return type question - operator oveloading

[Chris Liechti]

arthur.siegel at rsmi.com wrote in
news:mailman.1016322689.22392.python-list at python.org: 
> Given:
> 
> class Complex(complex):
>     def __mul__(self,other):
>        other=Complex(other)
>        t = complex.__mul__(self,other)
>        return Complex(t.real,t.imag)
>     __rmul__ = __mul__
> 
>     def __add__(self,other):
>        other=Complex(other)
>        return
>        Complex(self.real.__add__(other.real),
>            self.imag.__add__(other.imag)) 
>     __radd__ = __add__
> 
> Then:
> 
> print type(Complex(5,4) * 7)
>>><class '__main__.Complex'>
> print type(7 * Complex(5,4))
>>><class '__main__.Complex'>
> print type(Complex(5,4) + 7)
>>><class '__main__.Complex'>
> 
> But:
> 
> print type(7 + Complex(5,4))
>>><type 'complex'>

yes i get the same results (py 2.2).

> Multiple choice:
>   That the result at But is a surprise to me because I am missing:
>       1)Something obvious about __radd__ or general classic syntax
>       2)Something related to new style classes
>       3)Other

i vote for "3) Other". i think the integer class defines __add__ for 
complex numbers but not __mul__ for complex numbers. thus when adding
"7+Complex(5,4)" your __radd__ is never called cause the int class 
indles it.

chris

-- 
Chris <cliechti at gmx.net>