Re: [HACKERS] asynchronous execution

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От Amit Langote
Тема Re: [HACKERS] asynchronous execution
Дата
Msg-id 63a5a01c-2967-83e0-8bbf-c981404f529e@lab.ntt.co.jp
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Ответ на Re: [HACKERS] asynchronous execution  (Kyotaro HORIGUCHI <horiguchi.kyotaro@lab.ntt.co.jp>)
Ответы Re: [HACKERS] asynchronous execution  (Kyotaro HORIGUCHI <horiguchi.kyotaro@lab.ntt.co.jp>)
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Hi,

On 2017/06/29 13:45, Kyotaro HORIGUCHI wrote:
> Thank you for looking this.
> 
> At Wed, 28 Jun 2017 10:23:54 +0200, Antonin Houska wrote:
>> Kyotaro HORIGUCHI <horiguchi.kyotaro@lab.ntt.co.jp> wrote:
>>
>>> The patch got conflicted. This is a new version just rebased to
>>> the current master. Furtuer amendment will be taken later.
>>
>> Can you please explain this part of make_append() ?
>>
>> /* Currently async on partitioned tables is not available */
>> Assert(nasyncplans == 0 || partitioned_rels == NIL);
>>
>> I don't think the output of Append plan is supposed to be ordered even if the
>> underlying relation is partitioned. Besides ordering, is there any other
>> reason not to use the asynchronous execution?
> 
> It was just a developmental sentinel that will remind me later to
> consider the declarative partitions since I didn't have an idea
> of the differences (or the similarity) between appendrels and
> partitioned_rels. It is never to say the condition cannot
> make. I'll check it out and will support partitioned_rels sooner.
> Sorry for having left it as it is.

When making an Append for a partitioned table, among the arguments passed
to make_append(), 'partitioned_rels' is a list of RT indexes of
partitioned tables in the inheritance tree of which the aforementioned
partitioned table is the root.  'appendplans' is a list of subplans for
scanning the leaf partitions in the tree.  Note that the 'appendplans'
list contains no members corresponding to the partitioned tables, because
we don't need to scan them (only leaf relations contain any data).

The point of having the 'partitioned_rels' list in the resulting Append
plan is so that the executor can identify those relations and take the
appropriate locks on them.

Thanks,
Amit




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