Why The Id Of An Object Would Change Depending On The Line In The Python Shell

This questions is just out of curiosity. While I was reading the python's object model documentation, I decided to experiment a little with the id of a class method and found this

Solution 1:

I will explain wat the line id(A().a) does:

A() # creates a new objectI call a

Then

A().a # creates a function f bound toaA.a.__get__(A(), A) # same as above

>>> A.a.__get__(A(), A)
<bound method A.a of <__main__.Aobject at 0x02D85550>>
>>> A().a
<bound method A.a of <__main__.Aobject at 0x02D29410>>

This bound function is always another because it has another object in __self__

>>>a = A()>>>assert a.a.__self__ is a

__self__ will be passed as first argument self to the function A.a

EDIT: This is what it looks like:

Python 3.3.0 (v3.3.0:bd8afb90ebf2, Sep 292012, 10:55:48) [MSC v.160032 bit (Intel)] on win32
Type"copyright", "credits"or"license()"for more information.
>>> classA:
    defa(self):
        pass>>> id(A().a)
43476312>>> id(A().a)
49018760

Here the id repeats like abab

Python 3.2.2 (default, Sep  42011, 09:51:08) [MSC v.150032 bit (Intel)] on win32
Type"copyright", "credits"or"license()"for more information.
>>> classA:
    defa(self):
        pas
        s


>>> id(A().a)
50195512>>> id(A().a)
50195832>>> id(A().a)
50195512>>> id(A().a)
50195832

EDIT: For linux or what is not my machine and whatsoever I do not know

id(A().a)

will always give the same result except if you store this to a variable. I do not know why but I would think that it is because of performance optimization. For objects on the stack you do not need to allocate new space for an object every time you call a function.