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Re: swap-on-raidframe vs raidctl -P

>> 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....

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