What I'm figuring is that it should be fine for -Os, since that
optimises for size over speed, and -Os uses "or $-1, %rax" for setting a
register to -1 instead of "mov $-1, %rax". I copied this behaviour from
MSVC after finding a thread about the false dependency that OR causes
(Florian warned me of the false dependency when I proposed the
optimisation regardless of size or speed).
Outside of -Os, if the immediate previous instruction uses the same
register, then it's guaranteed to have been loaded and there shouldn't
be a pipeline stall, although out-of-order execution may cause some
problems. This one is going to be interesting to time.
In regards to how frequently it appears in code, if it turns out that
it's very infrequent (normally I compile and analyse the RTL for this
purpose, and sometimes the compiler itself), I likely won't code it,
although in the case of the "fast mod" algorithm, it appears very
frequently, so should probably be coded directly, especially as the
multiplication can take 3 cycles, depending on the processor, and is the
limiting factor.
Gareth aka. Kit
On 17/10/2021 23:53, Stefan Glienke via fpc-devel wrote:
According to compiler explorer clang, gcc and msvc compile this to the
same code with -O3 as FPC does. So I would assume that is fine.
Am 17.10.2021 um 13:25 schrieb J. Gareth Moreton via fpc-devel:
Hi everyone,
While reading up on some algorithms, I came across a recommendation
of using a shorter arithmetic function to change the value of a
constant in a register rather than loading the new value directly.
However, the algorithm assumes a RISC-like processor, so I'm not sure
if it applies to an Intel x86-64 processor. Consider the following:
movq $0xaaaaaaaaaaaaaaab,%rax
imulq %rax,%rcx
movq $0x5555555555555555,%rax
cmpq %rax,%rcx
setle %al
This algorithm sets %al to 1 if %rcx is divisible by 3, and 0 if it's
not, and was compiled from the following Pascal code (under -O3, but
-O1 produces almost exactly the same):
function IsDivisible3(Numerator: QWord): Boolean;
begin
Result := (Numerator * $AAAAAAAAAAAAAAAB) <= $5555555555555555;
end;
(One of my merge requests produces this code from "Result := (x mod
3) = 0")
My question is this: can "movq $0x5555555555555555,%rax" be replaced
with "shrq $0x1,%rax" without incurring an additional pipeline
stall? The MOV instruction takes 10 bytes to store, while "SHR 1"
takes only 3. Given that %rax is used beforehand and the CMP
instruction has to wait until the IMUL instruction has finished
executing, logic tells me that I can get away with it here, but I'm
not sure if the metric to go by is the execution speed of IMUL (i.e.
the IMUL instruction is the limiting factor before CMP can be
executed), or the simple fact that the previous value of %rax was
used and will be loaded with $AAAAAAAAAAAAAAAB by the time it comes
to load it with a new value.
Gareth aka. Kit
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