RE: [PATCH 1/1] riscv: prevent pipeline stall in __asm_to/copy_from_user

[David Laight <David.Laight@ACULAB.COM>]

From: Palmer Dabbelt
> Sent: 12 June 2021 05:05
...
> > I don't know the architecture, but unless there is a stunning
> > pipeline delay for memory reads a simple interleaved copy
> > may be fast enough.
> > So something like:
> > 	a = src[0];
> > 	do {
> > 		b = src[1];
> > 		src += 2;
> > 		dst[0] = a;
> > 		dst += 2;
> > 		a = src[0];
> > 		dst[-1] = b;
> > 	} while (src != src_end);
> > 	dst[0] = a;
> >
> > It is probably worth doing benchmarks of the copy loop
> > in userspace.
> 
> I also don't know this microarchitecture, but this seems like a pretty
> wacky load-use delay.

It is quite sane really.

While many cpu can use the result of the ALU in the next clock
(there is typically special logic to bypass the write to the
register file) this isn't always true for memory (cache) reads.
It may even be that the read itself takes more than one cycle
(probably pipelined so they can happen every cycle).

So a simple '*dest = *src' copy loop suffers the 'memory read'
penalty ever iteration.
At out-of-order execution unit that uses register renames
(like most x86) will just defer the writes until the data
is available - so isn't impacted.

Interleaving the reads and writes means you issue a read
while waiting for the value from the previous read to
get to the register file - and be available for the
write instruction.

Moving the 'src/dst += 2' into the loop gives a reasonable
chance that they are executed in parallel with a memory
access (on in-order superscaler cpu) rather than bunching
them up at the end where the start adding clocks.

If your cpu can only do one memory read or one memory write
per clock then you only need it to execute two instructions
per clock for the loop above to run at maximum speed.
Even with a 'read latency' of two clocks.
(Especially since riscv has 'mips like' 'compare and branch'
instructions that probably execute in 1 clock when predicted
taken.)

If the cpu can do a read and a write in one clock then the
loop may still run at the maximum speed.
For this to happen you do need he read data to be available
next clock and to run load, store, add and compare instructions
in a single clock.
Without that much parallelism it might be necessary to add
an extra read/write interleave (an maybe a 4th to avoid a
divide by three).

The 'elephant in the room' is a potential additional stall
on reads if the previous cycle is a write to the same cache area.
For instance the nios2 (a soft cpu for altera fpga) can do
back to back reads or back to back writes, but since the reads
are done speculatively (regardless of the opcode!) they have to
be deferred when a write is using the memory block.

	David

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