En Wed, 28 Oct 2009 23:12:53 -0300, Jess Austin <jess.aus...@gmail.com>
escribió:
class mySet(set):
... def __eq__(self, other):
... print "called mySet.__eq__()!"
... if isinstance(other, (set, frozenset)):
... return True
... return set.__eq__(self, other)
...
Now I want the builtin set and frozenset types to use the new
__eq__() with mySet symmetrically.
mySet() == set([1])
called mySet.__eq__()!
True
mySet() == frozenset([1])
called mySet.__eq__()!
True
set([1]) == mySet()
called mySet.__eq__()!
True
frozenset([1]) == mySet()
False
frozenset doesn't use mySet.__eq__() because mySet is not a subclass
of frozenset as it is for set. [...failed attempts to inherit from both
set and frozenset...]
I must redefine __eq__(), and I'd like to be able to compare
instances of the class to both set and frozenset instances.
We know the last test fails because the == logic fails to recognize mySet
(on the right side) as a "more specialized" object than frozenset (on the
left side), because set and frozenset don't have a common base type
(although they share a lot of implementation)
I think the only way would require modifying tp_richcompare of
set/frozenset objects, so it is aware of subclasses on the right side.
Currently, frozenset() == mySet() effectively ignores the fact that mySet
is a subclass of set.
--
Gabriel Genellina
--
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