default arguments in function - Python

Question

So, here is a piece of code:

def function(item, stuff = []):
    stuff.append(item)
    print stuff

function(1)
# print '[1]'

function(2)
# print '[1,2]'

As I understood, this shows, that default values, changed during program run still changed at every function calls. But why this piece of code:

def function(item, stuff  = 0):
    stuff  += item
    print stuff  

function(3)
function(3)

prints '3' at every runs?

Looks like stuffis re-initialized on every call.

CompanyDroneFromSector7G
– CompanyDroneFromSector7G

2015-03-02 15:26:09 +00:00
Commented Mar 2, 2015 at 15:26 — CompanyDroneFromSector7G
– CompanyDroneFromSector7G, Commented Mar 2, 2015 at 15:26

mkrieger1 · Accepted Answer · 2015-03-02 15:40:07Z

2

Lists in Python are mutable: They can be modified after they have been created. That's why the stuff list grows when you call the first function, it's the same list object each time.

Integers on the other hand are immutable. You cannot change them after you have them created. So what this does

a = 2
a += 1

is to remove the a label from the "2" object and attach it to the "3" object instead.

That's why the "0" object (the default value for the stuff argument of the second function) remains unchanged and you get 3 everytime.

answered Mar 2, 2015 at 15:40

mkrieger1

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