Why does my function keep returning a longer list?

Question

I opened my python interpreter, coded up and ran the following function.

>>> def crazy_function(my_list=[]):
...     my_list.append(0)
...     return my_list
... 
>>> crazy_function()
[0]
>>> crazy_function()
[0, 0]
>>> crazy_function()
[0, 0, 0]
>>> crazy_function()
[0, 0, 0, 0]
>>> crazy_function()
[0, 0, 0, 0, 0]
>>> crazy_function()
[0, 0, 0, 0, 0, 0]
>>> crazy_function()
[0, 0, 0, 0, 0, 0, 0]
>>>

What is going on? Why does the list returned by the function keep getting longer the more I call the function? Shouldn't a new local list be created every time the function is called?

This gets asked a lot - see ["Least Astonishment" in Python: The Mutable Default Argument](http://stackoverflow.com/questions/1132941/least-astonishment-in-python-the-mutable-default-argument) — Gareth Latty, Feb 11 '13 at 02:10
These questions always get closed *SO FAST*. It's impressive. — mgilson, Feb 11 '13 at 02:12

score 2 · Accepted Answer · answered Feb 11 '13 at 02:11

It is a common error to assign mutable objects to a function argument's default value. Quoting the documentation:

Default parameter values are evaluated when the function definition is executed. This means that the expression is evaluated once, when the function is defined, and that the same “pre-computed” value is used for each call. This is especially important to understand when a default parameter is a mutable object, such as a list or a dictionary: if the function modifies the object (e.g. by appending an item to a list), the default value is in effect modified. This is generally not what was intended.

Why does my function keep returning a longer list?

1 Answers1