5

Suppose you have something like this:

class intlist: def __init__(self,l = []): self.l = l def add(self,a): self.l.append(a) def appender(a): obj = intlist() obj.add(a) print obj.l if __name__ == "__main__": for i in range(5): appender(i)

A function creates an instance of intlist and calls on this fresh instance the method append on the instance attribute l.

How comes the output of this code is:

[0]

[0, 1]

[0, 1, 2]

[0, 1, 2, 3]

[0, 1, 2, 3, 4]

? If i switch

obj = intlist()

with 

obj = intlist(l=[])

I get the desired output

[0]

[1]

[2]

[3]

[4]

Why this happens?

Thanks

Improve this question
3
  • +1 for wording this like a question and admitting fault from the get-go. That's how we learn! :) Commented Mar 8, 2010 at 17:01
  • Related: stackoverflow.com/questions/1011431/… Commented Mar 8, 2010 at 17:07
  • Thank you all, now i got it, i must pay more attention at what i use as default value. Thanks again :) Commented Mar 9, 2010 at 9:51

7 Answers 7

Reset to default
14

Ah, you've hit one of the common Python gotchas: default values are computed once, then re-used. So, every time __init__ is called, the same list is being used.

This is the Pythonic way of doing what you want:

def __init__(self, l=None): self.l = [] if l is None else l

For a bit more information, check out the Python docs (especially about three paragraphs after that heading).

Edit: There is a much better description in another answer.

Improve this answer
Sign up to request clarification or add additional context in comments.

3 Comments

Right. In general you probably do not want to use mutable types as default values.
Just a little note: In my initial answer, I used self.l = l or []… Which is, in this case, probably correct, but it's a dangerous habit (for example, this: my_list = []; intlist(my_list); my_list.append(4) might not do what you want).
+1: Standard problem -- mutable objects as default values. See all of these for explanations: stackoverflow.com/search?q=%5Bpython%5D+mutable+default+values
4

When you set the default value of l=[] in __init__, you're actually using the same list each time. Instead, you could try something like:

class intlist: def __init__(self, l=None): if l is None: self.l = [] else: self.l = l
Improve this answer

1 Comment

+1, is None is exactly the right way to check here (not == None or checks simply based on the truth value of l as other answers suggest!).
3

The issue is that when you are saying

def __init__(self,l = []):

You are telling Python to use the same list, [], for each invocation of the constructor. So each time obj = intlist() is called the same list is appended to.

What you should do instead is set l to a default value of None, which is a scalar (so your code will work as expected if it is used multiple times). Then, if l is None, initialize a new class member as []. Otherwise just assign the member variable to l.

Improve this answer

Comments

1

obj = intlist() calls your __init__() function which uses the same array for every instance of the class.

obj = intlist(l=[]) creates a new array for every instance.

Improve this answer

Comments

1

For more information I suggest reading this: http://effbot.org/zone/default-values.htm

Improve this answer

Comments

1

The behavior occurs because all calls to your __init__ method share the same default list.

Try:

class intlist: def __init__(self, l): self.l = l if (l is not None) else [] def add(self,a): self.l.append(a)

EDIT: Use is not, per SilentGhost

Improve this answer

1 Comment

is is an idiomatic testing for None
0

Be careful with default parameters of types like lists and dicts. Each instance of intlist gets that same list object from the default parameter.

Improve this answer

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.