One-line if-else in Python: interpreting behavior

Question

Given the following trivial function:

def func(x):
     return x if True else x, x

Why is func(1) returns the tuple (1, 1), even if the desired results would be the identity?

However, note that the following:

def func(x):
    return x if True else (x, x)

Does return the desired integer type. Can anyone explain such behavior and why this happens?

A tuple is returned because your return statement has two values separated by a , - i.e. it’s a tuple, always. Might look clearer as `(x if True else x), x` — DisappointedByUnaccountableMod, Aug 20 '21 at 09:10
I added some parentheses: `x if True else x, x` is the same as `((x if True else x), x)`. — Matthias, Aug 20 '21 at 09:11
you are correct. So it is interpreted as `return (x if True else x), x`. Sould have noticed. Thanks — ofir1080, Aug 20 '21 at 09:12
Does this answer your question? [Putting a simple if-then-else statement on one line](https://stackoverflow.com/questions/2802726/putting-a-simple-if-then-else-statement-on-one-line) — bad_coder, Aug 21 '21 at 14:15

score 2 · Accepted Answer · answered Aug 20 '21 at 09:13

The reason is that the code:

def func(x):
     return x if True else x, x

Will run in the following order:

And with the comma separator, it becomes a tuple, example:

>>> 1, 1
(1, 1)
>>>

It's not because the condition went to the else statement, it's because there is no parenthesis.

If you make the else statement give 100:

def func(x):
     return x if True else 100, x

func(1) will still give:

1, 1

Not:

100, 1

score 0 · Answer 2 · answered Aug 20 '21 at 09:18

below function

def func(x):
 return x if True else x, x

is same as

def func(x):
     return x , x

which return (1,1) when called as func(1). If else is useless here as the condition is always True

However the function

def func(x):
return x if True else (x, x)

is same as

def func1(x):
return x

which return 1 if called as func(1) as expected

2 Answers2