2. You calculate the cost to compare with jit_above_cost as:
plan->total_cost * plan->est_loops.
An alternative way might be to consider the rescan cost like cost_rescan. This should be closer for a final execution cost. However since it is hard to set a reasonable jit_above_cost, so I am feeling the current way is OK as well.
There are two observers after thinking more about this. a). due to the
rescan cost reason, plan->total_cost * plan->est_loops might be greater
than the whole plan's total_cost. This may cause users to be confused why
this change can make the plan not JITed in the past, but JITed now.
explain analyze select * from t1, t2 where t1.a = t2.a; QUERY PLAN ------------------------------------------------------------------------------------------------------------ Nested Loop (cost=0.00..154.25 rows=100 width=16) (actual time=0.036..2.618 rows=100 loops=1) Join Filter: (t1.a = t2.a) Rows Removed by Join Filter: 9900 -> Seq Scan on t1 (cost=0.00..2.00 rows=100 width=8) (actual time=0.015..0.031 rows=100 loops=1) -> Materialize (cost=0.00..2.50 rows=100 width=8) (actual time=0.000..0.010 rows=100 loops=100) -> Seq Scan on t2 (cost=0.00..2.00 rows=100 width=8) (actual time=0.007..0.023 rows=100 loops=1) Planning Time: 0.299 ms Execution Time: 2.694 ms (8 rows)
The overall plan's total_cost is 154.25, but the Materialize's JIT cost is 2.5 * 100 = 250.
b). Since the total_cost for a plan counts all the costs for its children, so if one
child plan is JITed, I think all its parents would JITed. Is this by design?
QUERY PLAN ---------------------------- Sort Sort Key: (count(*)) -> HashAggregate Group Key: a -> Seq Scan on t1
(If Seq Scan is JITed, both HashAggregate & Sort will be JITed.)
