__dict__ attribute for built-in types

[Steven D'Aprano]

On Fri, 28 Oct 2011 00:52:40 +0200, candide wrote:

> Le 28/10/2011 00:19, Steven D'Aprano a écrit :
>>
>> What, you think it goes against the laws of physics that nobody thought
>> to mention it in the docs?<wink>
> 
> 
> No but I'm expecting from Python documentation to mention the laws of
> Python ...

You seem to have missed my point. You said "I can't imagine" that the 
Python docs fail to mention that built-ins don't allow the addition of 
new attributes. I can, easily. The people writing the documentation are 
only human, and if they failed to mention it, oh well, perhaps they 
didn't think of it. This is hardly a surprise. Wanting to add arbitrary 
attributes to built-ins is not exactly an everyday occurrence.


>>> But beside this, how to recognise classes whose object doesn't have a
>>> __dict__ attribute ?
>>
>> The same way as you would test for any other attribute.
>>
>>>>> hasattr(42, '__dict__')
>> False
> 
> OK but I'm talking about classes, not instances  : 42 has no __dict__
> attribute but, may be, 43 _has_ such attribute, who knows in advance ?
> ;)

True, it is theoretically possible that (say) only odd numbers get a 
__dict__, or primes, or the smallest multiple of seventeen larger than 
the natural logarithm of a googol (10**100). But it's a safe bet that 
nothing so arbitrary will happen.

Dunder attributes ("Double UNDERscore") like __dict__ are reserved for 
use by Python, and __dict__ has known semantics. You can safely assume 
that either *all* instances of a type will have a __dict__, or *no* 
instances will have one. If some class violates that, oh well, your code 
can't be expected to support every badly-designed stupid class in the 
world.

Also, keep in mind the difference between a *class* __dict__ and an 
*instance* __dict__. 

>>> hasattr(int, '__dict__')  # Does the int class/type have a __dict__?
True
>>> hasattr(42, '__dict__')  # Does the int instance have a __dict__?
False



-- 
Steven