Note in-line

Regards

Jonathan Lewis
http://www.jlcomp.demon.co.uk

  The educated person is not the person
  who can answer the questions, but the
  person who can question the answers -- T. Schick Jr


Next public appearance2:
 March 2004 Hotsos Symposium - Keynote
 March 2004 Charlotte NC - OUG Tutorial
 April 2004 Iceland


One-day tutorials:
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Three-day seminar:
see http://www.jlcomp.demon.co.uk/seminar.html
____UK___February


The Co-operative Oracle Users' FAQ
http://www.jlcomp.demon.co.uk/faq/ind_faq.html


----- Original Message ----- 
To: "Multiple recipients of list ORACLE-L" <[EMAIL PROTECTED]>
Sent: Thursday, January 22, 2004 6:29 AM


Hi list,

1)Why fast commit generate no redo ?

It's called a fast commit BECAUSE it doesn't generate redo
(except for a tiny bit that describes the change to the transaction
table entry in the segment header block that marked the transaction
as active).

It doesn't need to generate redo because it's going to leave
(most of) the lock and change information on the blocks that
have been changed, and let some other visiter to the blocks
clean up the mess.

2)Is delayed cleanout generate redo?

Delayed "block cleanout" - where a later operation simply READS
a messy block and cleans it up (by referring back to the transaction
table to get the necessary commit details) will generate redo.

Delayed-logging "block cleanout" - which occurs when the first
transactions cleans out a few of the blocks it has dirtied but
does not log the cleanout - is effectively not going to generate
redo, as the next transaction to MODIFY the date will generate
some undo which looks as if it started from a clean block, rather
than the partly dirty block that is really there - so the cleanout is
effectively free.

3)In a block dump even after transactions commit why it shows lock 1 in ITL?

Because Oracle doesn't clean the block out properly, it will either
not revisit it at all (1), or just revisit the ITL and a couple of header
bytes (2).


Thanks in advance.
Syed


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