Home | History | Annotate | Line # | Download | only in kern
subr_time.c revision 1.1.10.4 
      1 /*	$NetBSD: subr_time.c,v 1.1.10.4 2008/01/21 09:46:22 yamt Exp $	*/
      2 
      3 /*
      4  * Copyright (c) 1982, 1986, 1989, 1993
      5  *	The Regents of the University of California.  All rights reserved.
      6  *
      7  * Redistribution and use in source and binary forms, with or without
      8  * modification, are permitted provided that the following conditions
      9  * are met:
     10  * 1. Redistributions of source code must retain the above copyright
     11  *    notice, this list of conditions and the following disclaimer.
     12  * 2. Redistributions in binary form must reproduce the above copyright
     13  *    notice, this list of conditions and the following disclaimer in the
     14  *    documentation and/or other materials provided with the distribution.
     15  * 3. Neither the name of the University nor the names of its contributors
     16  *    may be used to endorse or promote products derived from this software
     17  *    without specific prior written permission.
     18  *
     19  * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
     20  * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
     21  * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
     22  * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
     23  * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
     24  * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
     25  * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
     26  * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
     27  * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
     28  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
     29  * SUCH DAMAGE.
     30  *
     31  *	@(#)kern_clock.c	8.5 (Berkeley) 1/21/94
     32  *	@(#)kern_time.c 8.4 (Berkeley) 5/26/95
     33  */
     34 
     35 #include <sys/cdefs.h>
     36 __KERNEL_RCSID(0, "$NetBSD: subr_time.c,v 1.1.10.4 2008/01/21 09:46:22 yamt Exp $");
     37 
     38 #include <sys/param.h>
     39 #include <sys/kernel.h>
     40 #include <sys/timex.h>
     41 #include <sys/time.h>
     42 #include <sys/timetc.h>
     43 #include <sys/intr.h>
     44 
     45 /*
     46  * Compute number of hz until specified time.  Used to compute second
     47  * argument to callout_reset() from an absolute time.
     48  */
     49 int
     50 hzto(struct timeval *tvp)
     51 {
     52 	struct timeval now, tv;
     53 
     54 	tv = *tvp;	/* Don't modify original tvp. */
     55 	getmicrotime(&now);
     56 	timersub(&tv, &now, &tv);
     57 	return tvtohz(&tv);
     58 }
     59 
     60 /*
     61  * Compute number of ticks in the specified amount of time.
     62  */
     63 int
     64 tvtohz(struct timeval *tv)
     65 {
     66 	unsigned long ticks;
     67 	long sec, usec;
     68 
     69 	/*
     70 	 * If the number of usecs in the whole seconds part of the time
     71 	 * difference fits in a long, then the total number of usecs will
     72 	 * fit in an unsigned long.  Compute the total and convert it to
     73 	 * ticks, rounding up and adding 1 to allow for the current tick
     74 	 * to expire.  Rounding also depends on unsigned long arithmetic
     75 	 * to avoid overflow.
     76 	 *
     77 	 * Otherwise, if the number of ticks in the whole seconds part of
     78 	 * the time difference fits in a long, then convert the parts to
     79 	 * ticks separately and add, using similar rounding methods and
     80 	 * overflow avoidance.  This method would work in the previous
     81 	 * case, but it is slightly slower and assumes that hz is integral.
     82 	 *
     83 	 * Otherwise, round the time difference down to the maximum
     84 	 * representable value.
     85 	 *
     86 	 * If ints are 32-bit, then the maximum value for any timeout in
     87 	 * 10ms ticks is 248 days.
     88 	 */
     89 	sec = tv->tv_sec;
     90 	usec = tv->tv_usec;
     91 
     92 	if (usec < 0) {
     93 		sec--;
     94 		usec += 1000000;
     95 	}
     96 
     97 	if (sec < 0 || (sec == 0 && usec <= 0)) {
     98 		/*
     99 		 * Would expire now or in the past.  Return 0 ticks.
    100 		 * This is different from the legacy hzto() interface,
    101 		 * and callers need to check for it.
    102 		 */
    103 		ticks = 0;
    104 	} else if (sec <= (LONG_MAX / 1000000))
    105 		ticks = (((sec * 1000000) + (unsigned long)usec + (tick - 1))
    106 		    / tick) + 1;
    107 	else if (sec <= (LONG_MAX / hz))
    108 		ticks = (sec * hz) +
    109 		    (((unsigned long)usec + (tick - 1)) / tick) + 1;
    110 	else
    111 		ticks = LONG_MAX;
    112 
    113 	if (ticks > INT_MAX)
    114 		ticks = INT_MAX;
    115 
    116 	return ((int)ticks);
    117 }
    118 
    119 /*
    120  * Compute number of ticks in the specified amount of time.
    121  */
    122 int
    123 tstohz(struct timespec *ts)
    124 {
    125 	struct timeval tv;
    126 
    127 	/*
    128 	 * usec has great enough resolution for hz, so convert to a
    129 	 * timeval and use tvtohz() above.
    130 	 */
    131 	TIMESPEC_TO_TIMEVAL(&tv, ts);
    132 	return tvtohz(&tv);
    133 }
    134 
    135 /*
    136  * Check that a proposed value to load into the .it_value or
    137  * .it_interval part of an interval timer is acceptable, and
    138  * fix it to have at least minimal value (i.e. if it is less
    139  * than the resolution of the clock, round it up.)
    140  */
    141 int
    142 itimerfix(struct timeval *tv)
    143 {
    144 
    145 	if (tv->tv_sec < 0 || tv->tv_usec < 0 || tv->tv_usec >= 1000000)
    146 		return (EINVAL);
    147 	if (tv->tv_sec == 0 && tv->tv_usec != 0 && tv->tv_usec < tick)
    148 		tv->tv_usec = tick;
    149 	return (0);
    150 }
    151 
    152 int
    153 itimespecfix(struct timespec *ts)
    154 {
    155 
    156 	if (ts->tv_sec < 0 || ts->tv_nsec < 0 || ts->tv_nsec >= 1000000000)
    157 		return (EINVAL);
    158 	if (ts->tv_sec == 0 && ts->tv_nsec != 0 && ts->tv_nsec < tick * 1000)
    159 		ts->tv_nsec = tick * 1000;
    160 	return (0);
    161 }
    162