postgresql - 当我第二次使用完全相同的参数调用 IMMUTABLE 函数时，为什么计划时间加倍？

Question

相比:

在第一种情况下，我调用相同的 find_period具有相同的论点。因为函数是IMMUTABLE我想它的计划将被重用(相同的功能，相同的参数，相同的计划)，但似乎没有重复使用 为什么？
源函数:

CREATE OR REPLACE FUNCTION "find_period" (in _start timestamptz, in _interval interval, in _target timestamptz)
 RETURNS tstzrange
 LANGUAGE sql
 IMMUTABLE RETURNS NULL ON NULL INPUT
 AS $$
    SELECT CASE
    WHEN _interval = INTERVAL '00:00:00' THEN
      tstzrange( _start, _start, '[]' )
    ELSE (
      SELECT CASE WHEN max( date ) = _target
        THEN tstzrange( max( date ) -_interval, max( date ) )
        ELSE tstzrange( max( date ), max( date ) +_interval )
      END
      FROM generate_series( _start, _target, _interval ) t (date )
    ) END
    WHERE _start < _target  OR  _interval = INTERVAL '00:00:00'
$$

并查询:

EXPLAIN ANALYZE SELECT find_period( 
   '2020-04-03',  
   INTERVAL '1day', 
   '9999-01-01' 
)

UPD
与 EXPLAIN ( ANALYZE, buffers, timing )第一次调用的缓冲区是 12808 VS 6404第二:

db=> select version();
                                                     version                              
------------------------------------------------------------------------------------------
 PostgreSQL 13.1 (Debian 13.1-1.pgdg100+1) on x86_64-pc-linux-gnu, compiled by gcc (Debian
(1 row)

db=> EXPLAIN ( ANALYZE, buffers, timing ) SELECT find_period( 
   '2020-04-03',  
   INTERVAL '1day', 
   '9999-01-01' 
), find_period( 
db(>    '2020-04-03',  
db(>    INTERVAL '1day', 
db(>    '9999-01-01' 
db(> );
                                     QUERY PLAN                                      
-------------------------------------------------------------------------------------
 Result  (cost=0.00..0.01 rows=1 width=64) (actual time=0.003..0.003 rows=1 loops=1)
 Planning:
   Buffers: temp read=12808 written=12808
 Planning Time: 3215.465 ms
 Execution Time: 0.023 ms
(5 rows)

db=> EXPLAIN ( ANALYZE, buffers, timing ) SELECT find_period( 
db(>    '2020-04-03',  
db(>    INTERVAL '1day', 
db(>    '9999-01-01' 
db(> );
                                     QUERY PLAN                                      
-------------------------------------------------------------------------------------
 Result  (cost=0.00..0.01 rows=1 width=32) (actual time=0.002..0.002 rows=1 loops=1)
 Planning:
   Buffers: shared hit=97, temp read=6404 written=6404
 Planning Time: 1583.670 ms
 Execution Time: 0.017 ms
(5 rows)

Answer 1

与您期望的不同，PostgreSQL 两次调用该函数，因为它不会花费额外的计划工作来检查您是否使用相同的常量两次调用相同的函数。
做合理的事情:

SELECT x, x
FROM find_period('2020-04-03', INTERVAL '1day', '9999-01-01' ) AS f(x);