On 09:08, H. Peter Anvin wrote:
> Andre Noll wrote:
> >
> >> So back to my original question: Why does the kernel require 4 disks
> >> for a raid6 instead of allowing 3?
> >
> > Dunno. Maybe Dan, Neil or HPA can tell the reason for imposing this
> > limitation.
> >
>
> It's very simple: it avoids an ugly corner case in the RAID-6
> computation code. Rather than inserting special code (and verifying it,
> etc.) to deal with the 3-disk RAID-6 degenerate case, it was easier to
> just not permit it.
Out of curiosity, why is any special-casing needed for the 3-disk case?
As
P = D_0 + ... + D_{n-1}
coincides with
Q = g^0D_0 + ... + g^{n-1}D_{n-1}
in the case n=1 (because only the first addend is present), we have
D_0 = P = Q
i.e. raid6 with three disks is the same as raid1. If we lose D_0 and P,
the general formula for reconstruction, i.e.
D_x = g^{-x}(Q + Q_x)
still works for n=1: It reduces to D_0 = Q since x=0 and Q_0 = 0.
So I don't see why special-casing the n=1 case is necessary.
Regards
Andre
--
The only person who always got his work done by Friday was Robinson Crusoe
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