In article <895ntp$kni$1_at_soap.pipex.net>,
"Niall" <n-litchfield_at_audit-commission.gov.uk> wrote:
> "Kevin A Lewis" <Kevin_A_Lewis_at_Hotmail.com> wrote in message
> news:Bnts4.16$gY2.1753_at_newreader.ukcore.bt.net...
> > So which answer is the right one!
> >
> > Does anyone know the 'correct' answer?
> >
> > Does the question have a firm answer? Some obviously think so but
others
say
> > it all depends.
> >
> > I pity the poor guy who needs to know.
> >
> > Are all the respondents correct and I just can't reconcile the
answers?
>
> The guy who was right was the guy who said that the os block size
could take
> one of a range of values. In summary
>
> NTFS 512 Bytes (But can be varied by the user when formatting a disk)
> FAT 2-32K depending on drive size.
>
>
Amazing how such a simple question can be so difficult to answer. I've
spent the better part of three days trying to find and reconcile all the
conflicting information on Metalink, Deja, Technet, and Oracle's
documentation.
512 bytes does seem to be the correct answer. This is the size of a
sector (the smallest physical *storage* unit on the disk). You can not
vary this when formatting. What you can vary during formatting is the
cluster size, a cluster being the fundamental unit of disk
*allocation*). The only time the cluster size comes into play is
when you're creating/resizing the datafiles (and disk space is actually
allocated, as opposed to accessed). The default cluster size depends on
the size of the partition you're formatting.
Oracle uses unbuffered I/O, i.e. it totally bypasses the NT file system
cache (as it uses its own cache). Unbuffered I/O requests in NT must be
issued in multiples of the disk sector size, i.e. 512 bytes.
--
Ben Pomicter
Database Administrator
Digitas
--
Ben Pomicter
Database Administrator
Digitas
Sent via Deja.com http://www.deja.com/
Before you buy.
Received on Tue Mar 21 2000 - 00:00:00 CST
