On Tue, Sep 19, 2023 at 08:00:12AM +0000, David Laight wrote:
> ...
> > > So ending up with (something like):
> > > end = buff + length;
> > > ...
> > > while (++ptr < end) {
> > > csum += data;
> > > carry += csum < data;
> > > data = ptr[-1];
> > > }
> > > (Although a do-while loop tends to generate better code
> > > and gcc will pretty much always make that transformation.)
> > >
> > > I think that is 4 instructions per word (load, add, cmp+set, add).
> > > In principle they could be completely pipelined and all
> > > execute (for different loop iterations) in the same clock.
> > > (But that is pretty unlikely to happen - even x86 isn't that good.)
> > > But taking two clocks is quite plausible.
> > > Plus 2 instructions per loop (inc, cmp+jmp).
> > > They might execute in parallel, but unrolling once
> > > may be required.
> > >
> > It looks like GCC actually ends up generating 7 total instructions:
> > ffffffff808d2acc: 97b6 add a5,a5,a3
> > ffffffff808d2ace: 00d7b533 sltu a0,a5,a3
> > ffffffff808d2ad2: 0721 add a4,a4,8
> > ffffffff808d2ad4: 86be mv a3,a5
> > ffffffff808d2ad6: 962a add a2,a2,a0
> > ffffffff808d2ad8: ff873783 ld a5,-8(a4)
> > ffffffff808d2adc: feb768e3 bltu a4,a1,ffffffff808d2acc <do_csum+0x34>
> >
> > This mv instruction could be avoided if the registers were shuffled
> > around, but perhaps this way reduces some dependency chains.
>
> gcc managed to do 'data += csum' so had add 'csum = data'.
> If you unroll once that might go away.
> It might then be 10 instructions for 16 bytes.
> Although you then need slightly larger alignment code.
>
> David
>
> -
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> Registration No: 1397386 (Wales)
>
I messed with it a bit and couldn't get the mv to go away. I would expect
mv to be very cheap so it should be fine, and I would like to avoid adding
too much to the alignment code since it is already large, and I assume
that buff will be aligned more often than not.
Interestingly, the mv does not appear pre gcc 12, and does not appear on clang.
- Charlie
