背景：
在使用嵌入式汇编语言优化一些Pascal代码时，我注意到一条不必要的MOV指令，并将其删除。
令我惊讶的是，删除不必要的指令导致我的程序变慢。
我发现添加任意无用的MOV指令可以进一步提高性能。
效果是不稳定的，并且会根据执行顺序进行更改：由单行向上或向下转置的相同垃圾指令会产生减速。
我知道 CPU 做了各种优化和精简，但是，这似乎更像是黑魔法。
数据：
我的代码版本在运行时间的循环中间有条件地编译了三个垃圾操作。2**20==1048576（周围的程序只计算SHA-256哈希）。
在我相当旧的机器（Intel(R) Core(TM)2 CPU 6400 @ 2.13 GHz）上的结果：
avg time (ms) with -dJUNKOPS: 1822.84 ms
avg time (ms) without: 1836.44 ms
程序循环运行 25 次，每次运行顺序随机变化。
摘抄：
{$asmmode intel}
procedure example_junkop_in_sha256;
var s1, t2 : uint32;
begin
// Here are parts of the SHA-256 algorithm, in Pascal:
// s0 {r10d} := ror(a, 2) xor ror(a, 13) xor ror(a, 22)
// s1 {r11d} := ror(e, 6) xor ror(e, 11) xor ror(e, 25)
// Here is how I translated them (side by side to show symmetry):
asm
MOV r8d, a ; MOV r9d, e
ROR r8d, 2 ; ROR r9d, 6
MOV r10d, r8d ; MOV r11d, r9d
ROR r8d, 11 {13 total} ; ROR r9d, 5 {11 total}
XOR r10d, r8d ; XOR r11d, r9d
ROR r8d, 9 {22 total} ; ROR r9d, 14 {25 total}
XOR r10d, r8d ; XOR r11d, r9d
// Here is the extraneous operation that I removed, causing a speedup
// s1 is the uint32 variable declared at the start of the Pascal code.
//
// I had cleaned up the code, so I no longer needed this variable, and
// could just leave the value sitting in the r11d register until I needed
// it again later.
//
// Since copying to RAM seemed like a waste, I removed the instruction,
// only to discover that the code ran slower without it.
{$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;
自己试试：
如果您想自己尝试，代码在 GitHub 上在线。
我的问题：
- 为什么无用地将寄存器的内容复制到RAM会提高性能？
- 为什么相同的无用指令会在某些线路上提供加速，而在其他线路上提供减速？
- 这种行为是否可以被编译器可预测地利用？