1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
|
/* benchmark.c -- Benchmarks how quickly the computer does SHA512 hashing
*
* GPLv2 only - Copyright (C) 2009
* David Sommerseth <dazo@users.sourceforge.net>
*
* This program is free software; you can redistribute it and/or
* modify it under the terms of the GNU General Public License
* as published by the Free Software Foundation; version 2
* of the License.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU General Public License for more details.
*
* You should have received a copy of the GNU General Public License
* along with this program; if not, write to the Free Software
* Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA.
*
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <sys/time.h>
void benchmark_hashing(int rounds);
int timeval_subtract (result, x, y)
struct timeval *result, *x, *y;
{
/* Perform the carry for the later subtraction by updating y. */
if (x->tv_usec < y->tv_usec) {
int nsec = (y->tv_usec - x->tv_usec) / 1000000 + 1;
y->tv_usec -= 1000000 * nsec;
y->tv_sec += nsec;
}
if (x->tv_usec - y->tv_usec > 1000000) {
int nsec = (y->tv_usec - x->tv_usec) / 1000000;
y->tv_usec += 1000000 * nsec;
y->tv_sec -= nsec;
}
/* Compute the time remaining to wait.
tv_usec is certainly positive. */
result->tv_sec = x->tv_sec - y->tv_sec;
result->tv_usec = x->tv_usec - y->tv_usec;
/* Return 1 if result is negative. */
return x->tv_sec < y->tv_sec;
}
struct timeval *do_benchmark(int rounds) {
struct timeval start, end, *timediff = NULL;
timediff = (struct timeval *) malloc(sizeof(struct timeval)+2);
memset(timediff, 0, sizeof(struct timeval)+2);
gettimeofday(&start, NULL);
benchmark_hashing(rounds);
gettimeofday(&end, NULL);
timeval_subtract(timediff, &end, &start);
return timediff;
}
// This function will try to find the best minium and maximum rounds for
// the SHA512 hashing. The values are returned in the min and max variables
// and the thr_min and thr_max defines the minimum and maximum thresholds in
// milliseconds the hashing algorithm should use.
int benchmark(int *min, int *max, int thr_min, int thr_max) {
int i = 0, success = 0;
struct timeval *time = NULL, min_time, max_time ;
printf("Benchmarking ");
*min = 1000;
*max = 5000;
for( i = 1000; i < 100000000; i += 1000 ) {
printf(".");
fflush(stdout);
time = do_benchmark(i);
// printf("%i rounds: %ld sec %ld ms\n", i, time->tv_sec, time->tv_usec);
if( time->tv_usec > (thr_max * 1000) ) {
success = 1;
free(time);
break;
}
if( (*min == 1000) && (time->tv_usec > (thr_min * 1000)) ) {
*min = i;
min_time.tv_usec = time->tv_usec;
}
if( (time->tv_usec < (thr_max * 1000)) ) {
*max = i;
max_time.tv_usec = time->tv_usec;
}
free(time);
}
printf(" Done\n");
if( success == 1 ) {
printf("Suggested minimum rounds: %i (takes %ldms)\n", *min, min_time.tv_usec/1000);
printf("Suggested maximum rounds: %i (takes %ldms)\n", *max, max_time.tv_usec/1000);
} else {
*min = 0;
*max = 0;
printf("Could not find any good times, as your computer is too fast for this test.\n");
}
return success;
}
#if 0
int main() {
int min, max;
benchmark(&min, &max, 90, 220);
printf("---> %i - %i\n", min, max);
return 0;
}
#endif
|
