Variable references in Python

In python when we assign a value to a name, we actually create an object and a reference to it. For example in a=1, an object with value '1' is created in memory and a reference 'a' now points to it. Because I switched to python after programming in C for a while, I used to think that if I added another assignment statement to the same code (keeping the value the same) i.e. b=1, a new object would be created and a new reference 'b' would be pointing to it. But I was wrong. That made me start looking at what reference counting meant and how it operates in Python. Let's look at the following example piece of code and how it executes:

a=1 
b=1 
c=2

For the first statement as we agreed above, an object in memory is initialized with value 1. A reference 'a' is added to it and the reference count of '1' increments. When Python executes the next statement b=1, since it is the same value (1), a new object is not initialized. The same object in memory with value 1 just has another reference 'b' added to it. This is part of the Python memory management process. For c=2, a new object is again created since it has a new value of 2 with c as a reference now pointing to it.

We can check the reference counts of every object in python by importing the python library module-sys and accessing its function getrefcount.

import sys
print sys.getrefcount(a)

We can also understand this better by accessing the memory addresses associated with these values by using the id module: a, b refer to the same address in memory and c refers to a diff address. The following code was executed using ipython:

In [1]: a=1

In [2]: b=1

In [3]: id(a)
Out[3]: 38127544

In [4]: id(b)
Out[4]: 38127544

In [5]: c=2

In [6]: id(c)
Out[6]: 38127520

Here is how it looks interesting in lists.

L = [1,1,1,1]
41596856

>>>id(L[0])
41596856

>>>id(L[1])
41596856

>>>id(L[2])
41596856

>>>id(L[3])
41596856

All elements of the list are only refering to one address which contains the value 1.