# [PATCH] add slice by 8 algorithm to crc32.c

From: George Spelvin
Date: Mon Aug 08 2011 - 05:28:35 EST

Sorry I didn't see this when first posted.

The "slice by 8" terminology is pretty confusing. How about
"Extended Joakim Tjernlund's optimization from commit
836e2af92503f1642dbc3c3281ec68ec1dd39d2e to 8-way parallelism."

Which is essentally what you're doing. The renaming of tab[0] to t0_le
and t0_be, and removal of the DO_CRC4 macro just increases the diff size.

If you're looking at speeding up the CRC through larger tables, have
you tried using 10+11+11-bit tables? That would require 20K of tables
rather than 8K, but would reduce the number of table lookups per byte.

One more stunt you could try to increase parallelism: rather than maintain
the CRC in one register, maintain it in several, and only XOR and collapse
them at the end.

"q = *p32++" always loads 0, and therefore the whole block can be skipped.
(Since tab[0] = 0 for all CRC tables.)

This computes the CRC of the even words. Then do a second one in parallel
for the odd words into a separate CRC register. Then combine them at the end.
(Shift one up by 32 bits and XOR into the other.)

This would let you get away with 5K of tables: t4 through t7, and t0.
t1 through t3 could be skipped.

Ideally, I'd write all this code myself, but I'm a bit crunched at work
right now so wouldn't be able to get to it for a few days.

Another possible simplification to the startup code. There's no need
to compute init_bytes explicitly; just loop until the pointer is aligned:

while ((unsigned)buf & 3) {
if (!len--)
goto done;
#ifdef __LITTLE_ENDIAN
i0 = *buf++ ^ crc;
crc = t0_le[i0] ^ (crc >> 8);
#else
i0 = *buf++ ^ (crc >> 24);
crc = t0_le[i0] ^ (crc << 8);
#endif
}
p32 = (u32 const *)buf;
words = len >> 2;
end_bytes = len & 3;

... although I'd prefer to keep the DO_CRC() and DO_CRC4 macros, and
extend them to the 64-bit case, to avoid the nested #ifdefs. That would
make:

while ((unsigned)buf & 3) {
if (!len--)
goto done;
DO_CRC(*buf++);
}
p32 = (u32 const *)buf;
words = len >> 2;
end_bytes = len & 3;
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