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Re: fast commit

From: Jonathan Lewis <>
Date: Thu, 22 Jan 2004 01:29:24 -0800
Message-ID: <>

Note in-line


Jonathan Lewis

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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.

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Author: Jonathan Lewis

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Received on Thu Jan 22 2004 - 03:29:24 CST

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