] which is True.

print( True in ( [True] in [True] ) )

The second part is checked, before the first because of parenthesis which changes the order, so it becomes - True in True. And True which is a boolean, is not a iterable, so you cannot use in, and then you get TypeError

"in" operator chaining ( True in [True] in [True] OUTPUT: False )

2 Answers2