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Thomas Jollans thomas at jollybox.de
Wed Aug 11 22:24:35 CEST 2010


On Wednesday 11 August 2010, it occurred to Eric J. Van der Velden to exclaim:
> Hello,
> 
> I have these types,
> 
> class A:
>         def __init__(s):
>                 super().__init__()
>                 print("A")
> class B(A):
>         def __init__(s):
>                 super().__init__()
>                 print("B")
> class C(A):
>         def __init__(s):
>                 super().__init__()
>                 print("C")
> class D(B,C):
>         def __init__(s):
>                 super().__init__()
>                 print("D")
> 
> If I do (in 3.1)
> 
> >>> d=D()
> 
> A
> C
> B
> D
> 
> Why this order? I thought, first to D, then B, then A. He prints "A".
> He comes back in B and prints "B". He goes to C. Then somehow he
> doesn't go again to A. He prints "C". Then back to D and prints "D".

Think again about what you're seeing here. You're printing AFTER the call to 
super().__init__(). That means that it first walks the inheritance hierarchy, 
and then prints -- your trace is "the wrong way around"

So, what happens is: you call D(). In it, super() delegates to B(). Which has 
super delegate to C(), which then has super() delegate finally to A().

D, B, C, A

You say you were expecting D, B, A -- but what of C? You also imply that you 
would have expected two visits to A -- but that would defeat the point of 
super() -- you don't actually want one constructor to be called twice: the 
constructor almost certainly isn't written with that possibility in mind.

> 
> Thanks,
> 
> Eric J.



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