xref: /petsc/src/benchmarks/streams/BasicVersion.c (revision d1d3a73c9cae8f9d8a3fa669ac6c92ec8c5cdf31)

#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      200000
# 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();


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

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

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

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

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

  scalar = 3.0;
  for (k=0; k<NTIMES; k++)
  {

    times[0][k] = second();
/* should all these barriers be pulled outside of the time call? */

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

    times[1][k] = second();

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

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

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

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

# define        M        20

int checktick(void)
{
  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);
}