Re: [Help-gsl] Questions about the code of some functions in cblas implementation

[José Luis García Pallero]

El 23 de junio de 2009 18:40, Brian Gough <address@hidden> escribió:

> At Thu, 18 Jun 2009 23:42:57 +0200,
> José Luis García Pallero wrote:
>
> > No loop unrolling: 0.005 s
> > Loop unrolling: 0.6 s
> >
> >         for(i=0;i<n;i++)
> >         {
> >             a = i*i+i;
> >         }
> >     }
>
> I think the program below is probably more realistic for this
> case. Given the huge difference between the two results I suspect that
> the compiler is able to overoptimise the simple case above.  Maybe you
> could compare this or the actual function.
>
> #include <stdlib.h>
> #include <time.h>
> #include <stdio.h>
>
> int
> main (int argc, char *argv[])
> {
>  int n = 0, i = 0, j, m;
>  double *a, *x;
>  double t0, t1, t2;
>  double A = 3, B = 2;
>
>  n = atoi (argv[1]);
>  m = atoi (argv[1]);
>  a = malloc (sizeof (double) * n);
>  x = malloc (sizeof (double) * n);
>
>  t0 = clock ();
>  {
>    for (j = 0; j < m; j++)
>      for (i = 0; i < n; i++)
>        {
>          a[i] = A * x[i] + B;
>        }
>  }
>
>  t1 = clock ();
>
>  {
>    for (j = 0; j < m; j++)
>      for (i = 0; i < n; i += 4)
>        {
>          a[i] = A * x[i] + B;
>          a[i + 1] = A * x[i + 1] + B;
>          a[i + 2] = A * x[i + 2] + B;
>          a[i + 3] = A * x[i + 3] + B;
>        }
>  }
>
>  t2 = clock ();
>  printf ("operations = %g\n", (double) (n * m));
>  printf ("plain loop = %g\n", t1 - t0);
>  printf ("fancy loop = %g\n", t2 - t1);
>  return 0;
> }
>

OK, I've tried your code and you are right, the loop unrolled version is
faster. Thanks for the time that you spent in coding the example.

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José Luis García Pallero
address@hidden
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