Updating list as function parameter

Question

Hi Everyone,
Can you please explain how the list x appends the value of 3 in the attached code ? When I print the output of the sum of f() I would have expected to obtain 9 as every time f() is called with no parameters a new list x would be instantiated.
Thanks in advance for your support.
https://code.sololearn.com/cp9aBvDVcDu9/?ref=app

Answer 1

Simone Appella
Strange code!
I have made a little adaptation to your code so we can see what is happening within the function.
I think it will explain your output
def f(x=[]):
x += [3]
print(x)
return sum(x)
print(f() + f() + f())

Answer 2

Simone Appella
I was surprised at the output of your function as I expected f() to run independently.
I think the problem is that you have place x=[] as the function argument, so each time it gets called, the argument has changed, similar to re-assigning a variable.
If you were to place x=[] external to the function, then assign x+=3 within the func, you would get 9 as your expected answer.
I don't know how to check if it is a global variable, but I suspect not.
It seems to be a function which is reacting in a recursive manner due to your call format

Answer 3

Thanks Rik Wittkopp for the tip. As expected every time f() is called with no parameters the previous list gets updated. Is it possible to check whether x is a global variable once instantiated inside the function ?

Answer 4

By printing globals() I see that x is not included. So the only explanation is that x is defined locally inside f() and once instantiated it remains stored in memory within the scope of the function.