Re: Can self crush itself?
Why would you want to do that in the first place? 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
Ok ok Of course, it's a local name; -- just my silly slip. And seems it belongs to no dict[]... Just an internal volatile elf -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
On Thu, Nov 26, 2009 at 8:04 AM, Gregory Ewing greg.ew...@canterbury.ac.nzwrote: n00m wrote: 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. +1 QOTW Francesco -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
n00m wrote: Ok ok Of course, it's a local name; -- just my silly slip. And seems it belongs to no dict[]... Just an internal volatile elf 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. -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
In article 7ace3de0-4588-4b7d-9848-d97298717...@z41g2000yqz.googlegroups.com, n00m n...@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. -- Aahz (a...@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. -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
Then how can we destroy the 3rd instance, 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
On Wed, Nov 25, 2009 at 1:46 AM, n00m n...@narod.ru wrote: 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
n00m n...@narod.ru writes: 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
In article mailman.965.1259143133.2873.python-l...@python.org, Chris Rebert c...@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 (a...@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. -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
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__ = RESTART __main__ Traceback (most recent call last): File D:\Python25\zewrt.py, line 19, in module print __dict__ NameError: name '__dict__' is not defined -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
n00m wrote: 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. -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
On Wed, 25 Nov 2009 20:09:25 -0500, Terry Reedy wrote: n00m wrote: 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. 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
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)
--
http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
On Wed, 25 Nov 2009 18:39:09 -0800, n00m wrote: 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
n00m wrote: 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. -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
n00m wrote: 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 -- http://mail.python.org/mailman/listinfo/python-list
Can self crush itself?
Why does h instance stay alive? 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
n00m n...@narod.ru writes: 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Can self crush itself?
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 -- http://mail.python.org/mailman/listinfo/python-list
