On Sun, Dec 10, 2000 at 02:53:43PM -0500, Tom Lane wrote:
> > On my Celeron, the timing for those six opcodes is almost whopping 13
> > cycles per byte. Obviously there's some major performance hit to do the
> > memory instructions, because there's no more than 4 cycles worth of
> > dependant instructions in that snippet.
> Yes. It looks like we're looking at pipeline stalls for the memory
> reads.
In particular, for the single-byte memory read. By loading in 32-bit
words at a time, the cost drops to about 7 cycles per byte. I
imagine on a 64-bit CPU, loading 64-bit words at a time would drop the
cost even further.
word1 = *(unsigned long*)z;
while (c > 4)
{
z += 4;
ick = IUPDC32 (word1, ick); word1 >>= 8;
c -= 4;
ick = IUPDC32 (word1, ick); word1 >>= 8;
word1 = *(unsigned long*)z;
ick = IUPDC32 (word1, ick); word1 >>= 8;
ick = IUPDC32 (word1, ick);
}
I tried loading two words at a time, starting to load the second word
well before it's used, but that didn't actually reduce the time taken.
> As Nathan remarks nearby, this is just minutiae, but I'm interested
> anyway...
Yup.
--
Bruce Guenter <[EMAIL PROTECTED]> http://em.ca/~bruceg/
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