1 /* Get frequency of the system processor. powerpc/Linux version.
2 Copyright (C) 2000-2015 Free Software Foundation, Inc.
3 This file is part of the GNU C Library.
5 The GNU C Library is free software; you can redistribute it and/or
6 modify it under the terms of the GNU Lesser General Public
7 License as published by the Free Software Foundation; either
8 version 2.1 of the License, or (at your option) any later version.
10 The GNU C Library is distributed in the hope that it will be useful,
11 but WITHOUT ANY WARRANTY; without even the implied warranty of
12 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
13 Lesser General Public License for more details.
15 You should have received a copy of the GNU Lesser General Public
16 License along with the GNU C Library; if not, see
17 <http://www.gnu.org/licenses/>. */
24 #include <libc-internal.h>
26 #include <bits/libc-vdso.h>
29 __get_clockfreq (void)
31 /* We read the information from the /proc filesystem. /proc/cpuinfo
32 contains at least one line like:
34 We search for this line and convert the number into an integer. */
35 static hp_timing_t timebase_freq;
36 hp_timing_t result = 0L;
38 /* If this function was called before, we know the result. */
39 if (timebase_freq != 0)
42 /* If we can use the vDSO to obtain the timebase even better. */
44 INTERNAL_SYSCALL_DECL (err);
46 INTERNAL_VSYSCALL_NO_SYSCALL_FALLBACK (get_tbfreq, err, uint64_t, 0);
47 if (INTERNAL_SYSCALL_ERROR_P (timebase_freq, err)
48 && INTERNAL_SYSCALL_ERRNO (timebase_freq, err) == ENOSYS)
51 int fd = __open ("/proc/cpuinfo", O_RDONLY);
53 if (__glibc_likely (fd != -1))
55 /* The timebase will be in the 1st 1024 bytes for systems with up
56 to 8 processors. If the first read returns less then 1024
57 bytes read, we have the whole cpuinfo and can start the scan.
58 Otherwise we will have to read more to insure we have the
59 timebase value in the scan. */
63 n = __read (fd, buf, sizeof (buf));
64 if (n == sizeof (buf))
66 /* We are here because the 1st read returned exactly sizeof
67 (buf) bytes. This implies that we are not at EOF and may
68 not have read the timebase value yet. So we need to read
69 more bytes until we know we have EOF. We copy the lower
70 half of buf to the upper half and read sizeof (buf)/2
71 bytes into the lower half of buf and repeat until we
72 reach EOF. We can assume that the timebase will be in
73 the last 512 bytes of cpuinfo, so two 512 byte half_bufs
74 will be sufficient to contain the timebase and will
75 handle the case where the timebase spans the half_buf
77 const ssize_t half_buf = sizeof (buf) / 2;
80 memcpy (buf, buf + half_buf, half_buf);
81 n = __read (fd, buf + half_buf, half_buf);
87 if (__builtin_expect (n, 1) > 0)
89 char *mhz = memmem (buf, n, "timebase", 7);
91 if (__glibc_likely (mhz != NULL))
95 /* Search for the beginning of the string. */
96 while (mhz < endp && (*mhz < '0' || *mhz > '9')
100 while (mhz < endp && *mhz != '\n')
102 if (*mhz >= '0' && *mhz <= '9')
105 result += *mhz - '0';
111 timebase_freq = result;
117 return timebase_freq;