functions, classes, bound, unbound?

[Bruno Desthuilliers]

7stud a écrit :
> On Mar 25, 3:09 pm, Steven Bethard <steven.beth... at gmail.com> wrote:
> 
>>Here's another way of looking at it::
>>
>>     >>> class Test(object):
>>     ...     pass
>>     ...
>>     >>> def greet():
>>     ...     print 'Hello'
>>     ...
>>
>>>>Test.greet = greet
>>>>Test.greet
>>
>>  <unbound method Test.greet>
> 
> 
> Interesting.  After playing around with that example a bit and finally
> thinking I understood bound v. unbound, I found what appears to be an
> anomaly:
> ------------
> class Test(object):
>         pass
> 
> def greet(x):
>         print "hello"
> 
> Test.func = greet
> print Test.func
> 
> t = Test()
> print t.func
> 
> def sayBye(x):
>         print "bye"
> 
> t.bye = sayBye
> print t.bye
> ------------output:
> <unbound method Test.greet>
> <bound method Test.greet of <__main__.Test object at 0x6dc50>>
> <function sayBye at 0x624b0>
> 
> Why doesn't t.bye cause a method object to be created?
> 

(technical answer - I leave to some guru the care to explain the 
motivations for this behaviour)

Because t.bye is an attribute of instance t, not of class Test. A 
descriptor object has to be a class attribute for the descriptor 
protocol to be invoked.


Note that you can override __getattribute__ to change this - but this is 
somewhat hackish, and will probably slow down things quite a bit.

FWIW, if you want to dynamically add a method to an instance:
   inst.method = func.__get__(inst, inst.__class__)