blob: 0055a09b2802f58712ee2164ff505f9b37314e17 [file] [log] [blame]
sewardjb941be62005-01-08 18:26:47 +00001
2/* peach (7400, altivec supported, 450MHz, gcc -O)
3 m1 = 1.20000000000000018, exp = 1.19999999999999996
4 m2 = 1.19999999999998885, exp = 1.19999999999999996
5*/
6
7/* peach (7400, altivec supported, 450MHz, gcc)
8 m1 = 1.20000000000000018, exp = 1.19999999999999996
9 m2 = 1.19999999999998885, exp = 1.19999999999999996
10*/
11
12/* phoenix, gcc -O
13 m1 = 1.19999999999999996, exp = 1.19999999999999996
14 m2 = 1.19999999999999996, exp = 1.19999999999999996
15*/
16
17/* phoenix, icc -O
18 m1 = 1.19999999999999996, exp = 1.19999999999999996
19 m2 = 1.19999999999999996, exp = 1.19999999999999996
20*/
21
22/* phoenix, gcc -O, iropt-level=2
23 m1 = 1.20000000000000040, exp = 1.19999999999999996
24 m2 = 1.19999999999999440, exp = 1.19999999999999996
25*/
26
27/* phoenix, gcc -O, iropt-level=1/0
28 m1 = 1.20000000000000018, exp = 1.19999999999999996
29 m2 = 1.19999999999998885, exp = 1.19999999999999996
30*/
31
32
33#include <stdlib.h>
34#include <stdio.h>
35#include <math.h>
36
37#define NNN 1000
38
39
40double
41my_mean1 (const double data[], size_t stride, const size_t size)
42{
43 /* Compute the arithmetic mean of a dataset using the recurrence relation
44 mean_(n) = mean(n-1) + (data[n] - mean(n-1))/(n+1) */
45 long double mean = 0;
46 size_t i;
47
48 for (i = 0; i < size; i++)
49 {
50 mean += (data[i * stride] - mean) / (i + 1);
51 }
52 return mean;
53}
54
55double
56my_mean2 (const double data[], size_t stride, const size_t size)
57{
58 /* Compute the arithmetic mean of a dataset using the
59 obvious scheme. */
60 int i;
61 long double sum = 0;
62 for (i = 0; i < size; i++)
63 sum += data[i * stride];
64 return sum / (double)size;
65}
66
67
68int main (void)
69{
70 int i;
71 const size_t nacc2 = NNN+1;
72 double numacc2[NNN+1] ;
73
74 numacc2[0] = 1.2 ;
75
76 for (i = 1 ; i < NNN; i += 2)
77 numacc2[i] = 1.1 ;
78
79 for (i = 1 ; i < NNN; i += 2)
80 numacc2[i+1] = 1.3 ;
81
82#if 1
83 asm __volatile__("fninit");
84#endif
85
86 {
87 double m1 = my_mean1 (numacc2, 1, nacc2);
88 double m2 = my_mean2 (numacc2, 1, nacc2);
89 double expected_mean = 1.2;
90 printf("m1 = %19.17f, exp = %19.17f\n", m1, expected_mean);
91 printf("m2 = %19.17f, exp = %19.17f\n", m2, expected_mean);
92 }
93
94 return 0;
95}
96
97