using super

[Jean-Michel Pichavant]

Duncan Booth wrote:
> Jean-Michel Pichavant <jeanmichel at sequans.com> wrote:
>
>   
>>>>> class SubClass(Base):
>>>>>     colour = "Red"
>>>>>     def parrot(self):
>>>>>         """docstring for Subclass"""
>>>>>         return super(Subclass, self).parrot()
>>>>>           
>> I'm not a big fan of super, but I'm still wondering if
>>
>>     
>>>>> return super(self.__class__, self).parrot()
>>>>>           
>> would have made it.
>>     
>
> No, it wouldn't.
>
>   
>> What if Subclass has more than one base class ?
>>     
>
> super() will work in that case provided you specify the current class. It 
> never works correctly if you specify the class of self.
>
> Consider another level of subclasses and it may be clearer:
>
>   
>>>> class Base(object):
>>>>         
> 	def parrot(self):
> 		print "Base.parrot"
>
> 		
>   
>>>> class SubClass(Base):
>>>>         
> 	def parrot(self):
> 		print "SubClass.parrot"
> 		return super(SubClass, self).parrot()
>
> 	
>   
>>>> class SubSubClass(SubClass):
>>>>         
> 	def parrot(self):
> 		print "SubSubClass.parrot"
> 		return super(SubSubClass, self).parrot()
>
> 	
>   
>>>> SubSubClass().parrot()
>>>>         
> SubSubClass.parrot
> SubClass.parrot
> Base.parrot
>
>   
>>>> class SubClass(Base):
>>>>         
> 	def parrot(self):
> 		print "SubClass.parrot"
> 		return super(self.__class__, self).parrot()
>
> 	
>   
>>>> class SubSubClass(SubClass):
>>>>         
> 	def parrot(self):
> 		print "SubSubClass.parrot"
> 		return super(self.__class__, self).parrot()
>
>   
>>>> SubSubClass().parrot()
>>>>         
> SubClass.parrot
> SubClass.parrot
> SubClass.parrot
> SubClass.parrot
> SubClass.parrot
> SubClass.parrot
> SubClass.parrot
> SubClass.parrot
> SubClass.parrot
> ... (you're in an infinite loop now) ...
>   
I see.

Then is there a reason why
>>> return super(Subclass, self).parrot()
would be prefered over the "classic" 
>>> return Base.parrot(self) 
?

Or is it just a matter of preference ?

JM