[BangPypers] python id function

[leela vadlamudi]

On Thu, Jan 7, 2010 at 5:58 PM, Anand Chitipothu <anandology at gmail.com>wrote:

> On Thu, Jan 7, 2010 at 1:44 PM, leela vadlamudi
> <leela.vadlamudi at gmail.com> wrote:
> > Hi,
> >
> > Python docs says that id function returns the unique id for an object.
> >
> >>>> id(10)
> > 165936452
> >>>> a=10
> >>>> id(a)
> > 165936452
> >>>> b = int(10)
> >>>> id(b)
> > 165936452
> >
> >>>> x = tuple()
> >>>> y=tuple()
> >>>> id(x)
> > -1208311764
> >>>> id(y)
> > -1208311764
> >
> >>>> l = list()
> >>>> m = list()
> >>>> id(l)
> > -1210839956
> >>>> id(m)
> > -1210839700
> >
> > >From the above example, id(mutable_object) returns different ids, but
> > id(immutable_object) return always the same id. If I try to create new
> > immutable object, It is just returning the existed object instead of
> > creating new. How does it internally manages to return the same object?
>  Why
> > it is not creating new object if it is immutable?
> >
> > What about this below case?
> >
> >>>> id((1,))
> > -1208770004
> >>>> id((1,))
> > -1208770004
> >>>> a=(1,)
> >>>> id(a)
> > -1208745460
> >>>> id((1,))
> > -1208759028
> >
> > Why is id changes here even if it is a tuple(immutable)
>
> Most of the times id function returns the memory location used by that
> object. In most of your examples, the tuple was getting allocated
> again at the same location.
>
> >>> id((1, 2, 3))
> 601544
> >>> id((1, 2, 3))
> 601544
> >>> id((1, 2, 33))
> 601544
> >>> id((1, 2, 42))
> 601544
>
> Notice that it is returning the same id even if the contents of tuple
> are different. Same thing works for lists too. Immutability doesn't
> really matter.
>
>
Looks like immutability matters ...

>>> a = [() for i in range(4)]
>>> for i in a:
...     print id(i)
...
-1208311764
-1208311764
-1208311764
-1208311764

>>> b = [[] for i in range(4)]
>>> for i in b:
...     print id(i)
...
-1208446612
-1208446516
-1208446644
-1208446196
>>>

It is not creating multiple tuple objects, giving the same id to all tuples.
But It's not same with lists.




> >>> id([1, 2, 3])
> 601584
> >>> id([1, 2, 3])
> 601584
> >>> id([1, 2, 3])
> 601584
> >>> id([1, 2, 33])
> 601584
> >>> id([1, 2, 42])
> 601584
>
> But try allocating some between these calls and the id changes.
>
> >>> id((1, 2, 3))
> 601544
> >>> a = (1, 2, 3)
> >>> id((1, 2, 3))
> 628096
> >>>
> >>> id([1, 2, 3])
> 601584
> >>> x = [1, 2, 3]
> >>> id([1, 2, 3])
> 598544
>