v3: s-o-b comment, explanation of performance and descision for
the start/stop implementation

Implementing rmw functionality for RAID6 requires optimized syndrome
calculation. Up to now we can only generate a complete syndrome. The
target P/Q pages are always overwritten. With this patch we provide
a framework for inplace P/Q modification. In the first place simply
fill those functions with NULL values.

xor_syndrome() has two additional parameters: start & stop. These
will indicate the first and last page that are changing during a
rmw run. That makes it possible to avoid several unneccessary loops
and speed up calculation. The caller needs to implement the following
logic to make the functions work.

1) xor_syndrome(disks, start, stop, ...): "Remove" all data of source
blocks inside P/Q between (and including) start and end.

2) modify any block with start <= block <= stop

3) xor_syndrome(disks, start, stop, ...): "Reinsert" all data of
source blocks into P/Q between (and including) start and end.

Pages between start and stop that won't be changed should be filled
with a pointer to the kernel zero page. The reasons for not taking NULL
pages are:

1) Algorithms cross the whole source data line by line. Thus avoid
additional branches.

2) Having a NULL page avoids calculating the XOR P parity but still
need calulation steps for the Q parity. Depending on the algorithm
unrolling that might be only a difference of 2 instructions per loop.

The benchmark numbers of the gen_syndrome() functions are displayed in
the kernel log. Do the same for the xor_syndrome() functions. This
will help to analyze performance problems and give an rough estimate
how well the algorithm works. The choice of the fastest algorithm will
still depend on the gen_syndrome() performance.

With the start/stop page implementation the speed can vary a lot in real
life. E.g. a change of page 0 & page 15 on a stripe will be harder to
compute than the case where page 0 & page 1 are XOR candidates. To be not
to enthusiatic about the expected speeds we will run a worse case test
that simulates a change on the upper half of the stripe. So we do:

1) calculation of P/Q for the upper pages

2) continuation of Q for the lower (empty) pages
Signed-off-by: Markus Stockhausen <stockhausen@collogia.de>
Signed-off-by: NeilBrown <neilb@suse.de>