Question

我有以下代码，这是我的应用程序的一部分的瓶颈。我所做的只是从另一个中减去Array。这两个阵列都有大约100000个元素。我正试图找到一种方法来提高性能。

var
  Array1, Array2 : array of integer;

..... 
// Code that fills the arrays
.....

for ix := 0 to length(array1)-1
  Array1[ix] := Array1[ix] - Array2[ix];

end;

有人有建议吗？

Answer 1

在多个线程上运行它，使用那个大的数组将净线性加速。正如他们所说，这是令人尴尬的平行。

Answer 2

在更多线程上运行减法听起来不错，但100K整数sunstraction不占用大量CPU时间，所以也许是线程池...但是设置线程也有很多开销，所以短阵列的并行性能会降低线程比只有一个（主）线程！

您是否关闭了编译器设置，溢出和范围检查？

你可以尝试使用asm rutine，它非常简单......

类似的东西：

procedure SubArray(var ar1, ar2; length: integer);
asm
//length must be > than 0!
   push ebx
   lea  ar1, ar1 -4
   lea  ar2, ar2 -4
@Loop:
   mov  ebx, [ar2 + length *4]
   sub  [ar1 + length *4], ebx
   dec  length
//Here you can put more folloving parts of rutine to more unrole it to speed up.
   jz   @exit
   mov  ebx, [ar2 + length *4]
   sub  [ar1 + length *4], ebx
   dec  length
//
   jnz  @Loop
@exit:
   pop  ebx
end;


begin
   SubArray(Array1[0], Array2[0], length(Array1));

它可以更快......

编辑：使用SIMD说明添加了程序。 此过程请求SSE CPU支持。它可以在XMM寄存器中取4个整数并一次减去。也可以使用movdqa代替movdqu它更快，但您必须首先确保16字节对齐。您也可以像我的第一个asm案例一样解除XMM标准。（我对速度测量感兴趣。:)）

var
  array1, array2: array of integer;

procedure SubQIntArray(var ar1, ar2; length: integer);
asm
//prepare length if not rounded to 4
  push     ecx
  shr      length, 2
  jz       @LengthToSmall
@Loop:
  movdqu   xmm1, [ar1]          //or movdqa but ensure 16b aligment first
  movdqu   xmm2, [ar2]          //or movdqa but ensure 16b aligment first
  psubd    xmm1, xmm2
  movdqu   [ar1], xmm1          //or movdqa but ensure 16b aligment first
  add      ar1, 16
  add      ar2, 16
  dec      length
  jnz      @Loop
@LengthToSmall:
  pop      ecx
  push     ebx
  and      ecx, 3
  jz       @Exit
  mov      ebx, [ar2]
  sub      [ar1], ebx
  dec      ecx
  jz       @Exit
  mov      ebx, [ar2 + 4]
  sub      [ar1 + 4], ebx
  dec      ecx
  jz       @Exit
  mov      ebx, [ar2 + 8]
  sub      [ar1 + 8], ebx
@Exit:
  pop      ebx
end;

begin
//Fill arrays first!
  SubQIntArray(Array1[0], Array2[0], length(Array1));

Answer 3

在这个简单的案例中，我对速度优化非常好奇。所以我制作了6个简单的程序，测量数据块大小为100000的CPU时钟和时间;

使用编译器选项Range和Overflow Checking On
使用编译器选项Range和Overflow Check off的Pascal过程
Classic x86汇编程序。
使用SSE指令和未对齐的16字节移动

的汇编程序。
使用SSE指令和对齐16字节移动
的汇编程序。
汇编程序8次展开循环，带有SSE指令，对齐16字节移动。

检查图片和代码的结果以获取更多信息。

要获得16字节的内存对齐，请首先在文件'FastMM4Options.inc'指令中添加点{$ .define Align16Bytes} ！

program SubTest; {$APPTYPE CONSOLE} uses //In file 'FastMM4Options.inc' delite the dot in directive {$.define Align16Bytes} //to get 16 byte memory alignment! FastMM4, windows, SysUtils; var Ar1 :array of integer; Ar2 :array of integer; ArLength :integer; StartTicks :int64; EndTicks :int64; TicksPerMicroSecond :int64; function GetCpuTicks: int64; asm rdtsc end; {$R+} {$Q+} procedure SubArPasRangeOvfChkOn(length: integer); var n: integer; begin for n := 0 to length -1 do Ar1[n] := Ar1[n] - Ar2[n]; end; {$R-} {$Q-} procedure SubArPas(length: integer); var n: integer; begin for n := 0 to length -1 do Ar1[n] := Ar1[n] - Ar2[n]; end; procedure SubArAsm(var ar1, ar2; length: integer); asm //Length must be > than 0! push ebx lea ar1, ar1 - 4 lea ar2, ar2 - 4 @Loop: mov ebx, [ar2 + length * 4] sub [ar1 + length * 4], ebx dec length jnz @Loop @exit: pop ebx end; procedure SubArAsmSimdU(var ar1, ar2; length: integer); asm //Prepare length push length shr length, 2 jz @Finish @Loop: movdqu xmm1, [ar1] movdqu xmm2, [ar2] psubd xmm1, xmm2 movdqu [ar1], xmm1 add ar1, 16 add ar2, 16 dec length jnz @Loop @Finish: pop length push ebx and length, 3 jz @Exit //Do rest, up to 3 subtractions... mov ebx, [ar2] sub [ar1], ebx dec length jz @Exit mov ebx, [ar2 + 4] sub [ar1 + 4], ebx dec length jz @Exit mov ebx, [ar2 + 8] sub [ar1 + 8], ebx @Exit: pop ebx end; procedure SubArAsmSimdA(var ar1, ar2; length: integer); asm push ebx //Unfortunately delphi use first 8 bytes for dinamic array length and reference //counter, from that reason the dinamic array address should start with $xxxxxxx8 //instead &xxxxxxx0. So we must first align ar1, ar2 pointers! mov ebx, [ar2] sub [ar1], ebx dec length jz @exit mov ebx, [ar2 + 4] sub [ar1 + 4], ebx dec length jz @exit add ar1, 8 add ar2, 8 //Prepare length for 16 byte data transfer push length shr length, 2 jz @Finish @Loop: movdqa xmm1, [ar1] movdqa xmm2, [ar2] psubd xmm1, xmm2 movdqa [ar1], xmm1 add ar1, 16 add ar2, 16 dec length jnz @Loop @Finish: pop length and length, 3 jz @Exit //Do rest, up to 3 subtractions... mov ebx, [ar2] sub [ar1], ebx dec length jz @Exit mov ebx, [ar2 + 4] sub [ar1 + 4], ebx dec length jz @Exit mov ebx, [ar2 + 8] sub [ar1 + 8], ebx @Exit: pop ebx end; procedure SubArAsmSimdAUnrolled8(var ar1, ar2; length: integer); asm push ebx //Unfortunately delphi use first 8 bytes for dinamic array length and reference //counter, from that reason the dinamic array address should start with $xxxxxxx8 //instead &xxxxxxx0. So we must first align ar1, ar2 pointers! mov ebx, [ar2] sub [ar1], ebx dec length jz @exit mov ebx, [ar2 + 4] sub [ar1 + 4], ebx dec length jz @exit add ar1, 8 //Align pointer to 16 byte add ar2, 8 //Align pointer to 16 byte //Prepare length for 16 byte data transfer push length shr length, 5 //8 * 4 subtructions per loop jz @Finish //To small for LoopUnrolled @LoopUnrolled: //Unrolle 1, 2, 3, 4 movdqa xmm4, [ar2] movdqa xmm5, [16 + ar2] movdqa xmm6, [32 + ar2] movdqa xmm7, [48 + ar2] // movdqa xmm0, [ar1] movdqa xmm1, [16 + ar1] movdqa xmm2, [32 + ar1] movdqa xmm3, [48 + ar1] // psubd xmm0, xmm4 psubd xmm1, xmm5 psubd xmm2, xmm6 psubd xmm3, xmm7 // movdqa [48 + ar1], xmm3 movdqa [32 + ar1], xmm2 movdqa [16 + ar1], xmm1 movdqa [ar1], xmm0 //Unrolle 5, 6, 7, 8 movdqa xmm4, [64 + ar2] movdqa xmm5, [80 + ar2] movdqa xmm6, [96 + ar2] movdqa xmm7, [112 + ar2] // movdqa xmm0, [64 + ar1] movdqa xmm1, [80 + ar1] movdqa xmm2, [96 + ar1] movdqa xmm3, [112 + ar1] // psubd xmm0, xmm4 psubd xmm1, xmm5 psubd xmm2, xmm6 psubd xmm3, xmm7 // movdqa [112 + ar1], xmm3 movdqa [96 + ar1], xmm2 movdqa [80 + ar1], xmm1 movdqa [64 + ar1], xmm0 // add ar1, 128 add ar2, 128 dec length jnz @LoopUnrolled @FinishUnrolled: pop length and length, $1F //Do rest, up to 31 subtractions... @Finish: mov ebx, [ar2] sub [ar1], ebx add ar1, 4 add ar2, 4 dec length jnz @Finish @Exit: pop ebx end; procedure WriteOut(EndTicks: Int64; Str: string); begin WriteLn(Str + IntToStr(EndTicks - StartTicks) + ' Time: ' + IntToStr((EndTicks - StartTicks) div TicksPerMicroSecond) + 'us'); Sleep(5); SwitchToThread; StartTicks := GetCpuTicks; end; begin ArLength := 100000; //Set TicksPerMicroSecond QueryPerformanceFrequency(TicksPerMicroSecond); TicksPerMicroSecond := TicksPerMicroSecond div 1000000; // SetLength(Ar1, ArLength); SetLength(Ar2, ArLength); //Fill arrays //... //Tick time info WriteLn('CPU ticks per mikro second: ' + IntToStr(TicksPerMicroSecond)); Sleep(5); SwitchToThread; StartTicks := GetCpuTicks; //Test 1 SubArPasRangeOvfChkOn(ArLength); WriteOut(GetCpuTicks, 'SubAr Pas Range and Overflow Checking On, Ticks: '); //Test 2 SubArPas(ArLength); WriteOut(GetCpuTicks, 'SubAr Pas, Ticks: '); //Test 3 SubArAsm(Ar1[0], Ar2[0], ArLength); WriteOut(GetCpuTicks, 'SubAr Asm, Ticks: '); //Test 4 SubArAsmSimdU(Ar1[0], Ar2[0], ArLength); WriteOut(GetCpuTicks, 'SubAr Asm SIMD mem unaligned, Ticks: '); //Test 5 SubArAsmSimdA(Ar1[0], Ar2[0], ArLength); WriteOut(GetCpuTicks, 'SubAr Asm with SIMD mem aligned, Ticks: '); //Test 6 SubArAsmSimdAUnrolled8(Ar1[0], Ar2[0], ArLength); WriteOut(GetCpuTicks, 'SubAr Asm with SIMD mem aligned 8*unrolled, Ticks: '); // ReadLn; Ar1 := nil; Ar2 := nil; end.

...

最快的asm程序，8次展开的SIMD指令只需68us，比Pascal程序快4倍。

正如我们所看到的那样，Pascal循环过程可能并不重要，只需要大约277us（溢出和范围检查）在2,4GHz CPU上进行100000次减法。

所以这段代码不能成为瓶颈？

Answer 4

我不是汇编专家，但我认为如果不考虑SIMD指令或并行处理，以下几乎是最优的，后者可以通过将部分数组传递给函数来轻松完成。

像
Thread1：SubArray（ar1 [0]，ar2 [0]，50）;
Thread2：SubArray（ar1 [50]，ar2 [50]，50）;

procedure SubArray(var Array1, Array2; const Length: Integer);
var
  ap1, ap2 : PInteger;
  i : Integer;
begin
  ap1 := @Array1;
  ap2 := @Array2;
  i := Length;
  while i > 0 do
  begin
    ap1^ := ap1^ - ap2^;
    Inc(ap1);
    Inc(ap2);
    Dec(i);
  end;
end;

// similar assembly version
procedure SubArrayEx(var Array1, Array2; const Length: Integer);
asm
  // eax = @Array1
  // edx = @Array2
  // ecx = Length
  // esi = temp register for array2^
  push esi
  cmp ecx, 0
  jle @Exit
  @Loop:
  mov esi, [edx]
  sub [eax], esi
  add eax, 4
  add edx, 4
  dec ecx
  jnz @Loop
  @Exit:
  pop esi
end;


procedure Test();
var
  a1, a2 : array of Integer;
  i : Integer;
begin
  SetLength(a1, 3);
  a1[0] := 3;
  a1[1] := 1;
  a1[2] := 2;
  SetLength(a2, 3);
  a2[0] := 4;
  a2[1] := 21;
  a2[2] := 2;
  SubArray(a1[0], a2[0], Length(a1));

  for i := 0 to Length(a1) - 1 do
    Writeln(a1[i]);

  Readln;
end;

Answer 5

这不是你问题的真正答案，但是我会研究是否可以在某个时间进行减法，而用数值填充数组。我甚至可以选择在内存中考虑第三个数组来存储减法的结果。在现代计算中，内存的“成本”远低于对内存执行额外操作所花费的时间的“成本”。

理论上，当值仍然在寄存器或处理器缓存中时，您可以获得至少一点性能，但实际上您可能会偶然发现一些可以提高整个算法性能的技巧

从一个数组中减去另一个数组的最佳方法

5 个答案: