From: | Amit Khandekar <amitdkhan(dot)pg(at)gmail(dot)com> |
---|---|
To: | Pavel Stehule <pavel(dot)stehule(at)gmail(dot)com> |
Cc: | PostgreSQL Hackers <pgsql-hackers(at)postgresql(dot)org>, Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>, Peter Eisentraut <peter(dot)eisentraut(at)2ndquadrant(dot)com> |
Subject: | Re: calling procedures is slow and consumes extra much memory against calling function |
Date: | 2020-06-17 05:52:04 |
Message-ID: | CAJ3gD9dGg4oPTPX+Ksb8NNK4GJjqMRSrucBeG_j5__vJ=otCHg@mail.gmail.com |
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Lists: | pgsql-hackers |
On Wed, 10 Jun 2020 at 17:12, Pavel Stehule <pavel(dot)stehule(at)gmail(dot)com> wrote:
> st 10. 6. 2020 v 12:26 odesílatel Amit Khandekar <amitdkhan(dot)pg(at)gmail(dot)com> napsal:
>> Could you show an example testcase that tests this recursive scenario,
>> with which your earlier patch fails the test, and this v2 patch passes
>> it ? I am trying to understand the recursive scenario and the re-use
>> of expr->plan.
>
>
> it hangs on plpgsql tests. So you can apply first version of patch
>
> and "make check"
I could not reproduce the make check hang with the v1 patch. But I
could see a crash with the below testcase. So I understand the purpose
of the plan_owner variable that you introduced in v2.
Consider this recursive test :
create or replace procedure p1(in r int) as $$
begin
RAISE INFO 'r : % ', r;
if r < 3 then
call p1(r+1);
end if;
end
$$ language plpgsql;
do $$
declare r int default 1;
begin
call p1(r);
end;
$$;
In p1() with r=2, when the stmt "call p1(r+1)" is being executed,
consider this code of exec_stmt_call() with your v2 patch applied:
if (expr->plan && !expr->plan->saved)
{
if (plan_owner)
SPI_freeplan(expr->plan);
expr->plan = NULL;
}
Here, plan_owner is false. So SPI_freeplan() will not be called, and
expr->plan is set to NULL. Now I have observed that the stmt pointer
and expr pointer is shared between the p1() execution at this r=2
level and the p1() execution at r=1 level. So after the above code is
executed at r=2, when the upper level (r=1) exec_stmt_call() lands to
the same above code snippet, it gets the same expr pointer, but it's
expr->plan is already set to NULL without being freed. From this
logic, it looks like the plan won't get freed whenever the expr/stmt
pointers are shared across recursive levels, since expr->plan is set
to NULL at the lowermost level ? Basically, the handle to the plan is
lost so no one at the upper recursion level can explicitly free it
using SPI_freeplan(), right ? This looks the same as the main issue
where the plan does not get freed for non-recursive calls. I haven't
got a chance to check if we can develop a testcase for this, similar
to your testcase where the memory keeps on increasing.
-Amit
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