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117 lines
3.7 KiB
C
117 lines
3.7 KiB
C
/* Get frequency of the system processor. powerpc/Linux version.
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Copyright (C) 2000-2012 Free Software Foundation, Inc.
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This file is part of the GNU C Library.
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The GNU C Library is free software; you can redistribute it and/or
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modify it under the terms of the GNU Lesser General Public
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License as published by the Free Software Foundation; either
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version 2.1 of the License, or (at your option) any later version.
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The GNU C Library is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public
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License along with the GNU C Library; if not, see
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<http://www.gnu.org/licenses/>. */
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#include <ctype.h>
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#include <fcntl.h>
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#include <string.h>
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#include <unistd.h>
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#include <libc-internal.h>
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#include <sysdep.h>
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#include <bits/libc-vdso.h>
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hp_timing_t
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__get_clockfreq (void)
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{
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/* We read the information from the /proc filesystem. /proc/cpuinfo
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contains at least one line like:
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timebase : 33333333
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We search for this line and convert the number into an integer. */
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static hp_timing_t timebase_freq;
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hp_timing_t result = 0L;
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/* If this function was called before, we know the result. */
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if (timebase_freq != 0)
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return timebase_freq;
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/* If we can use the vDSO to obtain the timebase even better. */
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#ifdef SHARED
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INTERNAL_SYSCALL_DECL (err);
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timebase_freq = INTERNAL_VSYSCALL_NO_SYSCALL_FALLBACK (get_tbfreq, err, 0);
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if (INTERNAL_SYSCALL_ERROR_P (timebase_freq, err)
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&& INTERNAL_SYSCALL_ERRNO (timebase_freq, err) == ENOSYS)
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#endif
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{
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int fd = __open ("/proc/cpuinfo", O_RDONLY);
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if (__builtin_expect (fd != -1, 1))
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{
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/* The timebase will be in the 1st 1024 bytes for systems with up
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to 8 processors. If the first read returns less then 1024
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bytes read, we have the whole cpuinfo and can start the scan.
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Otherwise we will have to read more to insure we have the
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timebase value in the scan. */
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char buf[1024];
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ssize_t n;
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n = __read (fd, buf, sizeof (buf));
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if (n == sizeof (buf))
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{
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/* We are here because the 1st read returned exactly sizeof
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(buf) bytes. This implies that we are not at EOF and may
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not have read the timebase value yet. So we need to read
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more bytes until we know we have EOF. We copy the lower
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half of buf to the upper half and read sizeof (buf)/2
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bytes into the lower half of buf and repeat until we
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reach EOF. We can assume that the timebase will be in
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the last 512 bytes of cpuinfo, so two 512 byte half_bufs
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will be sufficient to contain the timebase and will
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handle the case where the timebase spans the half_buf
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boundry. */
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const ssize_t half_buf = sizeof (buf) / 2;
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while (n >= half_buf)
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{
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memcpy (buf, buf + half_buf, half_buf);
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n = __read (fd, buf + half_buf, half_buf);
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}
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if (n >= 0)
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n += half_buf;
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}
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if (__builtin_expect (n, 1) > 0)
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{
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char *mhz = memmem (buf, n, "timebase", 7);
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if (__builtin_expect (mhz != NULL, 1))
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{
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char *endp = buf + n;
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/* Search for the beginning of the string. */
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while (mhz < endp && (*mhz < '0' || *mhz > '9')
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&& *mhz != '\n')
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++mhz;
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while (mhz < endp && *mhz != '\n')
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{
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if (*mhz >= '0' && *mhz <= '9')
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{
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result *= 10;
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result += *mhz - '0';
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}
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++mhz;
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}
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}
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timebase_freq = result;
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}
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__close (fd);
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}
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}
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return timebase_freq;
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}
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