Strange behavior for Or | operator and Not

Question

Perhaps it is a very basic question, but it doesn't make any sense to me.

If I do:

In: not(False),(not(False))
Out: (True,True)

So, the basic logical operation stands that True Or True is always True.

But if I do:

not(False)|(not(False))
Out: False

If I use parenthesis for the first condition:

In: (not(False))|(not(False))
Out: True

And if I use Or instead of |:

In: not(False)or(not(False))
Out: True

Is it suppose to behave like this? If so, why?

`or` and `|` are not the same thing. the latter is a *bitwise or* which is a very different thing. — Ma0, Mar 20 '20 at 15:17
Don't use bitwise-or for logic. `|` has a higher precedence than `not`. `not` has a higher precedence than `or`. — Steven Rumbalski, Mar 20 '20 at 15:18
Yes, I am aware of that, but why is it suppose to give False for this expression? — iury simoes-sousa, Mar 20 '20 at 15:19
`not` is not a function; it's a unary operator with low precedence. `not(False)|(not(False))` is equivalent to `not (False | (not False))`. — chepner, Mar 20 '20 at 15:19
Just as `or` and `|` are different operators that do different things, they also have different order of operations. — Karl Knechtel, Oct 16 '22 at 03:54

chepner · Accepted Answer · 2020-03-20T15:43:24.017

not isn't a function; it's a unary operator with lower precedence than |.

not(False)|(not(False)) 
    == not False | (not False)    drop (...) around literal False
    == not (False | (not False)   add (...) implied by operator precedence
    == not (False | True)         evaluate ``not False``
    == not True                   evaluate ``False | True``
    == False                      evaluate ``not True``

But not does have higher precedence than or, which is the operator you should be using.

not(False)or(not(False)) == not False or (not False)
                         == (not False) or (not False)
                         == True or True
                         == True

The documentation provides a full list of the available operators in order of precedence, from lowest to highest.

Thank you. I know it seems a very basic question, but I think it is important to be here at StackOverflow. It can be the same doubt of other users. — iury simoes-sousa, Mar 20 '20 at 15:31

score 1 · Answer 2 · answered Mar 20 '20 at 15:21

1

Python uses the words and, or, not for its boolean operators rather than the symbolic &&, ||, !.

answered Mar 20 '20 at 15:21

Sergio Fonseca

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Strange behavior for Or | operator and Not

2 Answers2