The first query is quite clear: Oracle read all the table, sort all the rows. Memory usage depends on Oracle version as different version of Oracle uses different sort algorithms, as the column is of 6 bytes (I’m assuming 1 char=8 bit) then you have more or less 6*1.000.000, so 6M of RAM, but if some kind of “hash sort” is used, this can be lower.

As stated by many here Oracle (from 9 and up?) can do a “SORT ORDER BY STOPKEY” so in the second case, it will use an array of dimension 10 (the “rownum” we want in the example) and sort/replace only 10 rows (full table scan still apply). Memory used 10*6=60 bytes.

Bye,
Antonio

You are killing me. :)
I just tried to run the test and the results left me “shattered”… Here are the details:
First the spool file:
======================================================
SQL> select * from v$version ;

BANNER
—————————————————————-
Oracle Database 10g Enterprise Edition Release 10.2.0.1.0 – Prod
PL/SQL Release 10.2.0.1.0 – Production
CORE 10.2.0.1.0 Production
TNS for Linux: Version 10.2.0.1.0 – Production
NLSRTL Version 10.2.0.1.0 – Production

SQL> create table t (sortcode varchar2(10)) ;

Table created.

SQL> REM I am trying to populate the table with the specified criteria.
SQL> insert /*+ APPEND */ into t select dbms_random.string(‘a’,1)||ROUND(dbms_random.value(1,420000), 0) from dual connect by level commit ;

Commit complete.

SQL> select max(cnt) from (select count(*) cnt from t group by sortcode) ;

MAX(CNT)
———-
4

SQL> REM Now I will enable the trace as well as check V$MYSTAT for no. of rows sorted
SQL> alter session set events ’10046 trace name context forever, level 12′ ;

Session altered.

SQL> select b.name, a.value from V$MYSTAT a, V$STATNAME b where a.statistic# = b.statistic# and b.name like ‘sort%’ ;

NAME VALUE
—————————————————————- ———-
sorts (memory) 204
sorts (disk) 0
sorts (rows) 1159

SQL> REM first query in the quiz

SQL> select sortcode from t order by sortcode ;

– Output removed

SQL> select b.name, a.value from V$MYSTAT a, V$STATNAME b where a.statistic# = b.statistic# and b.name like ‘sort%’ ;

NAME VALUE
—————————————————————- ———-
sorts (memory) 205
sorts (disk) 0
sorts (rows) 1001159

SQL> REM now another query
SQL> select sortcode from (select sortcode from t order by sortcode) where rownum <= 10
;
– Output removed

SQL> select b.name, a.value from V$MYSTAT a, V$STATNAME b where a.statistic# = b.statistic# and b.name like ‘sort%’ ;

NAME VALUE
—————————————————————- ———-
sorts (memory) 206
sorts (disk) 0
sorts (rows) 2001159

SQL> spool off
=====================================================

The V$ views show that both the sqls sorted 1 million rows (but is it so?)

So then the trace details:
=====================================================
select sortcode
from
t order by sortcode

call count cpu elapsed disk query current rows
——- —— ——– ———- ———- ———- ———- ———-
Parse 1 0.00 0.00 0 1 0 0
Execute 1 0.00 0.00 0 0 0 0
Fetch 66668 5.70 5.09 0 1768 0 1000000
——- —— ——– ———- ———- ———- ———- ———-
total 66670 5.70 5.09 0 1769 0 1000000

Misses in library cache during parse: 1
Optimizer mode: ALL_ROWS
Parsing user id: 52

Rows Row Source Operation
——- —————————————————
1000000 SORT ORDER BY (cr=1768 pr=0 pw=0 time=9759508 us)
1000000 TABLE ACCESS FULL T (cr=1768 pr=0 pw=0 time=10000070 us)

Elapsed times include waiting on following events:
Event waited on Times Max. Wait Total Waited
—————————————- Waited ———- ————
SQL*Net message to client 66668 0.00 0.32
SQL*Net message from client 66668 0.20 127.59
********************************************************************************

select sortcode
from
(select sortcode from t order by sortcode) where rownum <= 10

call count cpu elapsed disk query current rows
——- —— ——– ———- ———- ———- ———- ———-
Parse 1 0.00 0.00 0 2 0 0
Execute 1 0.00 0.00 0 0 0 0
Fetch 2 0.34 0.34 0 1768 0 10
——- —— ——– ———- ———- ———- ———- ———-
total 4 0.34 0.34 0 1770 0 10

Misses in library cache during parse: 1
Optimizer mode: ALL_ROWS
Parsing user id: 52

Rows Row Source Operation
——- —————————————————
10 COUNT STOPKEY (cr=1768 pr=0 pw=0 time=343142 us)
10 VIEW (cr=1768 pr=0 pw=0 time=343058 us)
10 SORT ORDER BY STOPKEY (cr=1768 pr=0 pw=0 time=343025 us)
1000000 TABLE ACCESS FULL T (cr=1768 pr=0 pw=0 time=10000088 us)

Elapsed times include waiting on following events:
Event waited on Times Max. Wait Total Waited
—————————————- Waited ———- ————
SQL*Net message to client 2 0.00 0.00
SQL*Net message from client 2 0.00 0.00
=====================================================

The "Row-source Operation" section for second query shows "Sort order by stopkey" step producing only 10 rows. Does that not mean sort took place only on 10 rows?
But then which one is correct? :(

p.s. Sometimes, I think I should believe in "Don Burleson" and "Don Lewis" (from oracle forums) argument that Oracle is software and not a science…
:)

E.g. he says “I have a table with one million rows, there are no indexes on the table”, but what if the table was the index (IOT)?

Since SORTCODE “has no nulls” it could be the leading column of a composite primary key – so an IOT is a possible construct to avoid sort operations.

If you use now a suitable NLS_SORT = binary setting for the assumed CHAR(6) column, both statements could go along without any sort operation at all and no memory required for this.

What about an indexed cluster? Again the index is not on the table, but on the cluster… It’s a bit splitting hairs but just throwing out some ideas.

It is correct that with a heap table the second query can use a “SORT ORDER BY STOPKEY” operation that only needs to keep the top N rows as it goes along the unsorted row source and therefore requires a far smaller workarea size for the sort operation.

Randolf

PS: And no, my presentation above doesn’t (directly) cover this particular question

1) How many rows are sorted in each of these two queries

Query-1 will sort all 1M rows while the query-2 will sort ONLY 10 rows. Both the queries will read all 1M rows but query-2 will be able to limit number of rows to sort by using “SORT ORDER BY STOPKEY” step.

Rows Row Source Operation
——- —————————————————
10 COUNT STOPKEY (cr=1645 pr=701 pw=0 time=537221 us)
10 VIEW (cr=1645 pr=701 pw=0 time=537190 us)
10 SORT ORDER BY STOPKEY (cr=1645 pr=701 pw=0 time=537169 us)
1000000 TABLE ACCESS FULL T1 (cr=1645 pr=701 pw=0 time=4000117 us)

2) roughly how much memory would you expect Oracle to use ?

Query-1 uses around 12MB of memory while query-2 uses less than 1 MB. (from v$sesstat view)

Jonathan, this seems not fair. You are enjoying your holidays and making us work on this assignment. ;)

Damir Vadas
http://damir-vadas.blogspot.com

In the first case, all the 1 million rows will be sorted.

In the second case, the sort area will contain 10 rows sorted. Each row in the table will be traversed but only 10 rows will be sorted at a time, with the new row included or discarded,depending on the sortcode value.

Not sure about the exact calucualtion of memory needed but case 1 will take much . In case 2, sorting 10 rows in RAM will be far less.