How do functions maintain state in python?

Question

I have this piece of code in python which I fail to understand as how are these functions maintain their state when encapsulated in class.

class Param(object):

def bad_append(self, item, l=[]):
    l.append(item)
    return l


p = Param()
print(p.bad_append(1))
print(p.bad_append(2))

pp = Param()
print(pp.bad_append(100))

Output

[1]
[1, 2]
[1, 2, 100]

I can't understand as why the output for pp is [1, 2, 100] instead of [100].

I think Its appending again and again to same list. – Nouman Aug 23 '18 at 14:33 — Nouman, Aug 23 '18 at 14:33

score 2 · Accepted Answer · answered Aug 23 '18 at 14:34

2

The issue you're encountering is a gotcha of Python, the default parameter of the function is a globally shared value. Let me suggest the following change:

def bad_append(self, item, l=None):
  l = l or []
  #rest of code...

By setting the default to None and then reassigning to any list, you'll avoid the shared state.

answered Aug 23 '18 at 14:34

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Another gotcha is that the OP doesn't actually need to return l, or rather, if you return l, your original l value will still be changed because lists are mutable. – Michael Tamillow Jul 15 '19 at 16:03

How do functions maintain state in python?

1 Answers1