xref: /petsc/src/benchmarks/streams/BasicVersion.c (revision fe998a80077c9ee0917a39496df43fc256e1b478)
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 <limits.h>
23 # include <float.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      200000
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 
98 int main(int argc,char **args)
99 {
100   int           checktick(void);
101   register int j, k;
102   double       scalar, t, times[4][NTIMES],irate[4];
103 
104   /* --- SETUP --- determine precision and check timing --- */
105 
106   for (j=0; j<N; j++) {
107     a[j] = 1.0;
108     b[j] = 2.0;
109     c[j] = 0.0;
110   }
111 
112   t = second();
113   for (j = 0; j < N; j++) a[j] = 2.0E0 * a[j];
114   t = 1.0E6 * (second() - t);
115 
116   /*   --- MAIN LOOP --- repeat test cases NTIMES times --- */
117 
118   scalar = 3.0;
119   for (k=0; k<NTIMES; k++)
120   {
121 
122     times[0][k] = second();
123 /* should all these barriers be pulled outside of the time call? */
124 
125     for (j=0; j<N; j++) c[j] = a[j];
126     times[0][k] = second() - times[0][k];
127 
128     times[1][k] = second();
129 
130     for (j=0; j<N; j++) b[j] = scalar*c[j];
131     times[1][k] = second() - times[1][k];
132 
133     times[2][k] = second();
134     for (j=0; j<N; j++) c[j] = a[j]+b[j];
135     times[2][k] = second() - times[2][k];
136 
137     times[3][k] = second();
138     for (j=0; j<N; j++) a[j] = b[j]+scalar*c[j];
139     times[3][k] = second() - times[3][k];
140   }
141 
142   /*   --- SUMMARY --- */
143 
144   for (k=0; k<NTIMES; k++)
145     for (j=0; j<4; j++) mintime[j] = MIN(mintime[j], times[j][k]);
146 
147   for (j=0; j<4; j++) irate[j] = 1.0E-06 * bytes[j]/mintime[j];
148 
149   printf("Function      Rate (MB/s) \n");
150   for (j=0; j<4; j++) printf("%s%11.4f\n", label[j],irate[j]);
151   return 0;
152 }
153 
154 # define        M        20
155 
156 int checktick(void)
157 {
158   int    i, minDelta, Delta;
159   double t1, t2, timesfound[M];
160 
161 /*  Collect a sequence of M unique time values from the system. */
162 
163   for (i = 0; i < M; i++) {
164     t1 = second();
165     while (((t2=second()) - t1) < 1.0E-6) ;
166     timesfound[i] = t1 = t2;
167   }
168 
169 /*
170 * Determine the minimum difference between these M values.
171 * This result will be our estimate (in microseconds) for the
172 * clock granularity.
173 */
174 
175   minDelta = 1000000;
176   for (i = 1; i < M; i++) {
177     Delta    = (int)(1.0E6 * (timesfound[i]-timesfound[i-1]));
178     minDelta = MIN(minDelta, MAX(Delta,0));
179   }
180 
181   return(minDelta);
182 }
183 
184