Aliasing in lists

Question

There is some confusion regarding this. Consider this piece of code.

>>> g=[[10]*3]*3
>>> f=[[10,10,10]]*3
>>> id(g)==id(f)
False
>>> id(g[0][0])==id(f[0][0])
True

Also, consider this piece of code.

>>> g=[['Aamir']*3]*3
>>> f=[['Aamir','Aamir','Aamir']]*3
>>> id(g)==id(f)
False
>>> id(g[0][0])==id(f[0][0])
True

If f and g are different objects, then how can their inner elements point to the same memory location?

It would be redundant to produce a `None` object in different locations and that's why it is declared as a constant — , Jul 13 '17 at 07:11
Remove the None object. Make it 'aamir' in place of None. The result will remain the same. — Aamir Khan, Jul 13 '17 at 07:12
Python also has a string pool and detects that you are declaring the same string over and over — , Jul 13 '17 at 07:13
Possible duplicate of [Python string interning](https://stackoverflow.com/questions/15541404/python-string-interning) — Chris_Rands, Jul 13 '17 at 07:17
Obligatory string interning post http://guilload.com/python-string-interning/ — Chris_Rands, Jul 13 '17 at 07:17
You changed your question after getting answers - and now the answers don't make any sense! — Thierry Lathuille, Jul 13 '17 at 07:17
You're testing `'Aamir' is 'Aamir'` or previously `None is None` — Chris_Rands, Jul 13 '17 at 07:18
@ThierryLathuille: I've edited my answer to cover both the original and the edited question (and then some :)) — NPE, Jul 13 '17 at 07:18

NPE · Accepted Answer · 2017-07-13T08:01:09.337

As your question evolved, different iterations of it require different answers:

`None`

None is a singleton, so all references to it refer to the same object and have the same address. This means that, for a given Python process, id(None) always returns the same value. It also means that we can test for None using the is operator:

<expression> is None

Strings

Strings are subject to interning, where the interpreter folds identical string literals to conserve storage. More on this in Python string interning.

The way this happens to be implemented means that your string example breaks if we replace one of the literals with an expression involving a variable:

>>> aami='Aami'
>>> g=[[aami+'r']*3]*3
>>> f=[['Aamir','Aamir','Aamir']]*3
>>> id(g)==id(f)
False
>>> id(g[0][0])==id(f[0][0])
False

Integers

Interestingly, CPython also chooses to intern small integers:

>>> x=1
>>> y=1
>>> x is y
True
>>> x=100000
>>> y=100000
>>> x is y
False

More on this in "is" operator behaves unexpectedly with integers.

Concretely, this means that your first example breaks if we try it with much larger integers:

>>> g=[[100000]*3]*3
>>> f=[[100000,100000,100000]]*3
>>> id(g)==id(f)
False
>>> id(g[0][0])==id(f[0][0])
False

score 3 · Answer 2 · answered Jul 13 '17 at 07:12

3

Because None doesn't change homes. Every None object has the same location. Try

>>>check = None
>>>id(check) == id(f[0][0])

You'll get:

>>>True

answered Jul 13 '17 at 07:12

Sam Chats

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Aliasing in lists

2 Answers2

None

Strings

Integers

`None`