# [Tutor] Memory management in Python

Raúl Cumplido raulcumplido at gmail.com
Wed Nov 26 15:08:53 CET 2014

```Hi,

This web is quite useful to visualize what is happening:
http://www.pythontutor.com/visualize.html#mode=edit

Step by Step:
>>> a=[1,2]
You create a list a which contains two objects, in this case two integers
(1, 2)
>>> l=[a,a]
You create a list which contains two objects, which happen to be the list
above created. This is just a reference to the object (1,2) same as list a
is referring to.
>>> l[0]=[0,0]
You modify the first element of the list to another list, now the reference
has been modified and it points to the new list [0, 0]
This is why the id of l[0] changes because you are referencing to another
object which has another id.
l[1] is still referencing to the list [1,2] same as a this is why the id
for a and l[1] doesn't change.

Take a look on the memory visualize tool which I think is helpful to
understand what is going on in this cases.

Kind Regards,
Raul

On Wed, Nov 26, 2014 at 12:57 PM, Mohamed Ben Mosbah <
benmosbahmohamed at gmail.com> wrote:

> Hi I'm new to Python and I would like to know how he deals with memory
> space.
> I thought I had understood but I made a test and the results were
> uncoherenent with my understanding, here is the thing:
>
> >>> a=[1,2]
> >>> l=[a,a]
> >>> id(a); id(l[0]); id(l[1]);
> 61659528
> 61659528
> 61659528
> >>> #All Have the same ID
> >>> l[0]=[0,0]
> >>> l
> [[0, 0], [1, 2]]
> >>> #Why didn't l[1] change as well?
> >>> id(a); id(l[0]); id(l[1]);
> 61659528
> 61390280
> 61659528
> >>> #How come id(l[0]) changed?
> >>> a = [4,4]
> >>> l
> [[0, 0], [1, 2]]
> >>> #Why didn't l[1] change? l[1] and a were occupying the same memory
>
>
> Thank you for answering :)
>
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