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| Date | Sat, 2 Apr 2011 02:17:54 +1100 |
| Subject | Re: The Magick of __call__ (Or, Digging Deeper Than I Ought To) |
| From | Chris Angelico <rosuav@gmail.com> |
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On Sat, Apr 2, 2011 at 2:07 AM, Corey Richardson <kb1pkl@aim.com> wrote: > All callables (things you can foo(bar)) are really just objects that > implement the __call__ method, as far as I understand. Well then, that > would appear to make methods themselves callable, so let's do a little > playing around... Interesting. >>> def Foo(): pass >>> Foo <function Foo at 0x011A51F0> >>> Foo.__call__ <method-wrapper '__call__' of function object at 0x011A51F0> >>> Foo.__call__.__call__ <method-wrapper '__call__' of method-wrapper object at 0x011BF850> >>> Foo.__call__.__call__.__call__ <method-wrapper '__call__' of method-wrapper object at 0x011BF8F0> I'd have thought that it would simply use Foo.__call__ == Foo, for simplicity. It's not. >>> a=[Foo] >>> a.append(a[-1].__call__) # repeat above line as many times as desired - builds a list of the recursion >>> a[-1]==a[-2] False So at no point does it ever actually cycle onto itself. >>> import sys >>> [sys.getsizeof(q) for q in a] [60, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32] The function object takes 60 bytes, everything else 32. >>> [q.__hash__() for q in a] [18502128, 29287592, 10666040, 29287592, 10666040, 29287592, 10666040, 29287592, 10666040, 29287592, 10666040, 29287592, 10666040, 29287592] None of the elements compare equal to each other, but apparently there's only two hash values possible for them! Fascinating. I don't know that it's any use, but fascinating! ChrisA
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Re: The Magick of __call__ (Or, Digging Deeper Than I Ought To) Chris Angelico <rosuav@gmail.com> - 2011-04-02 02:17 +1100
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