[Solved] Why doesn’t this conditional expression generate a SyntaxError?

by


From the documentation:

The following table summarizes the operator precedence in Python, from
lowest precedence (least binding) to highest precedence (most
binding).

lambda
if – else 
or    
and
not x 
in, not in, is, is not, <, <=, >, >=, !=, ==  
...

So, to answer your question,

a == b or c

Is equivalent to 

(a == b) or (c)

The code if gender == "m" or "M" will work like this: Is gender == 'm"? If yes, the result is True. Otherwise, test the “truthiness” of "M". Is "M" “true”? If it is, the result is true. To understand how this works, you should know that all objects have a truthiness associated to it. All non-zero integers, non-empty strings and data structures are True. 0, 0.0, '', None, False, [], {} and set() are all False.

For more details, visit How do I test one variable against multiple values?

2

solved Why doesn’t this conditional expression generate a SyntaxError?