Is __mul__ sufficient for operator '*'?
gagsl-py2 at yahoo.com.ar
Tue Oct 20 08:52:06 CEST 2009
En Tue, 20 Oct 2009 00:59:12 -0300, Mick Krippendorf <mad.mick at gmx.de>
> Gabriel Genellina schrieb:
>> __special__ methods are searched in the type, not in the instance
>> directly. x*y looks for type(x).__mul__ (among other things)
> So I thought too, but:
> class meta(type):
> def __mul__(*args):
> return 123
> class boo(object):
> __metaclass__ = meta
> print boo.__mul__
> b = boo()
> print b * 7
> also explodes. Or am I misinterpreting the word "type" here?
This is by design; see
Methods defined on the meta-type aren't considered methods of the type;
otherwise, every object would have the methods defined in `type` itself,
because this is the metatype of every other object.
When I said "x*y looks for type(x).__mul__" the search for '__mul__' is
done in type(x) itself, and all its base types along the MRO - NOT on the
metatype, and not using getattr. There is no way to express this exact
search in Python code (that I know of), but it's more or less like
searching for '__mul__' in dir(type(x)).
In particular, __mul__ is stored in two slots (a slot is a field in a
structure holding function pointers: nb_multiply in the tp_as_number
structure, *and* sq_repeat in tp_as_sequence); defining or assigning to
__mul__ "magically" updates those internal pointers. When executing x*y,
__mul__ is retrieved directly from those pointers -- it is *not* searched
In short, you have to define the __mul__ method on the type itself or any
of its bases.
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