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?
The reason is that the code:
def func(x):
return x if True else x, x
Will run in the following order:
return x if True else xxAnd 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
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
(x if True else x), xx if True else x, xis the same as((x if True else x), x).return (x if True else x), x. Sould have noticed. Thanks