Question

在Intel core2Duo上使用SSE 2。

在sse_add（）和normal_add（）中花费的时间在多次运行中并不是常数，实际上现在经过多次修改后总是显示为0.

程序基本上找到以下矩阵的每列的总和：

  1,2,10,13,15,160,6,19

  1,2,10,13,15,160,6,19

  1,2,10,13,15,160,6,19

  1,2,10,13,15,160,6,19

  1,2,10,13,15,160,6,19

  1,2,10,13,15,160,6,19

  1,2,10,13,15,160,6,19

  1,2,10,13,15,160,6,19

我已经验证了结果，这两个功能都正确：

 results= 8, 16, 80, 104, 120, 1280, 48, 152

为什么会这样？还是因为我没有正确地测量时间？你可以在你的机器上运行相同的代码并验证吗？

已更新根据建议，我已经在下面显示了一个for循环，但是时间仍然是0（显然我必须将总时间除以否。迭代得到正确的值，但为什么我得到总时间0？）：

 // variable declarations used for time calculation
 double elapTicks;
 double elapMilli ;
 double begin = BeginTimer();

   for(int i=0; i<1000000000;i++)

   {

  //sum of consecutive rows 
  __m128i t1=_mm_adds_epi16(    x1[0] ,   x2[0]  );
  __m128i t2=_mm_adds_epi16(    x3[0] ,   x4[0]  );
  __m128i t3=_mm_adds_epi16(    x5[0] ,   x6[0]  );
  __m128i t4=_mm_adds_epi16(    x7[0] ,   x8[0]  );

 //t5=t1+t2 & t6=t3 + t4
 __m128i t5=_mm_adds_epi16(  t1 ,t2  );
 __m128i t6=_mm_adds_epi16(  t3 ,t4  );


 ///t7=t6+t5, which is final answer 
 __m128i t7=_mm_adds_epi16(  t5 ,t6  );


 }


   printf ("Timer set to: %f\n", begin);
 // variable definitions  to calculate time taken
     elapTicks = EndTimer(begin)-begin;    // stop the timer,and calculate the time 
     taken
     elapMilli = elapTicks/1000;     // milliseconds from Begin to End
     printf ("Time  in SSE in Milliseconds : %f\n", elapMilli);

}

原始程序如下。

* 更新： 删除所有printf和malloc *

时序在不同的程序中逐个执行：

SSE版本

 clock_t BeginTimer()
 {
 //timer declaration
 clock_t Begin; //initialize Begin
 Begin = clock() * CLK_TCK; //start the timer
 return Begin;
 }
 clock_t EndTimer(clock_t begin)
 {
 clock_t End;
 End = clock() * CLK_TCK;   //stop the timer
 return End;
 }


 int  main( )
 {
   sse_add();
   getch();       
  return 0;
  }


void sse_add()
{

__declspec(align(16)) unsigned short a1[8]={1,2,10,13,15,160,6,19};   
    __declspec(align(16)) unsigned short a2[8]={1,2,10,13,15,160,6,19};
__declspec(align(16)) unsigned short a3[8]={1,2,10,13,15,160,6,19};
__declspec(align(16)) unsigned short a4[8]={1,2,10,13,15,160,6,19};
__declspec(align(16)) unsigned short a5[8]={1,2,10,13,15,160,6,19};
__declspec(align(16)) unsigned short a6[8]={1,2,10,13,15,160,6,19};
__declspec(align(16)) unsigned short a7[8]={1,2,10,13,15,160,6,19};
__declspec(align(16)) unsigned short a8[8]={1,2,10,13,15,160,6,19};

   //__m128i maps to the XMM[0-7] registers
   __m128i *x1 = (__m128i*) &a1[0];  
   __m128i *x2 = (__m128i*) &a2[0]; 
   __m128i *x3 = (__m128i*) &a3[0];   
   __m128i *x4 = (__m128i*) &a4[0];          
   __m128i *x5 = (__m128i*) &a5[0];   
   __m128i *x6 = (__m128i*) &a6[0];   
   __m128i *x7 = (__m128i*) &a7[0];   
   __m128i *x8 = (__m128i*) &a8[0];   

//  _mm_adds_epi16 : Adds the 8 signed 16-bit integers in a to the 8 signed \
    //16-bit  integers in b and saturates.


 // variable declarations used for time calculation
    float elapTicks;
    float elapMilli ;


   double begin = BeginTimer();
       printf ("Timer set to: %.2f\n", begin); // print the initialised timer (0)


  //sum of consecutive rows 
      __m128i t1=_mm_adds_epi16(    x1[0] ,   x2[0]  );
      __m128i t2=_mm_adds_epi16(    x3[0] ,   x4[0]  );
      __m128i t3=_mm_adds_epi16(    x5[0] ,   x6[0]  );
      __m128i t4=_mm_adds_epi16(    x7[0] ,   x8[0]  );

   //t5=t1+t2 & t6=t3 + t4
     __m128i t5=_mm_adds_epi16(  t1 ,t2  );
     __m128i t6=_mm_adds_epi16(  t3 ,t4  );


   ///t7=t6+t5, which is final answer 
   __m128i t7=_mm_adds_epi16(  t5 ,t6  );


// variable definitions  to calculate time taken
elapTicks = EndTimer(begin);    // stop the timer, and calculate the time taken
elapMilli = elapTicks/1000;     // milliseconds from Begin to End
printf ("Time  in SSE in Milliseconds : %.2f\n", elapMilli);

}

普通版

 clock_t BeginTimer()
 {
 //timer declaration
 clock_t Begin; //initialize Begin
 Begin = clock() * CLK_TCK; //start the timer
 return Begin;
 }

 clock_t EndTimer(clock_t begin)
{
clock_t End;
End = clock() * CLK_TCK;   //stop the timer
return End;
}


int  main( )
{

normal_add();
   getch();
  return 0;
}


  void normal_add()
  {

 unsigned short a1[8]={1,2,10,13,15,160,6,19};
 unsigned short a2[8]={1,2,10,13,15,160,6,19};
 unsigned short a3[8]={1,2,10,13,15,160,6,19};
 unsigned short a4[8]={1,2,10,13,15,160,6,19};
 unsigned short a5[8]={1,2,10,13,15,160,6,19};
 unsigned short a6[8]={1,2,10,13,15,160,6,19};
 unsigned short a7[8]={1,2,10,13,15,160,6,19};
 unsigned short a8[8]={1,2,10,13,15,160,6,19};

   unsigned short t1[8], t2[8], t3[8], t4[8],t5[8], t6[8], t7[8];   



   float elapTicks;
       float elapMilli ;


double begin1 = BeginTimer();
    printf ("Timer reset to: %f\n", begin1); // print the initialised timer (0)


 for(int i=0; i<8;i++)
 {
 t1[i]=a1[i] +a2[i];
 }


 for(int i=0; i<8;i++)
 {
 t2[i]=a3[i] +a4[i];
 }


 for(int i=0; i<8;i++)
 {
 t3[i]=a5[i] +a6[i];
 }

 for(int i=0; i<8;i++)
 {
 t4[i]=a7[i] +a8[i];
 }

 for(int i=0; i<8;i++)
         {
 t5[i]=t1[i] +t2[i];
 }

 for(int i=0; i<8;i++)
         {
 t6[i]=t3[i] +t4[i];
 }

 for(int i=0; i<8;i++)
 {
 t7[i]=t5[i] +t6[i];
 }

 elapTicks = EndTimer(begin1);    // stop the timer, and calculete the time taken
   elapMilli = elapTicks/1000;     // milliseconds from Begin to End
   printf ("Time spent in normal add in Milliseconds : %.2f\n", elapMilli);


  }

Answer 1

printf()和malloc()调用很容易控制这两个函数的执行时间，而且很可能也会出现时序变化的情况。

没有必要在那里调用malloc() - 实际上你有内存泄漏，因为你调用它然后立即覆盖它的返回值。

如果要分析这些功能，请将附加逻辑本身与打印逻辑分开，并多次调用前者而不是一次。这将分摊由外部事件引起的随机变化，如中断。

Answer 2

您没有提到您获得的时间变化有多大，但可能会有一些变化。由于OS调度和由其他正在运行的线程引起的内存瓶颈等原因，指令并不总是以相同的时间量执行，而是对CPU中的晶体管的电荷变化进行充电。涉及到许多变量，您无法确定精确的时间。您可以做的最好的事情通常是运行几百万次并在平均执行时间上设置一个置信区间现在，如果您的变化很大，可能还有其他事情发生，但如果没有实际数字，则很难判断。

SSE 2功能执行时序不是常数并且超过正常值

2 个答案: