New style classes and rich comparison

[Stian Søiland]

Could anyone explain to me why this does not work as intended?

class Oldstyle: .... pass
.... old = Oldstyle()
old == 1 False old.__eq__ = lambda x: True
old == 1 True

Nice and fine.

newbie = Newstyle()
newbie == 1 False newbie.__eq__ = lambda x: True
newbie == 1 False

Doesn't work!
It seems that __eq__ is fetched directly from newbie's class, not from
newbie.
Newstyle.__eq__ = lambda self, x: True
newbie == 1 True

Adding a method __getattribute__ that prints the key and then calls
object.__getattribute__ proves this - nothing is printed for comparison.

This goes even for hash():
newbie.__hash__ = lambda x: 1337
hash(newbie) 1077345744 Newstyle.__hash__ = lambda x: 15
hash(newbie) 15

Why is this? Where is this documented?

Here's my real goal, to create a Dummy class that supports anything:

from dummy.py:

class Dummy(object):
"""An object that can do anything in the world"""
def __call__(self, *args, **kwargs):
"""Can be called with any kind of parameters, returns
a new Dummy."""
return Dummy()
def __getattribute__(self, key):
"""Any attribute created and stored on demand. All
attributes are new Dummys."""
try:
value = object.__getattribute__(self, key)
except AttributeError:
value = self()
# remember it to next time! =)
setattr(self, key, value)
return value

This dummy can be used for many things:
from dummy import Dummy
a = Dummy()
a.b <dummy.Dummy object at 0x81e02dc> a.b == a.b 1 a.b == a 0 c = a.b.anything(123)
c.dot.something <dummy.Dummy object at 0x81f3764>
Now the trouble is that __operator__ is not resolved as I would expect.
a.__add__(a) <dummy.Dummy object at 0x81ad4dc> a + a Traceback (most recent call last):
File "<stdin>", line 1, in ?
TypeError: unsupported operand types for +: 'Dummy' and 'Dummy'
Why are operators looked up in the class instead of the object? How
can I modify my Dummy class to return 'anything' even directly in the
class?

Ie. what I probably want is something like this (simulated):
Dummy.asdads

<dummy.Dummy object at 0x81ad4dc>

That way Dummy.__add__ would return a new Dummy instance, and
Dummy() + Dummy() would work as expected.

My first attempt was to manually define different __operator__-methods
in the dummy, but this got rather boring. Can I use __metaclass__ or
something?
--
Stian Søiland Being able to break security doesn't make
Trondheim, Norway you a hacker more than being able to hotwire
http://stain.portveien.to/ cars makes you an automotive engineer. [ESR]

[Rainer Deyke]

Stian Søiland wrote:

newbie.__hash__ = lambda x: 1337
hash(newbie) 1077345744 Newstyle.__hash__ = lambda x: 15
hash(newbie)

15

Why is this?

Consider:

class A(object):
def __hash__(self): return 5
print hash(A) # Which function should this call?
--
Rainer Deyke - ra*****@eldwood.com - http://eldwood.com

[Stian Søiland]

* Rainer Deyke spake thusly:

Consider:

class A(object):
def __hash__(self): return 5
print hash(A) # Which function should this call?

The hash method of A's class, ie. A.__class__.__hash__, usually
type.__hash__.

Hmmm. I think I get your point. hash and cmp and so on should call
A.__class__.hash(A) to be generic enough.

It's still confusing.. =) And how should I resolve this issue with my
Dummy class? (it's ok for the Dummy class itself to be a Dummy, but it
didn't work well to make Dummy = Dummy())

--
Stian Søiland Being able to break security doesn't make
Trondheim, Norway you a hacker more than being able to hotwire
http://stain.portveien.to/ cars makes you an automotive engineer. [ESR]

[Michael Hudson]

st***@stud.ntnu.no (Stian Søiland) writes:

Could anyone explain to me why this does not work as intended?
[snippo]
It seems that __eq__ is fetched directly from newbie's class, not from
newbie.
Yup! One of the differences between new-style and old-style classes.

To see why something has to be a bit like this, consider the
difference between a unbound method for the instance and a bound
method for the *type* of the instance...

[biggo snippo]
Can I use __metaclass__ or something?

I think that's what you want, yes.

Cheers,
mwh

--
Ignoring the rules in the FAQ: 1" slice in spleen and prevention
of immediate medical care.
-- Mark C. Langston, asr