On Thu, Feb 4, 2016 at 3:08 AM, David Laight <david.lai...@aculab.com> wrote: > From: Tom Herbert >> Sent: 03 February 2016 19:19 > ... >> + /* Main loop */ >> +50: adcq 0*8(%rdi),%rax >> + adcq 1*8(%rdi),%rax >> + adcq 2*8(%rdi),%rax >> + adcq 3*8(%rdi),%rax >> + adcq 4*8(%rdi),%rax >> + adcq 5*8(%rdi),%rax >> + adcq 6*8(%rdi),%rax >> + adcq 7*8(%rdi),%rax >> + adcq 8*8(%rdi),%rax >> + adcq 9*8(%rdi),%rax >> + adcq 10*8(%rdi),%rax >> + adcq 11*8(%rdi),%rax >> + adcq 12*8(%rdi),%rax >> + adcq 13*8(%rdi),%rax >> + adcq 14*8(%rdi),%rax >> + adcq 15*8(%rdi),%rax >> + lea 128(%rdi), %rdi >> + loop 50b > > I'd need convincing that unrolling the loop like that gives any significant > gain. > You have a dependency chain on the carry flag so have delays between the > 'adcq' > instructions (these may be more significant than the memory reads from l1 > cache). > > I also don't remember (might be wrong) the 'loop' instruction being executed > quickly. > If 'loop' is fast then you will probably find that: > > 10: adcq 0(%rdi),%rax > lea 8(%rdi),%rdi > loop 10b > > is just as fast since the three instructions could all be executed in > parallel. > But I suspect that 'dec %cx; jnz 10b' is actually better (and might execute as > a single micro-op). > IIRC 'adc' and 'dec' will both have dependencies on the flags register > so cannot execute together (which is a shame here). > > It is also possible that breaking the carry-chain dependency by doing 32bit > adds (possibly after 64bit reads) can be made to be faster.
If nothing else reducing the size of this main loop may be desirable. I know the newer x86 is supposed to have a loop buffer so that it can basically loop on already decoded instructions. Normally it is only something like 64 or 128 bytes in size though. You might find that reducing this loop to that smaller size may improve the performance for larger payloads. - Alex
