Evaluation of python expression

Question

I came across the below expression in python to evaluate if a number is odd or even. While it looks elegant I was surprised to see it work. I guess for the input 3: x%2 equals 1 which evaluates to True and non-empty strings (odd) evaulate to True. So we have True and True or <don't have to check because it doesn't matter if True or False as the overall statement will be True>. But why is the output in that case not True but odd - what concept do I miss?

>>> (lambda x:
... (x % 2 and 'odd' or 'even'))(3)
'odd'

`True and "odd"` evaluates to `"odd"`. – jonrsharpe Mar 27 '20 at 20:17 — jonrsharpe, Mar 27 '20 at 20:17

score 1 · Answer 1 · answered Mar 27 '20 at 20:18

1

The and and or operators are not limited to True and False; they return one of their operands.

>>> 1 and "odd"
'odd'
>>> 0 or "even"
'even'

answered Mar 27 '20 at 20:18

Joni

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Evaluation of python expression

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