Python's method-resolution algorithm: An implementation question

[Evan Aad]

I don't see how, since the L(B*)'s are listed in order in the argument
list: L(B1), L(B2), ..., and each L(B*) starts with B*: L(B1) = <B1,
...>, L(B2) = <B2, ...>, ...

Could you please give a counter-example?

On Tue, Aug 15, 2017 at 9:44 PM, Ian Kelly <ian.g.kelly at gmail.com> wrote:
>
> On Tue, Aug 15, 2017 at 9:56 AM, Evan Aad <oddeveneven at gmail.com> wrote:
> > According to the description of Python's method resolution order (mro)
> > (https://www.python.org/download/releases/2.3/mro/), a.k.a. C3
> > linearization (see Wikipedia), the algorithm can be described as
> > follows:
> >
> > "the linearization of C is the sum of C plus the merge of the
> > linearizations of the parents and the list of the parents."
> >
> > Symbolically, the algorithm is defined recursively thus:
> >
> > L(O) = <O>
> > L(C) = <C> + merge(L(B1),..., L(Bn), <B1,...,Bn>)
> >
> > where
> >
> > * O is the class from which every class inherits.
> > * C is a class that inherits directly from B1, ..., Bn, in this order.
> > * < and > are list delimiters.
> > * + is the list-concatenation operator.
> > * 'merge' merges its list arguments into a single list with no
> > repetitions in the manner described in the next paragraph.
> >
> > Consider the head of the first list, i.e L(B1)[0]: if it is a good
> > head, i.e. if it is not in the proper tail of any of the other lists,
> > add it to the linearization of C, and remove it from all the lists in
> > the merge. Otherwise, consider the head of the next list, etc. Repeat
> > until no more classes, or no good heads. In the latter case, it is
> > impossible to construct the merge.
> >
> >
> >
> > Why is the list <B1,...,Bn> of parents given as an argument to
> > 'merge'? Will the algorithm produce different results if this argument
> > is omitted?
>
> It ensures that the parents are linearized in that order. Without that
> list, it would be possible for B2 to appear before B1 in the MRO of C.
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