setting a property via self.__dict__

Bengt Richter bokr at oz.net
Thu Oct 31 03:25:19 CET 2002


On 30 Oct 2002 12:51:47 -0800, fvondelft at syrrx.com (Frank von Delft) wrote:

>Duncan Booth <duncan at rcp.co.uk> wrote in message news:<Xns92B75C6BDF4Eduncanrcpcouk at rcp.co.uk>...
>
>> The property is an entry in the class's dictionary, not in the instance.
>
>
>Um...  after some mulling:  so what is the advantage of having the
>property in the class dictionary?  Because I realised it's quite happy
>in the instance dictionary, like such:
>
>>>> class A(object):
>	def __init__(self,**kwargs):
>		self.a = property(self._getA,self._setA)
>		self.__dict__.update(**kwargs)
>	def _getA(self): 
>		return self._a
>	def _setA(self,a):
>		self._a = a
>
>Really I should ask:  why do all the examples show it defined in the
>class dictionary?
>
Why don't you try your code and see how it behaves?

 >>> class A(object):
 ...     def __init__(self,**kwargs):
 ...         self.a = property(self._getA, self._setA)
 ...         self.__dict__.update(**kwargs)
 ...     def _getA(self):
 ...         return self._a
 ...     def _setA(self,a):
 ...         self._a = a
 ...
 >>> a=A(x=1,y='two')
 Traceback (most recent call last):
   File "<stdin>", line 1, in ?
   File "<stdin>", line 4, in __init__
 TypeError: update() takes no keyword arguments

Ok, minor typo, to give you the benefit.

 >>> class A(object):
 ...     def __init__(self,**kwargs):
 ...         self.a = property(self._getA, self._setA)
 ...         self.__dict__.update(kwargs)
 ...     def _getA(self):
 ...         return self._a
 ...     def _setA(self,a):
 ...         self._a = a
 ...
 >>> a=A(x=1,y='two')
 >>> vars(a)
 {'a': <property object at 0x007F51A0>, 'y': 'two', 'x': 1}
 >>> a.a
 <property object at 0x007F51A0>

Did you expect to trigger _getA there?

You probably didn't try this?


 

Regards,
Bengt Richter



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