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      1 /*	$NetBSD: subr_time_arith.c,v 1.7 2026/01/04 03:21:19 riastradh Exp $	*/
      2 
      3 /*-
      4  * Copyright (c) 2000, 2004, 2005, 2007, 2008, 2009, 2020
      5  *     The NetBSD Foundation, Inc.
      6  * All rights reserved.
      7  *
      8  * This code is derived from software contributed to The NetBSD Foundation
      9  * by Christopher G. Demetriou, by Andrew Doran, and by Jason R. Thorpe.
     10  *
     11  * Redistribution and use in source and binary forms, with or without
     12  * modification, are permitted provided that the following conditions
     13  * are met:
     14  * 1. Redistributions of source code must retain the above copyright
     15  *    notice, this list of conditions and the following disclaimer.
     16  * 2. Redistributions in binary form must reproduce the above copyright
     17  *    notice, this list of conditions and the following disclaimer in the
     18  *    documentation and/or other materials provided with the distribution.
     19  *
     20  * THIS SOFTWARE IS PROVIDED BY THE NETBSD FOUNDATION, INC. AND CONTRIBUTORS
     21  * ``AS IS'' AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED
     22  * TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR
     23  * PURPOSE ARE DISCLAIMED.  IN NO EVENT SHALL THE FOUNDATION OR CONTRIBUTORS
     24  * BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
     25  * CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
     26  * SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
     27  * INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
     28  * CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
     29  * ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE
     30  * POSSIBILITY OF SUCH DAMAGE.
     31  */
     32 
     33 /*
     34  * Copyright (c) 1982, 1986, 1989, 1993
     35  *	The Regents of the University of California.  All rights reserved.
     36  *
     37  * Redistribution and use in source and binary forms, with or without
     38  * modification, are permitted provided that the following conditions
     39  * are met:
     40  * 1. Redistributions of source code must retain the above copyright
     41  *    notice, this list of conditions and the following disclaimer.
     42  * 2. Redistributions in binary form must reproduce the above copyright
     43  *    notice, this list of conditions and the following disclaimer in the
     44  *    documentation and/or other materials provided with the distribution.
     45  * 3. Neither the name of the University nor the names of its contributors
     46  *    may be used to endorse or promote products derived from this software
     47  *    without specific prior written permission.
     48  *
     49  * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
     50  * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
     51  * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
     52  * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
     53  * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
     54  * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
     55  * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
     56  * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
     57  * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
     58  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
     59  * SUCH DAMAGE.
     60  *
     61  *	@(#)kern_clock.c	8.5 (Berkeley) 1/21/94
     62  *	@(#)kern_time.c 8.4 (Berkeley) 5/26/95
     63  */
     64 
     65 #include <sys/cdefs.h>
     66 __KERNEL_RCSID(0, "$NetBSD: subr_time_arith.c,v 1.7 2026/01/04 03:21:19 riastradh Exp $");
     67 
     68 #include <sys/types.h>
     69 
     70 #include <sys/errno.h>
     71 #include <sys/time.h>
     72 #include <sys/timearith.h>
     73 
     74 #if defined(_KERNEL)
     75 
     76 #include <sys/kernel.h>
     77 #include <sys/sdt.h>
     78 #include <sys/systm.h>
     79 
     80 #include <machine/limits.h>
     81 
     82 #elif defined(_TIME_TESTING)
     83 
     84 #include <assert.h>
     85 #include <limits.h>
     86 #include <stdbool.h>
     87 
     88 extern int hz;
     89 extern int tick;
     90 
     91 #define	KASSERT		assert
     92 #define	MIN(X, Y)	((X) < (Y) ? (X) : (Y))
     93 #define	SET_ERROR(E)	(E)
     94 
     95 #endif
     96 
     97 /*
     98  * Compute number of ticks in the specified amount of time.
     99  */
    100 int
    101 tvtohz(const struct timeval *tv)
    102 {
    103 	unsigned long ticks;
    104 	long sec, usec;
    105 
    106 	/*
    107 	 * If the number of usecs in the whole seconds part of the time
    108 	 * difference fits in a long, then the total number of usecs will
    109 	 * fit in an unsigned long.  Compute the total and convert it to
    110 	 * ticks, rounding up and adding 1 to allow for the current tick
    111 	 * to expire.  Rounding also depends on unsigned long arithmetic
    112 	 * to avoid overflow.
    113 	 *
    114 	 * Otherwise, if the number of ticks in the whole seconds part of
    115 	 * the time difference fits in a long, then convert the parts to
    116 	 * ticks separately and add, using similar rounding methods and
    117 	 * overflow avoidance.  This method would work in the previous
    118 	 * case, but it is slightly slower and assumes that hz is integral.
    119 	 *
    120 	 * Otherwise, round the time difference down to the maximum
    121 	 * representable value.
    122 	 *
    123 	 * If ints are 32-bit, then the maximum value for any timeout in
    124 	 * 10ms ticks is 248 days.
    125 	 */
    126 	sec = tv->tv_sec;
    127 	usec = tv->tv_usec;
    128 
    129 	KASSERT(usec >= 0);
    130 	KASSERT(usec < 1000000);
    131 
    132 	/* catch overflows in conversion time_t->int */
    133 	if (tv->tv_sec > INT_MAX)
    134 		return INT_MAX;
    135 	if (tv->tv_sec < 0)
    136 		return 0;
    137 
    138 	if (sec < 0 || (sec == 0 && usec == 0)) {
    139 		/*
    140 		 * Would expire now or in the past.  Return 0 ticks.
    141 		 * This is different from the legacy tvhzto() interface,
    142 		 * and callers need to check for it.
    143 		 */
    144 		ticks = 0;
    145 	} else if (sec <= (LONG_MAX / 1000000))
    146 		ticks = (((sec * 1000000) + (unsigned long)usec + (tick - 1))
    147 		    / tick) + 1;
    148 	else if (sec <= (LONG_MAX / hz))
    149 		ticks = (sec * hz) +
    150 		    (((unsigned long)usec + (tick - 1)) / tick) + 1;
    151 	else
    152 		ticks = LONG_MAX;
    153 
    154 	if (ticks > INT_MAX)
    155 		ticks = INT_MAX;
    156 
    157 	return ((int)ticks);
    158 }
    159 
    160 /*
    161  * Compute number of ticks in the specified amount of time.
    162  *
    163  * Round up, clamped to INT_MAX.  Return 0 iff ts <= 0.
    164  */
    165 int
    166 tstohz(const struct timespec *ts)
    167 {
    168 	struct timeval tv;
    169 
    170 	KASSERT(ts->tv_nsec >= 0);
    171 	KASSERT(ts->tv_nsec < 1000000000);
    172 
    173 	/*
    174 	 * usec has great enough resolution for hz, so convert to a
    175 	 * timeval and use tvtohz() above.
    176 	 */
    177 	tv.tv_sec = ts->tv_sec;
    178 	tv.tv_usec = (ts->tv_nsec + 999)/1000;
    179 	if (tv.tv_usec >= 1000000) {
    180 		if (__predict_false(tv.tv_sec == __type_max(time_t)))
    181 			return INT_MAX;
    182 		tv.tv_sec++;
    183 		tv.tv_usec -= 1000000;
    184 	}
    185 	return tvtohz(&tv);
    186 }
    187 
    188 /*
    189  * Check that a proposed value to load into the .it_value or
    190  * .it_interval part of an interval timer is acceptable, and
    191  * fix it to have at least minimal value (i.e. if it is less
    192  * than the resolution of the clock, round it up.). We don't
    193  * timeout the 0,0 value because this means to disable the
    194  * timer or the interval.
    195  */
    196 int
    197 itimerfix(struct timeval *tv)
    198 {
    199 
    200 	if (tv->tv_usec < 0 || tv->tv_usec >= 1000000)
    201 		return SET_ERROR(EINVAL);
    202 	if (tv->tv_sec < 0)
    203 		return SET_ERROR(ETIMEDOUT);
    204 	if (tv->tv_sec == 0 && tv->tv_usec != 0 && tv->tv_usec < tick)
    205 		tv->tv_usec = tick;
    206 	return 0;
    207 }
    208 
    209 int
    210 itimespecfix(struct timespec *ts)
    211 {
    212 
    213 	if (ts->tv_nsec < 0 || ts->tv_nsec >= 1000000000)
    214 		return SET_ERROR(EINVAL);
    215 	if (ts->tv_sec < 0)
    216 		return SET_ERROR(ETIMEDOUT);
    217 	if (ts->tv_sec == 0 && ts->tv_nsec != 0 && ts->tv_nsec < tick * 1000)
    218 		ts->tv_nsec = tick * 1000;
    219 	return 0;
    220 }
    221 
    222 /*
    223  * timespecaddok(tsp, usp)
    224  *
    225  *	True if tsp + usp can be computed without overflow, i.e., if it
    226  *	is OK to do timespecadd(tsp, usp, ...).
    227  */
    228 bool
    229 timespecaddok(const struct timespec *tsp, const struct timespec *usp)
    230 {
    231 	enum { TIME_MIN = __type_min(time_t), TIME_MAX = __type_max(time_t) };
    232 	time_t a = tsp->tv_sec;
    233 	time_t b = usp->tv_sec;
    234 	bool carry;
    235 
    236 	/*
    237 	 * Caller is responsible for guaranteeing valid timespec
    238 	 * inputs.  Any user-controlled inputs must be validated or
    239 	 * adjusted.
    240 	 */
    241 	KASSERT(tsp->tv_nsec >= 0);
    242 	KASSERT(usp->tv_nsec >= 0);
    243 	KASSERT(tsp->tv_nsec < 1000000000L);
    244 	KASSERT(usp->tv_nsec < 1000000000L);
    245 	__CTASSERT(1000000000L <= __type_max(long) - 1000000000L);
    246 
    247 	/*
    248 	 * Fail if a + b + carry overflows TIME_MAX, or if a + b
    249 	 * overflows TIME_MIN because timespecadd adds the carry after
    250 	 * computing a + b.
    251 	 *
    252 	 * Break it into two mutually exclusive and exhaustive cases:
    253 	 * I. a >= 0
    254 	 * II. a < 0
    255 	 */
    256 	carry = (tsp->tv_nsec + usp->tv_nsec >= 1000000000L);
    257 	if (a >= 0) {
    258 		/*
    259 		 * Case I: a >= 0.  If b < 0, then b + 1 <= 0, so
    260 		 *
    261 		 *	a + b + 1 <= a + 0 <= TIME_MAX,
    262 		 *
    263 		 * and
    264 		 *
    265 		 *	a + b >= 0 + b = b >= TIME_MIN,
    266 		 *
    267 		 * so this can't overflow.
    268 		 *
    269 		 * If b >= 0, then a + b + carry >= a + b >= 0, so
    270 		 * negative results and thus results below TIME_MIN are
    271 		 * impossible; we need only avoid
    272 		 *
    273 		 *	a + b + carry > TIME_MAX,
    274 		 *
    275 		 * which we will do by rejecting if
    276 		 *
    277 		 *	b > TIME_MAX - a - carry,
    278 		 *
    279 		 * which in turn is incidentally always false if b < 0
    280 		 * so we don't need extra logic to discriminate on the
    281 		 * b >= 0 and b < 0 cases.
    282 		 *
    283 		 * Since 0 <= a <= TIME_MAX, we know
    284 		 *
    285 		 *	0 <= TIME_MAX - a <= TIME_MAX,
    286 		 *
    287 		 * and hence
    288 		 *
    289 		 *	-1 <= TIME_MAX - a - 1 < TIME_MAX.
    290 		 *
    291 		 * So we can compute TIME_MAX - a - carry (i.e., either
    292 		 * TIME_MAX - a or TIME_MAX - a - 1) safely without
    293 		 * overflow.
    294 		 */
    295 		if (b > TIME_MAX - a - carry)
    296 			return false;
    297 	} else {
    298 		/*
    299 		 * Case II: a < 0.  If b >= 0, then since a + 1 <= 0,
    300 		 * we have
    301 		 *
    302 		 *	a + b + 1 <= b <= TIME_MAX,
    303 		 *
    304 		 * and
    305 		 *
    306 		 *	a + b >= a >= TIME_MIN,
    307 		 *
    308 		 * so this can't overflow.
    309 		 *
    310 		 * If b < 0, then the intermediate a + b is negative
    311 		 * and the outcome a + b + 1 is nonpositive, so we need
    312 		 * only avoid
    313 		 *
    314 		 *	a + b < TIME_MIN,
    315 		 *
    316 		 * which we will do by rejecting if
    317 		 *
    318 		 *	a < TIME_MIN - b.
    319 		 *
    320 		 * (Reminder: The carry is added afterward in
    321 		 * timespecadd, so to avoid overflow it is not enough
    322 		 * to merely reject a + b + carry < TIME_MIN.)
    323 		 *
    324 		 * It is safe to compute the difference TIME_MIN - b
    325 		 * because b is negative, so the result lies in
    326 		 * (TIME_MIN, 0].
    327 		 */
    328 		if (b < 0 && a < TIME_MIN - b)
    329 			return false;
    330 	}
    331 
    332 	return true;
    333 }
    334 
    335 /*
    336  * timespecsubok(tsp, usp)
    337  *
    338  *	True if tsp - usp can be computed without overflow, i.e., if it
    339  *	is OK to do timespecsub(tsp, usp, ...).
    340  */
    341 bool
    342 timespecsubok(const struct timespec *tsp, const struct timespec *usp)
    343 {
    344 	enum { TIME_MIN = __type_min(time_t), TIME_MAX = __type_max(time_t) };
    345 	time_t a = tsp->tv_sec, b = usp->tv_sec;
    346 	bool borrow;
    347 
    348 	/*
    349 	 * Caller is responsible for guaranteeing valid timespec
    350 	 * inputs.  Any user-controlled inputs must be validated or
    351 	 * adjusted.
    352 	 */
    353 	KASSERT(tsp->tv_nsec >= 0);
    354 	KASSERT(usp->tv_nsec >= 0);
    355 	KASSERT(tsp->tv_nsec < 1000000000L);
    356 	KASSERT(usp->tv_nsec < 1000000000L);
    357 	__CTASSERT(1000000000L <= __type_max(long) - 1000000000L);
    358 
    359 	/*
    360 	 * Fail if a - b - borrow overflows TIME_MIN, or if a - b
    361 	 * overflows TIME_MAX because timespecsub subtracts the borrow
    362 	 * after computing a - b.
    363 	 *
    364 	 * Break it into two mutually exclusive and exhaustive cases:
    365 	 * I. a < 0
    366 	 * II. a >= 0
    367 	 */
    368 	borrow = (tsp->tv_nsec - usp->tv_nsec < 0);
    369 	if (a < 0) {
    370 		/*
    371 		 * Case I: a < 0.  If b < 0, then -b - 1 >= 0, so
    372 		 *
    373 		 *	a - b - 1 >= a + 0 >= TIME_MIN,
    374 		 *
    375 		 * and, since a <= -1, provided that TIME_MIN <=
    376 		 * -TIME_MAX - 1 so that TIME_MAX <= -TIME_MIN - 1 (in
    377 		 * fact, equality holds, under the assumption of
    378 		 * two's-complement arithmetic),
    379 		 *
    380 		 *	a - b <= -1 - b = -b - 1 <= TIME_MAX,
    381 		 *
    382 		 * so this can't overflow.
    383 		 */
    384 		__CTASSERT(TIME_MIN <= -TIME_MAX - 1);
    385 
    386 		/*
    387 		 * If b >= 0, then a - b - borrow <= a - b < 0, so
    388 		 * positive results and thus results above TIME_MAX are
    389 		 * impossible; we need only avoid
    390 		 *
    391 		 *	a - b - borrow < TIME_MIN,
    392 		 *
    393 		 * which we will do by rejecting if
    394 		 *
    395 		 *	a < TIME_MIN + b + borrow.
    396 		 *
    397 		 * The right-hand side is safe to evaluate for any
    398 		 * values of b and borrow as long as TIME_MIN +
    399 		 * TIME_MAX + 1 <= TIME_MAX, i.e., TIME_MIN <= -1.
    400 		 * (Note: If time_t were unsigned, this would fail!)
    401 		 *
    402 		 * Note: Unlike Case I in timespecaddok, this criterion
    403 		 * does not work for b < 0, nor can the roles of a and
    404 		 * b in the inequality be reversed (e.g., -b < TIME_MIN
    405 		 * - a + borrow) without extra cases like checking for
    406 		 * b = TEST_MIN.
    407 		 */
    408 		__CTASSERT(TIME_MIN < -1);
    409 		if (b >= 0 && a < TIME_MIN + b + borrow)
    410 			return false;
    411 	} else {
    412 		/*
    413 		 * Case II: a >= 0.  If b >= 0, then
    414 		 *
    415 		 *	a - b <= a <= TIME_MAX,
    416 		 *
    417 		 * and, provided TIME_MIN <= -TIME_MAX - 1 (in fact,
    418 		 * equality holds, under the assumption of
    419 		 * two's-complement arithmetic)
    420 		 *
    421 		 *	a - b - 1 >= -b - 1 >= -TIME_MAX - 1 >= TIME_MIN,
    422 		 *
    423 		 * so this can't overflow.
    424 		 */
    425 		__CTASSERT(TIME_MIN <= -TIME_MAX - 1);
    426 
    427 		/*
    428 		 * If b < 0, then a - b >= a >= 0, so negative results
    429 		 * and thus results below TIME_MIN are impossible; we
    430 		 * need only avoid
    431 		 *
    432 		 *	a - b > TIME_MAX,
    433 		 *
    434 		 * which we will do by rejecting if
    435 		 *
    436 		 *	a > TIME_MAX + b.
    437 		 *
    438 		 * (Reminder: The borrow is subtracted afterward in
    439 		 * timespecsub, so to avoid overflow it is not enough
    440 		 * to merely reject a - b - borrow > TIME_MAX.)
    441 		 *
    442 		 * It is safe to compute the sum TIME_MAX + b because b
    443 		 * is negative, so the result lies in [0, TIME_MAX).
    444 		 */
    445 		if (b < 0 && a > TIME_MAX + b)
    446 			return false;
    447 	}
    448 
    449 	return true;
    450 }
    451 
    452 static bool
    453 timespec2nsok(const struct timespec *ts)
    454 {
    455 
    456 	return ts->tv_sec < INT64_MAX/1000000000 ||
    457 	    (ts->tv_sec == INT64_MAX/1000000000 &&
    458 		ts->tv_nsec <= INT64_MAX - (INT64_MAX/1000000000)*1000000000);
    459 }
    460 
    461 /*
    462  * itimer_transition(it, now, next, &overruns)
    463  *
    464  *	Given:
    465  *
    466  *	- it: the current state of an itimer (it_value = last expiry
    467  *	  time, it_interval = periodic rescheduling interval), and
    468  *
    469  *	- now: the current time on the itimer's clock;
    470  *
    471  *	compute:
    472  *
    473  *	- next: the next time the itimer should be scheduled for, and
    474  *	- overruns: the number of overruns if we're firing late.
    475  *
    476  *	XXX This should maybe also say whether the itimer should expire
    477  *	at all.
    478  */
    479 void
    480 itimer_transition(const struct itimerspec *restrict it,
    481     const struct timespec *restrict now,
    482     struct timespec *restrict next,
    483     int *restrict overrunsp)
    484 {
    485 	int64_t last_val, next_val, interval, remainder, now_ns;
    486 	int backwards;
    487 
    488 	/*
    489 	 * Zero the outputs so we can test assertions in userland
    490 	 * without undefined behaviour.
    491 	 */
    492 	timespecclear(next);
    493 	*overrunsp = 0;
    494 
    495 	/*
    496 	 * Paranoia: Caller should guarantee this.
    497 	 */
    498 	if (!timespecisset(&it->it_interval)) {
    499 		timespecclear(next);
    500 		return;
    501 	}
    502 
    503 	/* Did the clock wind backwards? */
    504 	backwards = (timespeccmp(&it->it_value, now, >));
    505 
    506 	/* Valid value and interval guaranteed by itimerfix. */
    507 	KASSERT(it->it_value.tv_sec >= 0);
    508 	KASSERT(it->it_value.tv_nsec < 1000000000);
    509 	KASSERT(it->it_interval.tv_sec >= 0);
    510 	KASSERT(it->it_interval.tv_nsec < 1000000000);
    511 
    512 	/* Nonnegative interval guaranteed by itimerfix.  */
    513 	KASSERT(it->it_interval.tv_sec >= 0);
    514 	KASSERT(it->it_interval.tv_nsec >= 0);
    515 
    516 	/* Handle the easy case of non-overflown timers first. */
    517 	if (__predict_true(!backwards)) {
    518 		if (__predict_false(!timespecaddok(&it->it_value,
    519 			    &it->it_interval)))
    520 			goto overflow;
    521 		timespecadd(&it->it_value, &it->it_interval, next);
    522 		if (__predict_true(timespeccmp(now, next, <)))
    523 			return;
    524 	}
    525 
    526 	/*
    527 	 * If we can't represent the input as a number of nanoseconds,
    528 	 * bail.  This is good up to the year 2262, if we start
    529 	 * counting from 1970 (2^63 nanoseconds ~ 292 years).
    530 	 */
    531 	if (__predict_false(!timespec2nsok(now)) ||
    532 	    __predict_false(!timespec2nsok(&it->it_value)) ||
    533 	    __predict_false(!timespec2nsok(&it->it_interval)))
    534 		goto overflow;
    535 
    536 	now_ns = timespec2ns(now);
    537 	last_val = timespec2ns(&it->it_value);
    538 	interval = timespec2ns(&it->it_interval);
    539 
    540 	KASSERT(now_ns >= 0);
    541 	KASSERT(last_val >= 0);
    542 	KASSERT(interval >= 0);
    543 
    544 	/*
    545 	 *            now [backwards]         overruns    now [forwards]
    546 	 *           |                      v    v    v  |
    547 	 * |--+----+-*--x----+----+----|----+----+----+--*-x----+-->
    548 	 *            \/               |               \/
    549 	 *         remainder        last_val        remainder
    550 	 *     (zero or negative)                (zero or positive)
    551 	 *
    552 	 * Set next_val to last_value + k*interval for some k.
    553 	 *
    554 	 * The interval is always positive, and division in C
    555 	 * truncates, so dividing a positive duration by the interval
    556 	 * always gives zero or a positive remainder, and dividing a
    557 	 * negative duration by the interval always gives zero or a
    558 	 * negative remainder.  Hence:
    559 	 *
    560 	 * - If now_ns < last_val -- which happens iff backwards, i.e.,
    561 	 *   the clock was wound backwards -- then remainder is zero or
    562 	 *   negative, so subtracting it stays in place or moves
    563 	 *   forward in time, and thus this finds the _earliest_ value
    564 	 *   that is not earlier than now_ns.  We will advance this by
    565 	 *   one more interval if we are already firing exactly on the
    566 	 *   interval to find the earliest value _after_ now_ns.
    567 	 *
    568 	 * - If now_ns > last_val -- which happens iff !backwards,
    569 	 *   i.e., the clock ran fast -- then remainder is zero or
    570 	 *   positive positive, so this finds the _latest_ value not
    571 	 *   later than now_ns.  We will always advance this by one
    572 	 *   more interval to find the earliest value _after_ now_ns.
    573 	 *   We will also count overflows.
    574 	 *
    575 	 * (now_ns == last_val is not possible at this point because it
    576 	 * only happens if the addition of struct timespec would
    577 	 * overflow, and that is only possible when timespec2ns would
    578 	 * also overflow for at least one of the inputs.)
    579 	 */
    580 	KASSERT(last_val != now_ns);
    581 	remainder = (now_ns - last_val) % interval;
    582 	next_val = now_ns - remainder;
    583 	KASSERT((last_val - next_val) % interval == 0);
    584 	if (backwards) {
    585 		/*
    586 		 * If the clock was wound back to an exact multiple of
    587 		 * the interval, so next_val = now_ns, don't demand to
    588 		 * fire again in the same instant -- advance to the
    589 		 * next interval.  Overflow is not possible; proof is
    590 		 * asserted.
    591 		 */
    592 		if (remainder == 0) {
    593 			KASSERT(now_ns < last_val);
    594 			KASSERT(next_val == now_ns);
    595 			KASSERT(last_val - next_val >= interval);
    596 			KASSERT(interval <= last_val - next_val);
    597 			KASSERT(next_val <= last_val - interval);
    598 			KASSERT(next_val <= INT64_MAX - interval);
    599 			next_val += interval;
    600 		}
    601 	} else {
    602 		/*
    603 		 * next_val is the largest integer multiple of interval
    604 		 * not later than now_ns.  Count the number of full
    605 		 * intervals that were skipped (division should be
    606 		 * exact here), not counting any partial interval
    607 		 * between next_val and now_ns, as the number of
    608 		 * overruns.  Advance by one interval -- unless that
    609 		 * would overflow.
    610 		 */
    611 		*overrunsp = MIN(INT_MAX, (next_val - last_val) / interval);
    612 		if (__predict_false(next_val > INT64_MAX - interval))
    613 			goto overflow;
    614 		next_val += interval;
    615 	}
    616 
    617 	next->tv_sec = next_val / 1000000000;
    618 	next->tv_nsec = next_val % 1000000000;
    619 	return;
    620 
    621 overflow:
    622 	next->tv_sec = 0;
    623 	next->tv_nsec = 0;
    624 }
    625