Thanks for working on the previous comments. The code really looks good now.

On Fri, Jun 23, 2017 at 2:29 PM, Amit Langote
<> wrote:
>> Don't we need an exclusive lock to
>> make sure that the constraints are not changed while we are validating those?
> If I understand your question correctly, you meant to ask if we don't need
> the strongest lock on individual partitions while looking at their
> constraints to prove that we don't need to scan them.  We do and we do
> take the strongest lock on individual partitions even today in the second
> call to find_all_inheritors().  We're trying to eliminate the second call
> here.

The comment seems to imply that we need strongest lock only when we
"scan" the table/s.

Some more comments on 0001
-     * Prevent circularity by seeing if rel is a partition of attachRel. (In
+     * Prevent circularity by seeing if rel is a partition of attachRel, (In
      * particular, this disallows making a rel a partition of itself.)
The sentence outside () doesn't have a full-stop. I think the original
construct was better.

+     * We want to avoid having to construct this list again, so we request the
"this list" is confusing here since the earlier sentence doesn't mention any
list at all. Instead we may reword it as "We will need the list of children
later to check whether any of those have a row which would not fit the
partition constraints. So, take the strongest lock ..."

     * XXX - Do we need to lock the partitions here if we already have the
     * strongest lock on attachRel?  The information we need here to check
     * for circularity cannot change without taking a lock on attachRel.
I wondered about this. Do we really need an exclusive lock to check whether
partition constraint is valid? May be we can compare this condition with ALTER
TABLE ... ADD CONSTRAINT since the children will all get a new constraint
effectively. So, exclusive lock it is.

        Assert(linitial_oid(attachRel_children) ==
        if (attachRel->rd_rel->relkind == RELKIND_PARTITIONED_TABLE)
            attachRel_children = list_delete_first(attachRel_children);
Is it necessary for this code to have OID of the relation being attached as the
first one? You could simply call list_delete_oid() instead of
list_delete_first(). If for any reason find_all_inheritors() changes the output
order, this assertion and code would need a change.\

>> Comments on 0002 patch.
>> Thanks for the refactoring. The code looks really good now.
> Thanks.
>> The name skipPartConstraintValidation() looks very specific to the case at
>> hand. The function is really checking whether existing constraints on the 
>> table
>> can imply the given constraints (which happen to be partition constraints). 
>> How
>> about PartConstraintImpliedByRelConstraint()? The idea is to pick a general
>> name so that the function can be used for purposes other than skipping
>> validation scan in future.
> I liked this idea, so done.

+ * skipPartConstraintValidation
+PartConstraintImpliedByRelConstraint(Relation partrel, List *partConstraint)
Different function names in prologue and the definition.

>>  * This basically returns if the partrel's existing constraints, which
>> returns "true". Add "otherwise returns false".
>>     if (constr != NULL)
>>     {
>>         ...
>>     }
>>     return false;
>> May be you should return false when constr == NULL (I prefer !constr, but
>> others may have different preferences.). That should save us one level of
>> indentation. At the end just return whatever predicate_implied_by() returns.
> Good suggestion, done.

+    if (predicate_implied_by(partConstraint, existConstraint, true))
+        return true;
+    /* Tough luck. */
+    return false;
why not to just return the return value of predicate_implied_by() as is?

With this we can actually handle constr == NULL a bit differently.
+    if (constr == NULL)
+        return false;
To be on safer side, return false when partConstraint is not NULL. If both the
relation constraint and partConstraint are both NULL, the first does imply the
other. Or may be leave that decision to predicate_implied_by(), which takes
care of it right at the beginning of the function.

+         * For each leaf partition, check if it we can skip the validation
An extra "it".

+         * Note that attachRel's OID is in this list.  Since we already
+         * determined above that its validation scan cannot be skipped, we
+         * need not check that again in the loop below.  If it's partitioned,
I don't see code to skip checking whether scan can be skipped for relation
being attached. The loop below this comments executes for every unpartitioned
table in the list of OIDs returned. Thus for an unpartitioned relation being
attached, it will try to compare the constraints again. Am I correct?

+         * comparing it to similarly-processed qual clauses, and may fail
There are no "qual clauses" here only constraints :).

The testcase looks good to me.
Best Wishes,
Ashutosh Bapat
EnterpriseDB Corporation
The Postgres Database Company

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