Re: row filtering for logical replication

On Tue, Jan 18, 2022 at 2:31 PM Amit Kapila <amit(dot)kapila16(at)gmail(dot)com> wrote:
>
> On Tue, Jan 18, 2022 at 8:41 AM Greg Nancarrow <gregn4422(at)gmail(dot)com> wrote:
> >
> > On Tue, Jan 18, 2022 at 2:31 AM houzj(dot)fnst(at)fujitsu(dot)com
> > <houzj(dot)fnst(at)fujitsu(dot)com> wrote:
> > >
> > > > (2) GetTopMostAncestorInPublication
> > > > Is there a reason why there is no "break" after finding a
> > > > topmost_relid? Why keep searching and potentially overwrite a
> > > > previously-found topmost_relid? If it's intentional, I think that a
> > > > comment should be added to explain it.
> > >
> > > The code was moved from get_rel_sync_entry, and was trying to get the
> > > last oid in the ancestor list which is published by the publication. Do you
> > > have some suggestions for the comment ?
> > >
> >
> > Maybe the existing comment should be updated to just spell it out like that:
> >
> > /*
> > * Find the "topmost" ancestor that is in this publication, by getting the
> > * last Oid in the ancestors list which is published by the publication.
> > */
> >
>
> I am not sure that is helpful w.r.t what Peter is looking for as that
> is saying what code is doing and he wants to know why it is so? I
> think one can understand this by looking at get_partition_ancestors
> which will return the top-most ancestor as the last element. I feel
> either we can say see get_partition_ancestors or maybe explain how the
> ancestors are stored in this list.
>

(note: I asked the original question about why there is no "break", not Peter)
Maybe instead, an additional comment could be added to the
GetTopMostAncestorInPublication function to say "Note that the
ancestors list is ordered such that the topmost ancestor is at the end
of the list". Unfortunately the get_partition_ancestors function
currently doesn't explicitly say that the topmost ancestors are
returned at the end of the list (I guess you could conclude it by then
looking at get_partition_ancestors_worker code which it calls).
Also, this leads me to wonder if searching the ancestors list
backwards might be better here, and break at the first match? Perhaps
there is only a small gain in doing that ...

Regards,
Greg Nancarrow
Fujitsu Australia