I see the same behaviour, moreover when I change class Quantity to a classic class (removing '(object)'), it works as expected. (i.e. Quanitity.__add__() is called after the fourth print. I run Python 2.6.2 on Vista.
On Sun, Oct 18, 2009 at 7:50 AM, Darren Dale <dsdal...@gmail.com> wrote: > According to http://docs.python.org/reference/datamodel.html , the > reflected operands functions like __radd__ "are only called if the > left operand does not support the corresponding operation and the > operands are of different types. [3] For instance, to evaluate the > expression x - y, where y is an instance of a class that has an > __rsub__() method, y.__rsub__(x) is called if x.__sub__(y) returns > NotImplemented." > > Consider the following simple example: > > ========================== > class Quantity(object): > > def __add__(self, other): > return '__add__ called' > > def __radd__(self, other): > return '__radd__ called' > > class UnitQuantity(Quantity): > > def __add__(self, other): > return '__add__ called' > > def __radd__(self, other): > return '__radd__ called' > > print 'Quantity()+Quantity()', Quantity()+Quantity() > print 'UnitQuantity()+UnitQuantity()', UnitQuantity()+UnitQuantity() > print 'UnitQuantity()+Quantity()', UnitQuantity()+Quantity() > print 'Quantity()+UnitQuantity()', Quantity()+UnitQuantity() > ========================== > > The output should indicate that __add__ was called in all four trials, > but the last trial calls __radd__. Interestingly, if I comment out the > definition of __radd__ in UnitQuantity, then the fourth trial calls > __add__ like it should. > > I think this may be an important bug. I'm running Python 2.6.4rc1 > (r264rc1:75270, Oct 13 2009, 17:02:06) an ubuntu Karmic. Is it a known > issue, or am I misreading the documentation? > > Thanks, > Darren > _______________________________________________ > Python-Dev mailing list > Python-Dev@python.org > http://mail.python.org/mailman/listinfo/python-dev > Unsubscribe: > http://mail.python.org/mailman/options/python-dev/ehsanamiri%40gmail.com >
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