On Fri, 2020-06-19 at 13:27 +0530, Srinivasa T N wrote: > > > How can I see the output of hash function that is used internally? > > > > In the case of "integer", the hash function is "pg_catalog"."hashint4". > > I guess output formatting is wrong, any help? > > postgres=# select pg_catalog.hashint4(7); > hashint4 > ------------ > -978793473 > (1 row) No, that is fine. Just take the result mod 10 if that is how hash partitioning was defined: select pg_catalog.hashint4(7) - floor(pg_catalog.hashint4(7) / 10.0) * 10; ?column? ---------- 7 (1 row) So that should end up in the eighth partition. You have no choice which hash function to use for partitioning. Yours, Laurenz Albe -- Cybertec | https://www.cybertec-postgresql.com