On Wed, Sep 22, 2021 at 1:50 PM Amit Kapila <amit.kapil...@gmail.com> wrote: > > On Wed, Sep 22, 2021 at 6:42 AM Ajin Cherian <itsa...@gmail.com> wrote: > > > > Why do you think that the second assumption (if there is an old tuple > it will contain all RI key fields.) is broken? It seems to me even > when we are planning to include unchanged toast as part of old_key, it > will contain all the key columns, isn't that true?
Yes, I assumed wrongly. Just checked. What you say is correct. > > > I think we > > still need to deform both old tuple and new tuple, just to handle this case. > > > > Yeah, but we will anyway talking about saving that cost for later if > we decide to send that tuple. I think we can further try to optimize > it by first checking whether the new tuple has any toasted value, if > so then only we need this extra pass of deforming. Ok, I will go ahead with this approach. > > > There is currently logic in ReorderBufferToastReplace() which already > > deforms the new tuple > > to detoast changed toasted fields in the new tuple. I think if we can > > enhance this logic for our > > purpose, then we can avoid an extra deform of the new tuple. > > But I think you had earlier indicated that having untoasted unchanged > > values in the new tuple > > can be bothersome. > > > > I think it will be too costly on the subscriber side during apply > because it will update all the unchanged toasted values which will > lead to extra writes both for WAL and data. > Ok, agreed. regards, Ajin Cherian Fujitsu Australia
