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I am working on a large application for which I need to perform loop unrolling on subsequent dependent loops for a certain procedure. I have written below a small sample piece of code to replicate the larger version.

Consider the original code:

void main()
{

 int a[20] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20};
 int b[20] = {10,9,8,7,6,5,4,3,2,1,20,19,18,17,16,15,14,13,12,11};
 int i,j,k,l;
 int nab =4, vab =10;
 int dimi, dimj, dimij, dimk, diml, dimkl, dimijkl;
 int count = 0;

 for (i = nab+1; i< nab+vab; i++) 
 {
   dimi = a[i];
   for (j = i; j< nab+vab; j++)
   {
    dimj = b[j];
    dimij = dimi*dimj;
    count = count +1;

    for (k = nab+1; k< nab+vab; k++)
    {
     dimk = a[k-1];
     for (l =k; l< nab+vab; l++)
     {
      diml = a[l-1];
      dimkl = dimk*diml;
      dimijkl = dimij * dimkl;
     }
    }
   }
  }
 printf ("Final dimension:%d \n ", dimijkl);
 printf ("Count:%d \n ", count);
}

Now I am unrolling the loop i by a factor of 2:

for (i = nab+1; i< nab+vab; i+=2)
{
  dimi = a[i];
  for (j = i; j< nab+vab; j++)
  {
   dimj = b[j];
   dimij = dimi*dimj;
   count = count +1;

   for (k = nab+1; k< nab+vab; k++)
   {
     dimk = a[k-1];
     for (l =k; l< nab+vab; l++)
     {
      diml = a[l-1];
      dimkl = dimk*diml;
      dimijkl = dimij * dimkl;
     }
    }
  }

  dimi = a[i+1];
  for (j = i+1; j< nab+vab; j++)
  {
    dimj = b[j];
    dimij = dimi*dimj;
    count = count +1;

     for (k = nab+1; k< nab+vab; k++)
     {
      dimk = a[k-1];
      for (l =k; l< nab+vab; l++)
      {
        diml = a[l-1];
        dimkl = dimk*diml;
        dimijkl = dimij * dimkl;
      }    
     }
    }
   }
   printf ("Final dimension:%d \n ", dimijkl);
   printf ("Count:%d \n ", count);

Now I wish to unroll the loop i and j by a factor of 2, but since loop j depends on loop i, I am a bit unsure as to how I should approach writing it. How can I rewrite the code to unroll both i and j by a factor of 2. Also the code will become increasingly clumsier as i increase the unroll factor. Is there a clever way to unroll it manually, without the code becoming too ugly.

I cannot use compiler flags (example:-funroll-loops) in this particular case. I want approach it by manual loop unrolling.

Thank you for your time.

4

1 回答 1

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也许不是最优雅的解决方案，但您可以使用宏：

#define INNER_L_LOOP(l) \
    { \
        diml = a[l-1]; \
        dimkl = dimk*diml; \
        dimijkl = dimij * dimkl; \
    } 

#define INNER_K_LOOP(k) \
    { \
        dimk = a[k-1]; \
        for (l =k; l< nab+vab; l++) \
        { \
            INNER_L_LOOP(l) \
        } \
    } 

#define INNER_J_LOOP(j) \
    { \
        dimj = b[j]; \
        dimij = dimi*dimj; \
        count = count +1; \
        \
        for (k = nab+1; k< nab+vab; k += 2) \
        { \
            INNER_K_LOOP(k) \
            if (k+1 < nab+vab) \
                INNER_K_LOOP(k+1) \
        } \
    } 

#define INNER_I_LOOP(i) \
    { \
        dimi = a[i]; \
        for (j = i; j< nab+vab; j+= 2) \
        { \
            INNER_J_LOOP(j) \
            if (j+1 < nab+vab) \
                INNER_J_LOOP(j+1) \
        } \
    } 

 for (i = nab+1; i< nab+vab; i+=2) 
 {
     INNER_I_LOOP(i)
     INNER_I_LOOP(i+1)
 }

在这里，我将 i、j 和 k 循环分别展开了 2 倍，您会注意到最内层循环的代码不必重复 8 次（这是您试图避免的，如果我理解正确）。

您还会注意到，对于 j 和 k 循环，我正在检查索引是否超出数组边界。此检查对性能有轻微影响，我将由您自行优化。;)

于 2013-06-12T16:40:17.573 回答