Re: swap-on-raidframe vs raidctl -P

[der Mouse <mouse%Rodents.Montreal.QC.CA@localhost>]

>> How can it compute P' without having B on hand?  I must be missing
>> something.
> In the normal case, P is equal to A XOR B.
> With a change from A to A', P' is computed as P' = P XOR A XOR A'.

Oh, right; this isn't full sideways ECC, just simple XOR.  (And me
quoting and writing "parity" all over the place, duh!)  Never mind....

/~\ The ASCII                           der Mouse
\ / Ribbon Campaign
 X  Against HTML               mouse%rodents.montreal.qc.ca@localhost
/ \ Email!           7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B