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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