Re: DB_FILE_MULTIBLOCK_READ_COUNT question

I want to recall my message. OS block size is not a trivial topic after reading
http://x55.deja.com/[ST_rn=ps]/getdoc.xp?AN=666487447&CONTEXT=969389130.2122 711133&hitnum=1 (and other messages in the thread) http://www.ixora.com.au/tips/creation/block_size.htm (look at the last sentence in the first paragraph)
and
http://www.sunworld.com/sunworldonline/swo-05-1999/swol-05-filesystem_p.html (and its sequel)

I believe the correct way to determine it is by df -g, not by creating a directory and see its size. Also, df -g, which implements statvfs(2), also reports fragment size, a concept in Solaris (or maybe all System V R4). But in any case, we're mostly interested in filesystem buffer size. This (as I understand) should match the block size reported by df -g; on Solaris 2.6 UFS it's 8K, matching memory pagesize, incidentally. To absolutely correctly find out what chunk size you're reading/writing in, in Solaris you may issue the command truss -p theprocessnumberofyourapplication and look at the 3rd argument of read, write (or pwrite) system calls.

I'm still studying this subject.

Yong Huang
yhuang_at_indigopool.com

Yong Huang <yhuang_at_indigopool.com> wrote in message news:8q62i0$dm2$1_at_news.sinet.slb.com...
> Determine the block size like this:
>
> lurch% mkdir aaa
> lurch% ls -ld aaa
> drwxr-xr-x 2 oracle8 oinstall 512 Sep 18 14:35 aaa
>
> So it's 512 bytes. The benefit of a small size is better disk space usage.
> I'm sure there're other benefits.
>
> Yong Huang
> yhuang_at_indigopool.com
Received on Wed Sep 20 2000 - 00:00:00 CDT