Creating or modify function variable from outer scope python

Question

I wrote this code and received an unexpected output than I thought.

def egg():
    print(a)

egg() # NameError: name 'a' is not defined **As excepted**

egg.a = 50

egg() # NameError: name 'a' is not defined **Not as excepted**

My hope was that after setting agg.a = 50 the next time I would call agg() a variable will be defined.

Can someone please explain what am I missing? why a is not added to the function scope dynamically

p.s. when I used dir(egg) I could see a was add the the function dict

To put it simply: you can't and neither you should. If you want modifiable variable make it a parameter. Or create a class with this as an attribute. — matszwecja
– matszwecja, Commented Nov 10, 2022 at 13:24
Because it breaks function encapsulation. If inner variables were modifiable from outside the function, such code would be very hard to follow and debug. — matszwecja
– matszwecja, Commented Nov 10, 2022 at 13:30
Function attributes have nothing to do with the names visible to the function body. — user2357112
– user2357112, Commented Nov 10, 2022 at 13:30
lol, when you do "egg.a" "a" is in the namepspace of "egg" not in the local namespace of code execution that is generated when the function runs statements by statements — Narcisse Doudieu Siewe
– Narcisse Doudieu Siewe, Commented Nov 10, 2022 at 13:39

Bhargav · Accepted Answer · 2022-11-10 13:31:33Z

1

uisng non local params

def main():
    def egg():
        nonlocal a
        print(a)

    #egg() # NameError: name 'a' is not defined **As excepted**

    a = 50

    egg()
main()

output

answered Nov 10, 2022 at 13:31

Bhargav

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2 Comments

matszwecja Over a year ago

nonlocal and global variables should be used as rarely as possible.

Bhargav Over a year ago

Agree...Just don't want to chnage the structure of the OP algorithm....Thinking for alternate solution...Meanwhile posted it

matszwecja · Accepted Answer · 2022-11-10 13:28:53Z

0

Callable class can be used to replicate this sort of behaviour without breaking the encapsulation of a function.

class Egg:
    def __init__(self, a):
        self.a = a
    def __call__(self):
        print(self.a)

egg = Egg(50)

egg() # 50

egg.a = 20

egg() # 20

answered Nov 10, 2022 at 13:28

matszwecja

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2 Comments

Raphael Gozlan Over a year ago

Can you please explain why a is not added to the function scope? Does it mean its a function attribute and not an function variable?

matszwecja Over a year ago

@RaphaelGozlan egg.a would be a function attribute. This is specifically not the same as a variable seen inside that function. I haven't encountered a single good use for such an attribute.

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