Friday, March 26, 2010

Difference between Session and Process

Someone asked me the other day the difference between process and session in Oracle. Are they the same thing? Well, why then there are two parameters in the initialization parameter file – sessions and processes? Is it possible that a process does not have a corresponding session? Conversely, is it possible to have a session without a process?

And, furthermore, the processes parameter is larger than the sessions. So, are there more processes than sessions? To answer that, she got the count from v$session and v$process:

SQL> select count(1) from v$process;

COUNT(1)
----------
23

SQL> select count(1) from v$session;

COUNT(1)
----------
20

At least the answer was pretty clear – there are more processes than sessions (23 versus 20). But then, someone pointed to the view V$RESOURCE_LIMIT, which records the current usage of the various definable limits. Checking the limits for two resources – sessions and processes:

select resource_name, current_utilization, max_utilization, limit_value
from v$resource_limit
where resource_name in ( 'sessions', 'processes')
/

RESOURCE_NAME CURRENT_UTILIZATION MAX_UTILIZATION LIMIT_VALUE
------------- ------------------- --------------- -----------
processes 23 23 40
sessions 26 26 49
This shows that the highest possible numbers for sessions and processes are 40 and 49 respectively. [Note, I have set the processes parameter in init.ora to 49 to reduce the limit artificially]. It shows that currently there are 26 sessions and 23 processes. Finally it shows the sessions and processes have touched a high number of 26 and 23 respectively.

This contradicts what we saw earlier. The processes usage matches – 23 processes seen from the resource limit view and count from v$process; but the session count differs. From the resource limit, we see that there are 26 sessions; but v$session shows only 20. Why there is a difference?

Processes without Sessions

Let’s see the first one – processes without sessions. We can easily find them:

select pid, spid, addr, program
from v$process
where addr not in (
select paddr from v$session
)

PID SPID ADDR PROGRAM
---------- ---------- -------- ------------------------------
1 4116A2B8 PSEUDO
18 23904 411758AC oracle@oradba1 (D000)
19 23906 41176360 oracle@oradba1 (S000)


Other than the first one, these are real processes. The processes refer to the dispatcher process and the shared server processes. These are not background processes; so they don’t appear in V$BGPROCESS.

Sessions without Processes

Now what about the second case – the sessions without processes? That one is harder to visualize. A session in Oracle is a connection to the database. As we know from the two-task architecture, a client process of a session kicks off a server process which in turn does the heavy lifting from the Oracle database. Keeping that in mind, it might be difficult to envision a session without a process. With a twist of the same concept, is it possible for two sessions to share a single process? Once again, that defies understanding as well. A session is an exclusive conduit to the database; how can that be shared? Even in case of share servers, each shared server session has its own process.

It is possible, under some special circumstances. They are used by Recursive Database Sessions, persistent for a very small amount of time; but they are sessions. When you create a table, alter an index or perform other tasks, Oracle must perform certain tasks on your behalf, which are called recursive actions. One example may make it clear. When you create a table, Oracle must determine that you have sufficient privileges to do that and the determination is made through a recursive session. Let’s see a demonstration of the sessions in action.

First connect to the database and get your own SID:

select sid, paddr, spid
from v$session s, v$process p
where s.audsid = userenv('SESSIONID')
and p.addr = s.paddr
/

SID PADDR SPID
---------- -------- ------------------------
17 4117F938 24042

This shows that my own session has the SID 17, the OS process id of 24042 and the address of the server process is 4117F938. I know, I know; this is a 32-bit system, so the process address is small.

If you look at v$process, you can get the details on that process as well, using the paddr value. Conversely, if you select the sessions with that paddr value:

SQL> select sid from v$session where paddr = '4117F938';

SID
---
17

There is exactly one session. If you check it for any process paddr, you will see exactly one SID. But, that’s where the mystery lies. The v$session view does not tell the whole story; it does not show all the sessions. Here is the definition of the view:

SQL> select view_definition from v$fixed_view_definition where view_name = 'GV$SESSION';

VIEW_DEFINITION
----------------------------------------------------------------------------------
select s.inst_id,s.addr,s.indx,s.ksuseser,s.ksuudses,s.ksusepro,s.ksuudlui,s.ksuudlna,s.ksuudoct,s.ksusesow, decode(s.ksusetrn,hextoraw('00'),null,s.ksusetr
n),decode(s.ksqpswat,hextoraw('00'),null,s.ksqpswat),decode(bitand(s.ksuseidl,11),1,'ACTIVE',0,decode(bitand(s.ksuseflg,4096),0,'INACTIVE','CACHED'),2,'SNIP
ED',3,'SNIPED', 'KILLED'),decode(s.ksspatyp,1,'DEDICATED',2,'SHARED',3,'PSEUDO','NONE'), s.ksuudsid,s.ksuudsna,s.ksuseunm,s.ksusepid,s.ksusemnm,s.ksusetid,
s.ksusepnm, decode(bitand(s.ksuseflg,19),17,'BACKGROUND',1,'USER',2,'RECURSIVE','?'), s.ksusesql, s.ksusesqh, s.ksusesqi, decode(s.ksusesch, 65535, to_numbe
r(null), s.ksusesch), s.ksusesesta, decode(s.ksuseseid, 0, to_number(null), s.ksuseseid), s.ksusepsq, s.ksusepha, s.ksusepsi, decode(s.ksusepch, 65535,
to_number(null), s.ksusepch), s.ksusepesta, decode(s.ksusepeid, 0, to_number(null), s.ksusepeid), decode(s.ksusepeo,0,to_number(null),s.ksusepeo), decod
e(s.ksusepeo,0,to_number(null),s.ksusepes), decode(s.ksusepco,0,to_number(null),s.ksusepco), decode(s.ksusepco,0,to_number(null),s.ksusepcs), s.ksuseapp,
s.ksuseaph, s.ksuseact, s.ksuseach, s.ksusecli, s.ksusefix, s.ksuseobj, s.ksusefil, s.ksuseblk, s.ksuseslt, s.ksuseltm, s.ksusectm,decode(bitand(s.ksusepxo
pt, 12),0,'NO','YES'),decode(s.ksuseft, 2,'SESSION', 4,'SELECT',8,'TRANSACTIONAL','NONE'),decode(s.ksusefm,1,'BASIC',2,'PRECONNECT',4,'PREPARSE','NONE'),dec
ode(s.ksusefs, 1, 'YES', 'NO'),s.ksusegrp,decode(bitand(s.ksusepxopt,4),4,'ENABLED',decode(bitand(s.ksusepxopt,8),8,'FORCED','DISABLED')),decode(bitand(s.ks
usepxopt,2),2,'FORCED',decode(bitand(s.ksusepxopt,1),1,'DISABLED','ENABLED')),decode(bitand(s.ksusepxopt,32),32,'FORCED',decode(bitand(s.ksusepxopt,16),16,'
DISABLED','ENABLED')), s.ksusecqd, s.ksuseclid, decode(s.ksuseblocker,4294967295,'UNKNOWN', 4294967294, 'UNKNOWN',4294967293,'UNKNOWN',4294967292,'NO HOLD
ER', 4294967291,'NOT IN WAIT','VALID'),decode(s.ksuseblocker, 4294967295,to_number(null),4294967294,to_number(null), 4294967293,to_number(null), 4294967292
,to_number(null),4294967291, to_number(null),bitand(s.ksuseblocker, 2147418112)/65536),decode(s.ksuseblocker, 4294967295,to_number(null),4294967294,to_numb
er(null), 4294967293,to_number(null), 4294967292,to_number(null),4294967291, to_number(null),bitand(s.ksuseblocker, 65535)),w.kslwtseq,w.kslwtevt,e.kslednam, e.ksledp1,w.kslwtp1,w.kslwtp1r, e.ksledp2,w.kslwtp2,w.kslwtp2r,e.ksledp3,w.kslwtp3,w.kslwtp3r, e.ksledclassid,e.ksledclass#,e.ksledclass, decode(w.kslwtin
wait, 0,decode(bitand(w.kslwtflags,256), 0,-2, decode(round(w.kslwtstime/10000), 0,-1,
round(w.kslwtstime/10000))), 0), decode(w.kslwtinwait,0,round((w.kslwtstime+w.kslwtltime)/1000000), round(w.kslwtstime/1000000)), de
code(w.kslwtinwait,1,'WAITING', decode(bitand(w.kslwtflags,256),0,'WAITED UNKNOWN TIME', decode(round(w.kslwtstime/10000),0,'WAITED SHORT TIME', 'WAIT
ED KNOWN TIME'))),w.kslwtstime, decode(w.kslwtinwait,0,to_number(null), decode(bitand(w.kslwtflags,64),64,0,w.kslwttrem)), , decode(
bitand(s.ksuseflg2,32),32,'ENABLED','DISABLED'),decode(bitand(s.ksuseflg2,64),64,'TRUE','FALSE'),decode(bitand(s.ksuseflg2,128),128,'TRUE','FALSE'),decode(b
itand(s.ksuseflg2,65536) + bitand(s.ksuseflg2,131072),65536,'ALL EXEC',131072,'NEVER',0,'FIRST EXEC'),s.ksuudsae,s.ksusecre,s.ksusecsn from x$ksuse s, x$ksled e, x$kslwt w where bitand(s.ksspaflg,1)!=0 and bitand(s.ksuseflg,1)!=0 and s.indx=w.kslwtsid and w.kslwtevt=e.indx

Note the section shown in red. The ksuseflg column in x$ksuse contains a lot of information, which the v$session view extracts and presents in a readable manner. Let’s use exactly the same SQL directly from the prompt. Noting that the ksusepro and indx and ksuudsna columns show the process address (paddr), the SID and the username respectively, we can extract the information:

select indx, decode(bitand(ksuseflg,19),17,'BACKGROUND',1,'USER',2,'RECURSIVE','UNKNOWN'),
ksuudsna user_name
from x$ksuse
where ksusepro = '41182408'
/
INDX DECODE(BIT USER_NAME
---------- ---------- ------------------------------
7 USER ARUP

It shows the same information that v$session would have shown; but with a big difference. Note the view definition again – there is a join condition at the end with two other views: x$ksled and x$kslwt. Not all the session information is available on the other views. So the join filters out some rows from x$ksuse. To examine when it does that and that sessions get created automatically, you have to conduct an experiment.

First you have to look at a view called V$RESOURCE_LIMIT, which shows the various definable limits, how much is being used and – most important – how much has been the high water mark usage of the limit. We are interested in only 2 limits – sessions and processes.

select resource_name, current_utilization, max_utilization, limit_value
from v$resource_limit
where resource_name in ( 'sessions', 'processes')
/

RESOURCE_NAME CURRENT_UTILIZATION MAX_UTILIZATION LIMIT_VALUE
------------- ------------------- --------------- -----------
processes 23 23 40
sessions 26 26 49

This shows that the highest possible numbers for sessions and processes are 40 and 49 respectively. [Note, I have set the processes parameter in init.ora to 49 to reduce the limit artificially]. It shows that currently there are 26 sessions and 23 processes. Finally it shows the sessions and processes have touched a high number of 26 and 23 respectively.

In this experiment, keep the above session connected to SYS. On a different OS prompt, connect to database using SQL*Plus as user ARUP. At the SQL> prompt, type host, which will show an OS prompt. From that shell, connect to the database again as ARUP. From the prompt go to host prompt and connect using SQL*Plus. Do it over and over so that the number of sessions builds up as shown below:

SQL> host
oracle@oradba1 ~/arup# sqlplus arup/arup

SQL*Plus: Release 11.1.0.7.0 - Production on Wed Mar 24 21:54:56 2010

Copyright (c) 1982, 2008, Oracle. All rights reserved.


Connected to:
Oracle Database 11g Enterprise Edition Release 11.1.0.7.0 - Production
With the Partitioning, Oracle Label Security, OLAP, Data Mining
and Real Application Testing options

SQL> host
oracle@oradba1 ~/arup# sqlplus arup/arup
.....
and so on ....

After some time you will receive an error like this:

ERROR:
ORA-00020: maximum number of processes (%s) exceeded

From session 1 (where you are connected as sysdba), check the resource limits:

select resource_name, current_utilization, max_utilization, limit_value
from v$resource_limit
where resource_name in ( 'sessions', 'processes')
/

RESOURCE_NAME CURRENT_UTILIZATION MAX_UTILIZATION LIMIT_VALUE
------------- ------------------- --------------- -----------
processes 39 40 40
sessions 45 47 49

This makes perfect sense. Note the current number of processes – 39, and the limit is 40. The very last session tried to create a session; and it failed, since it would have pushed the limit above 40.

Now, in the session of ARUP, execute a PL/SQL code that creates and drops a lot of objects in succession.

begin
for i in 1..1000 loop
execute immediate 'drop table t';
execute immediate 'create table t as select * from all_objects where 1=2';
end loop;
end;
/

This simply creates and drops the table called t in succession, 1000 times. While this code is running, check for the session from the x$ksuse view directly. You already know the process address:

select indx, decode(bitand(ksuseflg,19),17,'BACKGROUND',1,'USER',2,'RECURSIVE','UNKNOWN'),
ksuudsna user_name
from x$ksuse
where ksusepro = '41182408'
/

INDX DECODE(BIT USER_NAME
---------- ---------- ------------------------------
7 USER ARUP
10 RECURSIVE SYS

This is interesting. There are two sessions – SIDs 7 and 10 against the same process with address 41182408. Why is that? SID 7 is the actual session which you connected. If you look at the decoded value, it shows RECURSIVE for the SID 10, which shows that for the user SYS. This session was kicked off by Oracle for the recursive actions, which pushed the session count up without a real user session. This is a situation where a single process serves 2 sessions.

Another interesting point is to know what this recursive session’s waiting on? To know that you have to check the view X$KSLWT. First set the null display to “?” so that null values show up as “?”:

SQL> set null ?
Now execute the query to get the wait event:

select s.indx, decode(bitand(ksuseflg,19),17,'BACKGROUND',1,'USER',2,'RECURSIVE','UNKNOWN'),
ksuudsna user_name, w.kslwtevt
from x$ksuse s, x$kslwt w
where ksusepro = '41182408'
and s.indx=w.kslwtsid (+)
/


INDX DECODE(BIT USER_NAME KSLWTEVT
---------- ---------- ------------------------------ ----------
4 RECURSIVE SYS ?
7 USER ARUP 72

Note, I used the outer join. Why? It’s because the session will not be found in the wait events view, which is confirmed by the null output. As you can glean from the above output, the wait event is not recorded for the recursive sessions. This is why a corresponding row was not found in the view V$SESSION which joins x$ksuse and x$kslwt without an outer join.

Now you must be curious about other such sessions where they share the same process. Let’s check:

SQL> select ksusepro, count(1)
2 from x$ksuse
3 group by ksusepro
4 having count(1) > 1;

KSUSEPRO COUNT(1)
-------- ----------
00 2
4117AE4C 2
41174344 2
41178E30 2
4117EE84 2
4117D91C 2
41174DF8 2
4117B900 2
4117837C 2
4117CE68 2
4116CD88 2
4117A398 2

We got several processes with addresses. Next, let’s find out the exact program names.

SQL> with proc_tab as
2 (
3 select ksusepro addr
4 from x$ksuse
5 group by ksusepro
6 having count(1) > 1
7 )
8 select v.addr, v.program
9 from proc_tab p, v$process v
10 where v.addr = p.addr;

ADDR PROGRAM
-------- ------------------------------------------------
4117AE4C oracle@oradba1 (FBDA)
41174344 oracle@oradba1 (MMON)
41178E30 oracle@oradba1 (ARC1)
4117EE84 oracle@oradba1 (W000)
4117D91C oracle@oradba1 (CJQ0)
41174DF8 oracle@oradba1 (MMNL)
4117B900 oracle@oradba1 (QMNC)
4117837C oracle@oradba1 (ARC0)
4117CE68 oracle@oradba1 (q001)
4116CD88 oracle@oradba1 (DBRM)
4117A398 oracle@oradba1 (ARC3)

These are the sessions related to the background processes – Flashback Data Archiver, Memory Monitor and so on. Let’s dive down even further and look into the sessions these processes serve:

SQL> with proc_tab as
2 (
3 select ksusepro addr
4 from x$ksuse
5 group by ksusepro
6 having count(1) > 1
7 )
8 select p.addr, indx sid, decode(bitand(ksuseflg,19),17,'BACKGROUND',1,'USER',2,'RECURSIVE','UNKNOWN'),
9 ksuudsna user_name
10 from x$ksuse x, proc_tab p
11 where ksusepro = p.addr
12 order by 1,2;

ADDR SID DECODE(BIT USER_NAME
-------- ---------- ---------- ------------------------------
00 1 RECURSIVE SYS
00 10 RECURSIVE SYS
4116CD88 23 RECURSIVE SYS
4116CD88 45 BACKGROUND SYS
41174344 30 RECURSIVE SYS
41174344 34 BACKGROUND SYS
41174DF8 20 RECURSIVE SYS
41174DF8 33 BACKGROUND SYS
4117837C 3 RECURSIVE SYS
4117837C 29 BACKGROUND SYS
41178E30 2 RECURSIVE SYS
41178E30 28 BACKGROUND SYS
4117A398 25 RECURSIVE SYS
4117A398 26 BACKGROUND SYS
4117AE4C 21 RECURSIVE SYS
4117AE4C 22 BACKGROUND SYS
4117B900 19 RECURSIVE SYS
4117B900 24 BACKGROUND SYS
4117CE68 15 RECURSIVE SYS

Note, for each of the background processes has a corresponding recursive session, which is counted towards the total session count in V$RESOURCE_LIMIT, even if you don’t actually create user sessions. These recursive sessions are not visible in V$SESSION.

Takeaways

1. The first lesson to learn is a process not always ≠ a session. In many cases there is a one-to-one relationship; but not always. When Oracle has to run a recursive SQL as a result of some user commands, it creates the session objects behind the scenes to fulfill that request.
2. Second, these recursive sessions count towards the overall sessions limit. To know the current sessions connected, use the V$RESOURCE_LIMIT view instead of selecting the sum from V$SESSION.

4 comments:

Franck Pachot said...

Hi,
There is another case where a process can be used by several sessions. In the two-task architecture, the server process handles the connection, but several session can use the same connection (and then the same process). OCI can do that, and sqlplus uses that when we set 'autotrace': it opens another session in order to get the statistics of the first one (query on v$sesstat), without interfering in those statistics.
And in addition to that, wen we disconnect, the session is closed but the process is still there, so we have a process without a session.
Regards,
Franck.

Baskar said...

Hi Arup,

Excellent article. I ran the the last sql in my db where all SIDs have same process.
It shows username as unknown
SQL> with proc_tab as
2 (
3 select ksusepro addr
4 from x$ksuse
5 group by ksusepro
6 having count(1) > 1
7 )
8 select p.addr, indx sid, decode(bitand(ksuseflg,19),17,'BACKGROUND',1,'USER',2,'RECURSIVE','UNKNOWN'),
9 ksuudsna user_name
10 from x$ksuse x, proc_tab p
11 where ksusepro = p.addr
12 order by 1,2;

ADDR SID DECODE(BIT USER_NAME
-------- ---------- ---------- ------------------------------
00 1 UNKNOWN
00 2 UNKNOWN
00 3 UNKNOWN
00 4 UNKNOWN
00 5 UNKNOWN
00 6 UNKNOWN
00 7 UNKNOWN
00 8 UNKNOWN
00 9 UNKNOWN
00 10 UNKNOWN
00 11 UNKNOWN
00 12 UNKNOWN
00 13 UNKNOWN
00 14 UNKNOWN
00 15 UNKNOWN
00 16 UNKNOWN
00 17 UNKNOWN
00 18 UNKNOWN
00 19 UNKNOWN
00 20 UNKNOWN
00 21 UNKNOWN

I bounced the tomcat application yet these sessions are not cleared.

thanks,
baskar.l

sap upgrade said...

Thanks Arup for sharing this clean post. I am also confused between these two terms and thinks that they are almost similar. Your explanation made this point clear that these both are totally different from each other.

Sara said...

Youve got a great deal to say about this subject, and so much knowledge. Check out drivers download for windows 7, thanks!

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