Method resolution for super(Class, obj).

ddtl this at is.invalid
Thu Sep 7 21:05:10 CEST 2006


On 7 Sep 2006 10:42:54 -0700, in comp.lang.python you wrote:

>Let's examine what the mro order is for class D:
>>>> D.mro()
>[<class '__main__.D'>, <class '__main__.B'>, <class '__main__.C'>,
><class '__mai
>n__.A'>, <type 'object'>]
>
>When you call d.met(), the call dispatches to the D.met() method.
>After printing out 'D.met', you use super() to get the next class in
>the mro order, and call that class's met method.
>
>As shown with the mro(), the class after D is B.  So B.met() is called.
> Normally, we would be done.  But take a look at B's method!
>
>> class B(A):
>>     def met(self):
>>         print 'B.met'
>>         super(B,self).met()
>
>B.met calls super, and invokes the next met method!  So, the code does
>exactly what you've asked it to do, and searches for the next class
>after B in the mro list: class C.

But when super(B,self).met() is invoked, isn't it supposed to look
at MRO order of *B*, which is:

(<class '__main__.B'>, <class '__main__.A'>, <type 'object'>)

and B doesn't have any relation with C, that is: A's met() is the to
be called as a result. In effect, what you say impies that a call to
super() is context dependant - if super(B,self).met() is invoked as
a result of a call to D().met(), the effect is different from the effect 
of a call to B().met(). But a documentation of a super() doesn't mention
anything like that (or at least I didn't find it), and it seems a 
significant piece of information. Doesn't it imply that there should 
be another explanation?

ddtl.



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