xref: /petsc/src/benchmarks/streams/BasicVersion.c (revision 6f2b61bcedfb8ddbb24d22b198f01777d248bd01)

#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 <limits.h>
# include <float.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();
  register int j, k;
  double       scalar, t, times[4][NTIMES],irate[4],rate[4];
  int          rank,size;

  MPI_Init(&argc,&args);
  MPI_Comm_rank(MPI_COMM_WORLD,&rank);
  MPI_Comm_size(MPI_COMM_WORLD,&size);
  if (!rank) printf("Number of MPI processes %d\n",size);

  /* --- 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);
}