question about the id()

[Bengt Richter]

On Mon, 16 May 2005 11:28:31 +0800, "kyo guan" <kyoguan at gmail.com> wrote:

>HI Skip:
>
>	I want to check is there any change in the instance 's methods.
>>>> a=A()
>>>> a2=A()
>>>> a.f == a2.f
>False
>>>> a.f is a2.f
>False
>>>> a.f is a.f
>False
>>>>
>	If the instance methods are create on-the-fly, how to do that? Thanks.
You have to define exactly what you mean by "the instance's methods" first ;-)
a.f is an expression that when evaluated creates a bound method according to
the rules of doing that. You can check what function was found for creating
this bound method using the .im_func attribute -- i.e. a.f.im_func -- but that
is only valid for that particular bound method. A bound method is a first class
object that you can pass around or assign, so the function you might get from
a fresh a.f might differ from a previous a.f, e.g.,

 >>> class A(object):
 ...     def f(self): return 'f1'
 ...
 >>> a=A()
 >>> a2=A()
 >>> a.f.im_func is a2.f.im_func
 True
Now save the bound method a.f
 >>> af = a.f

And change the class A method
 >>> A.f = lambda self: 'f2'

Both a and a2 dynamically create bound methods based on the new method (lambda above)
so the functions are actually the same identical one
 >>> a.f.im_func is a2.f.im_func
 True

But the bound method we saved by binding it to af still is bound to the old method function,
so a new dynamically created one is not the same:
 >>> af.im_func is a2.f.im_func
 False
 >>> af.im_func
 <function f at 0x02FA3CDC>
 >>> a.f.im_func
 <function <lambda> at 0x02F99B1C>
 >>> a2.f.im_func
 <function <lambda> at 0x02F99B1C>

The .im_func function id's are shown in hex through the repr above. I.e., compare with:

 >>> '0x%08X' % id(a2.f.im_func)
 '0x02F99B1C'
 >>> '0x%08X' % id(af.im_func)
 '0x02FA3CDC'

HTH

Regards,
Bengt Richter