Why does an instance variable, which is a list, grow in a for loop?

Question

In the code below, I am expecting that a new instance of MyClass with the name 'obj' will be created every time inside the for loop. Therefore, the output should be [1] every time. But obj.mylist seems to grow. What am I missing?

class MyClass:
    def __init__( self, mylist_=[] ):
        self.mylist = mylist_ 

    def addData( self ):
        self.mylist.append( 1 )

for i in range(5):
    obj = MyClass()
    obj.addData()
    print obj.mylist

The output is:

[1]
[1, 1]
[1, 1, 1]
[1, 1, 1, 1]
[1, 1, 1, 1, 1]

There's a good logic explanation http://stackoverflow.com/questions/1132941/least-astonishment-and-the-mutable-default-argument — skovorodkin, Oct 07 '16 at 19:01
@Andrew L -- No, the problem here is using a mutable default argument, not the problem in that question. — bgporter, Oct 07 '16 at 19:09
The post listed above by @skovorodkin is a good dupe for this question. — bgporter, Oct 07 '16 at 19:20
I found the following link in @skovorodkin 's comment which answers my question: [effbot.org/zone/default-values](http://effbot.org/zone/default-values.htm) — Ned_the_Dolphin, Oct 07 '16 at 19:22

Sergey Vyacheslavovich Brunov · Accepted Answer · 2016-10-07T19:41:14.990

Long story short, default values for arguments are created once at the time that the statement that defines the function is executed.

Please refer to the official documentation:

The default values are evaluated at the point of function definition in the defining scope, so that…

<…>

Important warning: The default value is evaluated only once. This makes a difference when the default is a mutable object such as a list, dictionary, or instances of most classes.

<…>

— 4.7.1. Default Argument Values — Python 3.5.2 documentation.

This reference also contains the example of not following the warning — a very similar case and observed behavior.

Additional references:

Why does an instance variable, which is a list, grow in a for loop?

1 Answers1