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Hi Mindaugas,
On 07/23/2012 12:55 AM, Mindaugas Rasiukevicius wrote:
> Hello Lars,
>
> Lars Heidieker wrote:
>> slight improvement on kmem(9):
>>
>> splitting the lookup table into two parts - this reduces the size
>> of the lookup table -> touchin
Hello Lars,
Lars Heidieker wrote:
> slight improvement on kmem(9):
>
> splitting the lookup table into two parts
> - this reduces the size of the lookup table -> touching less cache-lines
> (e.g. on 64bit down from 4096b to 1024b + 64b)
Why do you think the split would result in touching fe
>> This is `raw' in that it bypasses partitioning allowing access to
>> the whole disk regardless of partitioning.
> It looks that I remember incorrectly what ws@ taught me back in the
> days how the ``full disc'' partition works.
> What I remember is that (let's assume sparc) partition ``c'' was
On Tue, Jul 17, 2012 at 07:11:43PM +0100, Mindaugas Rasiukevicius wrote:
> > If changes are incrememntal (which this one probably is), then there
> > is no real problem applying each change to cvs separately - the system
> > should still build after each.
>
> Heh. It is a bit more than "it s
On Tue, Jul 17, 2012 at 06:00:34PM +, Taylor R Campbell wrote:
>> Creating CVS branches does not help with this sort of thing; it just
>> makes development slower. It also makes it less likely that the
>> changes will get tested before the final branch merge, at which point
>
So to correct myself:
> It looks that I remember incorrectly what ws@ taught me back in the days
> how the ``full disc'' partition works.
I just called him on the phone and the correct answer is that the code reading
the disklabel enforces the ``full disc'' partition to be the full disc.
Also that
> As to the question about how to replace a failed component with a
> slightly smaller one, the answeris you don't. If you use 3TB disks as your
> raid components, you'd better have 3TB disks as replacements.
Yes, but for what precise value of ``3TB''?
The ones I presently have are 58605331
> There are two meanings of `raw' as applied to disk partitions.
Ah, I see. Thanks.
> This is `raw' in that it bypasses partitioning
> allowing access to the whole disk regardless of partitioning.
It looks that I remember incorrectly what ws@ taught me back in the days
how the ``full disc'' partit
