a question on __slots__ (and example from Nutshell)

Porky Pig Jr porky_pig_jr at my-deja.com
Wed Jun 2 23:58:08 CEST 2004


Hello, I"m still learning Python, but going through the Ch 5 OOP of
Nutshell book. There is discussion on __slots__, and my understanding
from reading this section is that if I have a class Rectangle (as
defined in some prior sections), and then I provide definition

class OptimizedRectangle(Rectangle):
     __slots__ = 'width', 'heigth'

I can use the instance of OptimizedRectangle, say, x, with 'width' and
'heigth', but (quoting the book) 'any attempt to bind on x any
attribute whose name is not in C.__slots__ raises an exception.

Alas, something goes wrong here, but I *can* bind any arbitrary
attribute, and seems like the instance of OptimizedRectangle does have
its own __dict__ where those attributes are kept. I'm using the latest
stable version of Python (2.3.4, I believe). Here is a session from
IDLE:

Here I'm defining the class with __slots__:
 
>>> class OptimizedRectangle(Rectangle):
	__slots__ = 'width', 'heigth'

	
and create its instance, 'ropt':
>>> ropt = OptimizedRectangle(4,5)

Note that __dict__ is still there:
>>> ropt.__dict__
{}

so I can define some arbitrary variable, say, newarea:

>>> ropt.newarea = 15

which goes into the dictionary

>>> ropt.__dict__
{'newarea': 15}

whereas width and heigth are still kept in __slots__:
>>> ropt.__slots__
('width', 'heigth')

My impression is that once __slots__ are defined, __dict__ is
effectively disabled - but in this example it's alive and well. Am I
missing anything here?

TIA.



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