xref: /petsc/src/benchmarks/streams/BasicVersion.c (revision df4a11de7f42103c952bf4d7573e839493e4f44b) !

#include <sys/time.h>
/* int gettimeofday(struct timeval *tp, struct timezone *tzp); */

double second()
{
/* struct timeval { long	tv_sec;
	    long	tv_usec;	};

struct timezone { int	tz_minuteswest;
	     int	tz_dsttime;	 };	*/

	struct timeval tp;
	struct timezone tzp;
	int i;

	i = gettimeofday(&tp,&tzp);
	return ( (double) tp.tv_sec + (double) tp.tv_usec * 1.e-6 );
}
# include <stdio.h>
# include <math.h>
# include <float.h>
# include <limits.h>
# include <sys/time.h>

/*
* Program: Stream
* Programmer: Joe R. Zagar
* Revision: 4.0-BETA, October 24, 1995
* Original code developed by John D. McCalpin
*
* This program measures memory transfer rates in MB/s for simple
* computational kernels coded in C.  These numbers reveal the quality
* of code generation for simple uncacheable kernels as well as showing
* the cost of floating-point operations relative to memory accesses.
*
* INSTRUCTIONS:
*
*	1) Stream requires a good bit of memory to run.  Adjust the
*          value of 'N' (below) to give a 'timing calibration' of
*          at least 20 clock-ticks.  This will provide rate estimates
*          that should be good to about 5% precision.
*/

# define N	2000000
# define NTIMES	50
# define OFFSET	0

/*
*	3) Compile the code with full optimization.  Many compilers
*	   generate unreasonably bad code before the optimizer tightens
*	   things up.  If the results are unreasonably good, on the
*	   other hand, the optimizer might be too smart for me!
*
*         Try compiling with:
*               cc -O stream_d.c second.c -o stream_d -lm
*
*         This is known to work on Cray, SGI, IBM, and Sun machines.
*
*
*	4) Mail the results to mccalpin@cs.virginia.edu
*	   Be sure to include:
*		a) computer hardware model number and software revision
*		b) the compiler flags
*		c) all of the output from the test case.
* Thanks!
*
*/

# define HLINE "-------------------------------------------------------------\n"

# ifndef MIN
# define MIN(x,y) ((x)<(y)?(x):(y))
# endif
# ifndef MAX
# define MAX(x,y) ((x)>(y)?(x):(y))
# endif

static double	a[N+OFFSET],
		b[N+OFFSET],
		c[N+OFFSET];
/*double *a,*b,*c;*/

static double mintime[4] = {FLT_MAX,FLT_MAX,FLT_MAX,FLT_MAX};

static const char     *label[4] = {"Copy:      ", "Scale:     ", "Add:       ", "Triad:     "};

static double   bytes[4] = {
   2 * sizeof(double) * N,
   2 * sizeof(double) * N,
   3 * sizeof(double) * N,
   3 * sizeof(double) * N
   };

extern double second();

#include <mpi.h>

int main(int argc,char **args)
   {
   int          quantum, checktick();
   int          BytesPerWord;
   register int j, k;
   double       scalar, t, times[4][NTIMES],irate[4],rate[4];
   int          rank;

   MPI_Init(&argc,&args);
   MPI_Comm_rank(MPI_COMM_WORLD,&rank);

   /* --- SETUP --- determine precision and check timing --- */

   if (!rank) {
     printf(HLINE);
     printf("Array size = %d, Offset = %d\n" , N, OFFSET);
     printf("Total memory required = %.1f MB.\n", (3 * N * BytesPerWord) / 1048576.0);
     printf("Each test is run %d times, but only\n", NTIMES);
     printf("the *best* time for each is used.\n");
     printf(HLINE);
   }

   /* Get initial value for system clock. */

   /*  a = malloc(N*sizeof(double));
   b = malloc(N*sizeof(double));
   c = malloc(N*sizeof(double));*/
   for (j=0; j<N; j++) {
        a[j] = 1.0;
        b[j] = 2.0;
        c[j] = 0.0;
        }

   if (!rank) {
     if  ( (quantum = checktick()) >= 1)
       printf("Your clock granularity/precision appears to be %d microseconds.\n", quantum);
     else
        printf("Your clock granularity appears to be less than one microsecond.\n");
   }

   t = second();
   for (j = 0; j < N; j++)
        a[j] = 2.0E0 * a[j];
   t = 1.0E6 * (second() - t);

   if (!rank) {
     printf("Each test below will take on the order of %d microseconds.\n", (int) t  );
     printf("   (= %d clock ticks)\n", (int) (t/quantum) );
     printf("Increase the size of the arrays if this shows that\n");
     printf("you are not getting at least 20 clock ticks per test.\n");
     printf(HLINE);
   }


   /*   --- MAIN LOOP --- repeat test cases NTIMES times --- */

   scalar = 3.0;
   for (k=0; k<NTIMES; k++)
        {
   MPI_Barrier(MPI_COMM_WORLD);
        times[0][k] = second();
   /* should all these barriers be pulled outside of the time call? */
   MPI_Barrier(MPI_COMM_WORLD);
        for (j=0; j<N; j++)
            c[j] = a[j];
   MPI_Barrier(MPI_COMM_WORLD);
        times[0][k] = second() - times[0][k];

        times[1][k] = second();
   MPI_Barrier(MPI_COMM_WORLD);
        for (j=0; j<N; j++)
            b[j] = scalar*c[j];
   MPI_Barrier(MPI_COMM_WORLD);
        times[1][k] = second() - times[1][k];

        times[2][k] = second();
   MPI_Barrier(MPI_COMM_WORLD);
        for (j=0; j<N; j++)
            c[j] = a[j]+b[j];
   MPI_Barrier(MPI_COMM_WORLD);
        times[2][k] = second() - times[2][k];

        times[3][k] = second();
   MPI_Barrier(MPI_COMM_WORLD);
        for (j=0; j<N; j++)
            a[j] = b[j]+scalar*c[j];
   MPI_Barrier(MPI_COMM_WORLD);
        times[3][k] = second() - times[3][k];
     }

   /*   --- SUMMARY --- */

   for (k=0; k<NTIMES; k++){
        for (j=0; j<4; j++) {
           mintime[j] = MIN(mintime[j], times[j][k]);
        }
      }

   for (j=0; j<4; j++) {
     irate[j] = 1.0E-06 * bytes[j]/mintime[j];
   }
   MPI_Reduce(irate,rate,4,MPI_DOUBLE,MPI_SUM,0,MPI_COMM_WORLD);

   if (!rank) {
     printf("Function      Rate (MB/s) \n");
     for (j=0; j<4; j++) {
        printf("%s%11.4f\n", label[j],rate[j]);
     }
   }
   MPI_Finalize();
   return 0;
}

# define	M	20

int
checktick()
   {
   int		i, minDelta, Delta;
   double	t1, t2, timesfound[M];

/*  Collect a sequence of M unique time values from the system. */

   for (i = 0; i < M; i++) {
	t1 = second();
	while( ((t2=second()) - t1) < 1.0E-6 )
	    ;
	timesfound[i] = t1 = t2;
	}

/*
* Determine the minimum difference between these M values.
* This result will be our estimate (in microseconds) for the
* clock granularity.
*/

   minDelta = 1000000;
   for (i = 1; i < M; i++) {
	Delta = (int)( 1.0E6 * (timesfound[i]-timesfound[i-1]));
	minDelta = MIN(minDelta, MAX(Delta,0));
	}

   return(minDelta);
   }