背景：

在使用嵌入式汇编语言优化某些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速度减慢。

我发现添加任意、无用的MOV指令会进一步提高性能。

效果是不稳定的，并且基于执行顺序的更改：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会进行各种优化和精简，但这更像是黑魔法。

数据：

我的代码的一个版本在一个循环的中间有条件地编译了三个垃圾操作，该循环运行了1048576次。（周围的程序只计算SHA-256散列）。

我相当旧的机器（Intel（R）Core（TM）2 CPU 6400@2.13）上的结果

avg time (ms) with -dJUNKOPS: 1822.84 ms
avg time (ms) without:        1836.44 ms

这些程序循环运行25次，每次运行顺序随机变化。

摘录：

{$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：

如果你想亲自试用，代码可以在GitHub上在线下载。

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