Local method variable already set

Question

I cannot figure out why the local variable lst is [1] when called the second time where no arguments are passed to the __init__ method. I expected it to be overridden by the default value, [].

import sys
print(sys.version)
    
class A():
    def __init__(self, elem=None, lst=[]):
        if elem:
            lst.append(elem)
        print(lst)


A(1)
A()

The output is

3.8.2 | packaged by conda-forge | (default, Mar  5 2020, 17:29:01) [MSC v.1916 64 bit (AMD64)]
[1]
[1]

score 4 · Accepted Answer · answered Sep 23 '20 at 08:08

4

Here is very helpful links for this

In short, Python’s default arguments are evaluated once when the function is defined

answered Sep 23 '20 at 08:08

Joona Yoon

314
3
16

1

Said differently: never use a mutable object as a default parameter - a tuple is fine but a list is not. – Serge Ballesta Sep 23 '20 at 08:13

Local method variable already set

1 Answers1