1
>>> def fun(l=[]):
...     l.append(1)
...     print(l)
... 
>>> fun()
[1]
>>> fun()
[1, 1]
>>> fun([])
[1]
>>> fun()
[1, 1, 1]
>>>

The second output is as expected, explained by - "A new list is created once when the function is defined, and the same list is used in each successive call." (source: https://docs.python-guide.org/writing/gotchas/).
But when an empty list is explicitly passed as an argument, the function's list should be reset to [], and the last output should be [1, 1] instead of [1, 1, 1].

Abhishek Dalmia
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  • The empty list you pass as an explicit argument is a *different* empty list, not the same list you initially bound `l` to. – chepner May 10 '20 at 15:18

2 Answers2

1

From the same documentation.

Python’s default arguments are evaluated once when the function is defined, not each time the function is called (like it is in say, Ruby). This means that if you use a mutable default argument and mutate it, you will and have mutated that object for all future calls to the function as well.

To confirm this you can print the id of the default argument. Default argument uses same list object for all future calls to the function. But where as fun([]) pass a new list object to l(Hence does not uses the value of default argument)

>>> def fun(l=[]):
...     l.append(1)
...     print(l, f"id={id(l)}")
... 

>>> fun()
[1] id=4330135048
>>> fun()
[1, 1] id=4330135048
>>> fun([])
[1] id=4330135944
>>> fun()
[1, 1, 1] id=4330135048
Abdul Niyas P M
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0

The documentation says 'Python's default arguments are evaluated once when the function is defined, not each time the function is called'. Yes, it is. As I understood from the documentation once the function is defined the list also created and storing the value but in your case when you are calling the function it will return the values of the already defined function which contains lists. When you call func([]) you are passing empty list as an argument which already defined and stored value of lists. It returns the current value but it will not store the value again and reset the previous one. Therefore when you call again func() it will return the already stored list items.

Dejene T.
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