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
|
// SPDX-License-Identifier: GPL-2.0-or-later
/*
* Copyright 2015, Cyril Bur, IBM Corp.
*
* This test attempts to see if the FPU registers change across a syscall (fork).
*/
#include <stdio.h>
#include <unistd.h>
#include <sys/syscall.h>
#include <sys/time.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <stdlib.h>
#include "utils.h"
extern int test_fpu(double *darray, pid_t *pid);
double darray[] = {0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0,
1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0,
2.1};
int syscall_fpu(void)
{
pid_t fork_pid;
int i;
int ret;
int child_ret;
for (i = 0; i < 1000; i++) {
/* test_fpu will fork() */
ret = test_fpu(darray, &fork_pid);
if (fork_pid == -1)
return -1;
if (fork_pid == 0)
exit(ret);
waitpid(fork_pid, &child_ret, 0);
if (ret || child_ret)
return 1;
}
return 0;
}
int test_syscall_fpu(void)
{
/*
* Setup an environment with much context switching
*/
pid_t pid2;
pid_t pid = fork();
int ret;
int child_ret;
FAIL_IF(pid == -1);
pid2 = fork();
/* Can't FAIL_IF(pid2 == -1); because already forked once */
if (pid2 == -1) {
/*
* Couldn't fork, ensure test is a fail
*/
child_ret = ret = 1;
} else {
ret = syscall_fpu();
if (pid2)
waitpid(pid2, &child_ret, 0);
else
exit(ret);
}
ret |= child_ret;
if (pid)
waitpid(pid, &child_ret, 0);
else
exit(ret);
FAIL_IF(ret || child_ret);
return 0;
}
int main(int argc, char *argv[])
{
return test_harness(test_syscall_fpu, "syscall_fpu");
}
|
