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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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