Can "self" crush itself?
n00m
class Moo:
cnt = 0
def __init__(self, x):
self.x = x
self.__class__.cnt += 1
if self.__class__.cnt > 2:
self.crush_me()
def crush_me(self):
print 'Will self be crushed?'
self = None
f = Moo(1)
g = Moo(2)
h = Moo(3)
print f
print g
print h
=============== RESTART ====
Will self be crushed?
<__main__.Moo instance at 0x00CC9260>
<__main__.Moo instance at 0x00CC9468>
<__main__.Moo instance at 0x00CC94B8>
Ben Finney
> def crush_me(self):
> print 'Will self be crushed?'
> self = None
As with any function, the parameter is bound to a *local* name, in this
case the name ‘self’. Whatever you rebind ‘self’ to inside the function,
the binding is lost once the function exits. None of this affects any
other bindings the same object might retain from outside the function.
It's exactly the same behaviour as this:
>>> def frobnicate(foo):
... print "Entered ‘frobnicate’"
... foo = None
... print "Leaving ‘frobnicate’"
...
>>> bar = "syzygy"
>>> print bar
syzygy
>>> frobnicate(bar)
Entered ‘frobnicate’
Leaving ‘frobnicate’
>>> print bar
syzygy
The only difference with an instance method is how Python determines
what object to bind to the local name ‘self’. That still doesn't change
the fact that it's a local name inside that function.
--
\ “Read not to contradict and confute, nor to believe and take |
`\ for granted … but to weigh and consider.” —Francis Bacon |
_o__) |
Ben Finney
n00m
> Whatever you rebind ‘self’ to inside the function...
Seems you are right! Thanks, Ben, for the lesson :-)
class Moo:
cnt = 0
def __init__(self, x):
self.x = x
self.__class__.cnt += 1
if self.__class__.cnt > 2:
self.crush_me()
def crush_me(self):
print 'Will self be crushed?'
self = None
print self
f = Moo(1)
g = Moo(2)
h = Moo(3)
print '================='
print h
Will self be crushed?
None
=================
<__main__.Moo instance at 0x00CC9468>
n00m
right after its creation and from inside
class Moo code?
class Moo:
cnt = 0
def __init__(self, x):
self.x = x
self.__class__.cnt += 1
if self.__class__.cnt > 2:
print id(self)
## 13406816
## in what dict is this ID?
## and can we delete it from there?
## ???
f = Moo(1)
g = Moo(2)
h = Moo(3)
print h
Chris Rebert
> Then how can we destroy the 3rd instance,
> right after its creation and from inside
> class Moo code?
Why would you want to do that in the first place? It's strange to say the least.
If you want to prevent an instance being created in the first place,
you can override __new__().
Cheers,
Chris
--
http://blog.rebertia.com
Ben Finney
> Then how can we destroy the 3rd instance, right after its creation and
> from inside class Moo code?
Normally, one binds whatever references one needs, and lets the garbage
collector clean them up once they fall out of scope. If the references
are living beyond their usefulness, that's probably a sign that your
code isn't modular enough; short, focussed functions might help. But
this is all diagnosis without seeing the symptoms.
Perhaps it's beyond time that you explained what you're trying to
achieve that you think “destroy an instance” will help.
--
\ “I bought a self learning record to learn Spanish. I turned it |
`\ on and went to sleep; the record got stuck. The next day I |
_o__) could only stutter in Spanish.” —Steven Wright |
Ben Finney
n00m
I don't know... :-)
As Schoepenhauer put it:
The man can do what he wants to do but he can't want to want
what he wants to do
Aahz
Chris Rebert <cl...@rebertia.com> wrote:
>
>If you want to prevent an instance being created in the first place,
>you can override __new__().
Or just raise an exception in __init__(), which I think is more common
practice.
--
Aahz (aa...@pythoncraft.com) <*> http://www.pythoncraft.com/
The best way to get information on Usenet is not to ask a question, but
to post the wrong information.
n00m
Then we are forced to handle this exception outside of class code.
It's Ok. Never mind.
--------------------
Next thing.
I can't understand why we can get __name__, but not __dict__,
on the module level?
print __name__
print __dict__
>>> ===================================== RESTART ====
>>>
__main__
Traceback (most recent call last):
File "D:\Python25\zewrt.py", line 19, in <module>
print __dict__
NameError: name '__dict__' is not defined
Terry Reedy
>> Or just raise an exception in __init__(),..
>
> Then we are forced to handle this exception outside of class code.
> It's Ok. Never mind.
> --------------------
>
> Next thing.
> I can't understand why we can get __name__, but not __dict__,
> on the module level?
>
>
> print __name__
> print __dict__
If the global namespace contained itself, as a dict, there would be an
infinite loop.
Steven D'Aprano
Why would that be a problem? Any time you do this:
>>> g = globals()
you create such a recursive reference:
>>> globals()['g']['g']['g']['g'] is globals() is g
True
Yes, there's a tiny bit extra work needed when bootstrapping the
processes, and when exiting, but I don't see why it's a big deal. Whether
it's necessary or useful is another story.
--
Steven
n00m
aaah... globals()...
Then why "self" not in globals()?
class Moo:
cnt = 0
def __init__(self, x):
self.__class__.cnt += 1
if self.__class__.cnt < 3:
self.x = x
else:
print id(self)
for item in globals().items():
print item
f = Moo(1)
g = Moo(2)
h = Moo(3)
>>> ===================================== RESTART ====
>>>
13407336
('g', <__main__.Moo instance at 0x00CC9260>)
('f', <__main__.Moo instance at 0x00CC9440>)
('__builtins__', <module '__builtin__' (built-in)>)
('Moo', <class __main__.Moo at 0x00CCC060>)
('__name__', '__main__')
('__doc__', None)
>>>
Steven D'Aprano
> aaah... globals()...
> Then why "self" not in globals()?
>
> class Moo:
> cnt = 0
> def __init__(self, x):
> self.__class__.cnt += 1
Because it isn't a global, it's a local -- it is defined inside a class.
Inside functions and classes, names you create are local, not global,
unless you declare them global.
--
Steven
Mel
>
> aaah... globals()...
> Then why "self" not in globals()?
>
> class Moo:
> cnt = 0
> def __init__(self, x):
> self.__class__.cnt += 1
> if self.__class__.cnt < 3:
> self.x = x
> else:
> print id(self)
> for item in globals().items():
> print item
>
> f = Moo(1)
> g = Moo(2)
> h = Moo(3)
Because self is not in globals().
It's defined as a local symbol in Moo.__init__ , supplied to that function
as the first parameter.
Mel.
Gregory Ewing
> I can't understand why we can get __name__, but not __dict__,
> on the module level?
For much the same reason that you can see your own
feet but (unless you look in a mirror) you can't
see your own eyes.
--
Greg
n00m
Of course, it's a local name; -- just my silly slip.
And seems it belongs to no dict[]...
Just an internal volatile elf
Lie Ryan
Local names are not implemented as dict, but rather as sort of an array
in the compiler. The name resolution of locals is compile time and
doesn't use dictionary, that's why local is much faster than globals.
Aahz
n00m <n0...@narod.ru> wrote:
>> Or just raise an exception in __init__(),..
>
>Then we are forced to handle this exception outside of class code.
Absolutely! That's the whole point. If you can't construct the class,
you *should* raise an error. The only other workable option is to leave
the target set to None, which is uglier because you don't track the
error.