c - SSE 2 函数执行时间不稳定并且比正常情况要多

Question

在 Intel core2Duo 上使用 SSE 2。

sse_add()和normal_add()在多次运行中花费的时间并不是恒定的，事实上现在经过多次修改后总是为0。

该程序基本上求出以下矩阵每一列的总和:

  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，这两个函数都是正确的:

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

为什么会发生这种情况？还是因为我没有正确测量时间？您能否在您的计算机上运行相同的代码并进行验证？

已更新根据建议，我放置了一个 for 循环，如下所示，但时间仍然为 0(显然我必须将总时间除以编号)。迭代以获得正确的值，但为什么我得到的总时间为 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);

}

原程序如下。

*更新:删除了所有 printf 和 malloc*

在单独的程序中一一计时功能:

上交所版本

 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() 调用很容易主导这两个函数的执行时间，这也可能是时间变化的来源。

没有必要在那里调用 malloc() - 事实上，您有内存泄漏，因为您调用它，然后立即覆盖它的返回值。

如果您想分析这些函数，请将加法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(例如中断)引起的随机变化。