Why would introducing useless MOV instructions speed up a tight loop in x86_64 assembly?
Background:
While optimizing some Pascal code with embedded assembly language, I noticed an unnecessary MOV
instruction, and removed it.
To my surprise, removing the un-necessary instruction caused my program to slow down.
I found that adding arbitrary, useless MOV
instructions increased performance even further.
The effect is erratic, and changes based on execution order: the same junk instructions transposed up or down by a single line produce a slowdown.
I understand that the CPU does all kinds of optimizations and streamlining, but, this seems more like black magic.
The data:
A version of my code conditionally compiles three junk operations in the middle of a loop that runs 2**20==1048576
times. (The surrounding program just calculates SHA-256 hashes).
The results on my rather old machine (Intel(R) Core(TM)2 CPU 6400 @ 2.13 GHz):
avg time (ms) with -dJUNKOPS: 1822.84 ms
avg time (ms) without: 1836.44 ms
The programs were run 25 times in a loop, with the run order changing randomly each time.
Excerpt:
{$asmmode intel}
procedure example_junkop_in_sha256;
var s1, t2 : uint32;
begin
// Here are parts of the SHA-256 algorithm, in Pascal:
// s0 {r10d} := ror(a, 2) xor ror(a, 13) xor ror(a, 22)
// s1 {r11d} := ror(e, 6) xor ror(e, 11) xor ror(e, 25)
// Here is how I translated them (side by side to show symmetry):
asm
MOV r8d, a ; MOV r9d, e
ROR r8d, 2 ; ROR r9d, 6
MOV r10d, r8d ; MOV r11d, r9d
ROR r8d, 11 {13 total} ; ROR r9d, 5 {11 total}
XOR r10d, r8d ; XOR r11d, r9d
ROR r8d, 9 {22 total} ; ROR r9d, 14 {25 total}
XOR r10d, r8d ; XOR r11d, r9d
// Here is the extraneous operation that I removed, causing a speedup
// s1 is the uint32 variable declared at the start of the Pascal code.
//
// I had cleaned up the code, so I no longer needed this variable, and
// could just leave the value sitting in the r11d register until I needed
// it again later.
//
// Since copying to RAM seemed like a waste, I removed the instruction,
// only to discover that the code ran slower without it.
{$IFDEF JUNKOPS}
MOV s1, r11d
{$ENDIF}
// The next part of the code just moves on to another part of SHA-256,
// maj { r12d } := (a and b) xor (a and c) xor (b and c)
mov r8d, a
mov r9d, b
mov r13d, r9d // Set aside a copy of b
and r9d, r8d
mov r12d, c
and r8d, r12d { a and c }
xor r9d, r8d
and r12d, r13d { c and b }
xor r12d, r9d
// Copying the calculated value to the same s1 variable is another speedup.
// As far as I can tell, it doesn't actually matter what register is copied,
// but moving this line up or down makes a huge difference.
{$IFDEF JUNKOPS}
MOV s1, r9d // after mov r12d, c
{$ENDIF}
// And here is where the two calculated values above are actually used:
// T2 {r12d} := S0 {r10d} + Maj {r12d};
ADD r12d, r10d
MOV T2, r12d
end
end;
Try it yourself:
The code is online at GitHub if you want to try it out yourself.
My questions:
- Why would uselessly copying a register's contents to RAM ever increase performance?
- Why would the same useless instruction provide a speedup on some lines, and a slowdown on others?
- Is this behavior something that could be exploited predictably by a compiler?