1 2 #include <sys/time.h> 3 /* int gettimeofday(struct timeval *tp, struct timezone *tzp); */ 4 5 double second() 6 { 7 /* struct timeval { long tv_sec; 8 long tv_usec; }; 9 10 struct timezone { int tz_minuteswest; 11 int tz_dsttime; }; */ 12 13 struct timeval tp; 14 struct timezone tzp; 15 int i; 16 17 i = gettimeofday(&tp,&tzp); 18 return ( (double) tp.tv_sec + (double) tp.tv_usec * 1.e-6 ); 19 } 20 # include <stdio.h> 21 # include <math.h> 22 # include <float.h> 23 # include <limits.h> 24 # include <sys/time.h> 25 26 /* 27 * Program: Stream 28 * Programmer: Joe R. Zagar 29 * Revision: 4.0-BETA, October 24, 1995 30 * Original code developed by John D. McCalpin 31 * 32 * This program measures memory transfer rates in MB/s for simple 33 * computational kernels coded in C. These numbers reveal the quality 34 * of code generation for simple uncacheable kernels as well as showing 35 * the cost of floating-point operations relative to memory accesses. 36 * 37 * INSTRUCTIONS: 38 * 39 * 1) Stream requires a good bit of memory to run. Adjust the 40 * value of 'N' (below) to give a 'timing calibration' of 41 * at least 20 clock-ticks. This will provide rate estimates 42 * that should be good to about 5% precision. 43 */ 44 45 # define N 2000000 46 # define NTIMES 50 47 # define OFFSET 0 48 49 /* 50 * 3) Compile the code with full optimization. Many compilers 51 * generate unreasonably bad code before the optimizer tightens 52 * things up. If the results are unreasonably good, on the 53 * other hand, the optimizer might be too smart for me! 54 * 55 * Try compiling with: 56 * cc -O stream_d.c second.c -o stream_d -lm 57 * 58 * This is known to work on Cray, SGI, IBM, and Sun machines. 59 * 60 * 61 * 4) Mail the results to mccalpin@cs.virginia.edu 62 * Be sure to include: 63 * a) computer hardware model number and software revision 64 * b) the compiler flags 65 * c) all of the output from the test case. 66 * Thanks! 67 * 68 */ 69 70 # define HLINE "-------------------------------------------------------------\n" 71 72 # ifndef MIN 73 # define MIN(x,y) ((x)<(y)?(x):(y)) 74 # endif 75 # ifndef MAX 76 # define MAX(x,y) ((x)>(y)?(x):(y)) 77 # endif 78 79 static double a[N+OFFSET], 80 b[N+OFFSET], 81 c[N+OFFSET]; 82 /*double *a,*b,*c;*/ 83 84 static double mintime[4] = {FLT_MAX,FLT_MAX,FLT_MAX,FLT_MAX}; 85 86 static const char *label[4] = {"Copy: ", "Scale: ", "Add: ", "Triad: "}; 87 88 static double bytes[4] = { 89 2 * sizeof(double) * N, 90 2 * sizeof(double) * N, 91 3 * sizeof(double) * N, 92 3 * sizeof(double) * N 93 }; 94 95 extern double second(); 96 97 #include <mpi.h> 98 99 int main(int argc,char **args) 100 { 101 int quantum, checktick(); 102 int BytesPerWord; 103 register int j, k; 104 double scalar, t, times[4][NTIMES],irate[4],rate[4]; 105 int rank; 106 107 MPI_Init(&argc,&args); 108 MPI_Comm_rank(MPI_COMM_WORLD,&rank); 109 110 /* --- SETUP --- determine precision and check timing --- */ 111 112 if (!rank) { 113 printf(HLINE); 114 printf("Array size = %d, Offset = %d\n" , N, OFFSET); 115 printf("Total memory required = %.1f MB.\n", (3 * N * BytesPerWord) / 1048576.0); 116 printf("Each test is run %d times, but only\n", NTIMES); 117 printf("the *best* time for each is used.\n"); 118 printf(HLINE); 119 } 120 121 /* Get initial value for system clock. */ 122 123 /* a = malloc(N*sizeof(double)); 124 b = malloc(N*sizeof(double)); 125 c = malloc(N*sizeof(double));*/ 126 for (j=0; j<N; j++) { 127 a[j] = 1.0; 128 b[j] = 2.0; 129 c[j] = 0.0; 130 } 131 132 if (!rank) { 133 if ( (quantum = checktick()) >= 1) 134 printf("Your clock granularity/precision appears to be %d microseconds.\n", quantum); 135 else 136 printf("Your clock granularity appears to be less than one microsecond.\n"); 137 } 138 139 t = second(); 140 for (j = 0; j < N; j++) 141 a[j] = 2.0E0 * a[j]; 142 t = 1.0E6 * (second() - t); 143 144 if (!rank) { 145 printf("Each test below will take on the order of %d microseconds.\n", (int) t ); 146 printf(" (= %d clock ticks)\n", (int) (t/quantum) ); 147 printf("Increase the size of the arrays if this shows that\n"); 148 printf("you are not getting at least 20 clock ticks per test.\n"); 149 printf(HLINE); 150 } 151 152 153 /* --- MAIN LOOP --- repeat test cases NTIMES times --- */ 154 155 scalar = 3.0; 156 for (k=0; k<NTIMES; k++) 157 { 158 MPI_Barrier(MPI_COMM_WORLD); 159 times[0][k] = second(); 160 /* should all these barriers be pulled outside of the time call? */ 161 MPI_Barrier(MPI_COMM_WORLD); 162 for (j=0; j<N; j++) 163 c[j] = a[j]; 164 MPI_Barrier(MPI_COMM_WORLD); 165 times[0][k] = second() - times[0][k]; 166 167 times[1][k] = second(); 168 MPI_Barrier(MPI_COMM_WORLD); 169 for (j=0; j<N; j++) 170 b[j] = scalar*c[j]; 171 MPI_Barrier(MPI_COMM_WORLD); 172 times[1][k] = second() - times[1][k]; 173 174 times[2][k] = second(); 175 MPI_Barrier(MPI_COMM_WORLD); 176 for (j=0; j<N; j++) 177 c[j] = a[j]+b[j]; 178 MPI_Barrier(MPI_COMM_WORLD); 179 times[2][k] = second() - times[2][k]; 180 181 times[3][k] = second(); 182 MPI_Barrier(MPI_COMM_WORLD); 183 for (j=0; j<N; j++) 184 a[j] = b[j]+scalar*c[j]; 185 MPI_Barrier(MPI_COMM_WORLD); 186 times[3][k] = second() - times[3][k]; 187 } 188 189 /* --- SUMMARY --- */ 190 191 for (k=0; k<NTIMES; k++){ 192 for (j=0; j<4; j++) { 193 mintime[j] = MIN(mintime[j], times[j][k]); 194 } 195 } 196 197 for (j=0; j<4; j++) { 198 irate[j] = 1.0E-06 * bytes[j]/mintime[j]; 199 } 200 MPI_Reduce(irate,rate,4,MPI_DOUBLE,MPI_SUM,0,MPI_COMM_WORLD); 201 202 if (!rank) { 203 printf("Function Rate (MB/s) \n"); 204 for (j=0; j<4; j++) { 205 printf("%s%11.4f\n", label[j],rate[j]); 206 } 207 } 208 MPI_Finalize(); 209 return 0; 210 } 211 212 # define M 20 213 214 int 215 checktick() 216 { 217 int i, minDelta, Delta; 218 double t1, t2, timesfound[M]; 219 220 /* Collect a sequence of M unique time values from the system. */ 221 222 for (i = 0; i < M; i++) { 223 t1 = second(); 224 while( ((t2=second()) - t1) < 1.0E-6 ) 225 ; 226 timesfound[i] = t1 = t2; 227 } 228 229 /* 230 * Determine the minimum difference between these M values. 231 * This result will be our estimate (in microseconds) for the 232 * clock granularity. 233 */ 234 235 minDelta = 1000000; 236 for (i = 1; i < M; i++) { 237 Delta = (int)( 1.0E6 * (timesfound[i]-timesfound[i-1])); 238 minDelta = MIN(minDelta, MAX(Delta,0)); 239 } 240 241 return(minDelta); 242 } 243 244