Q? Calling nearest inherited method

Guido van Rossum guido at python.org
Fri May 19 05:36:08 CEST 2000


> Laurent POINTAL wrote:
> >
> > On Wed, 17 May 2000 18:37:00 GMT, Guido van Rossum <guido at python.org>
> > wrote:
> >
> > >: There's a better way.  (I believe Jeremy Hylton first suggested this; he saw
> > >: it as an idiom in another language:)
> > >:
> > >: class C(B):
> > >:     super_doit = B.doit
> > >:     def doit(self, arg):
> > >:         self.super_doit(arg)
> > >:
> > >: This avoids the need to explicitly pass self.
> >
> > Not sure its really better...

> > 2) The biggest problem, it can go into infinite recursion...
> [details omitted]

I replied:
> Argh!  Good catch.  My suggestion: use a naming convention
> where it's called B_doit instead of super_doit.  But this *does*
> make it less elegant, and you now need to know the base class name...

Jeremy just pointed out to me that he has a very elegant solution:
he uses __super_doit = B.doit.

This nicely makes it class-specific without working the class name
into the identifier explicitly.  I think this is more elegant than
calling B.doit(self, arg).

-- 
--Guido van Rossum (home page: http://www.python.org/~guido/)



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