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subr_time_arith.c revision 1.1
      1 /*	$NetBSD: subr_time_arith.c,v 1.1 2024/12/22 23:24:20 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 1.1 2024/12/22 23:24:20 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/systm.h>
     78 
     79 #include <machine/limits.h>
     80 
     81 #elif defined(_TIME_TESTING)
     82 
     83 #include <assert.h>
     84 #include <limits.h>
     85 #include <stdbool.h>
     86 
     87 extern int hz;
     88 extern long tick;
     89 
     90 #define	KASSERT		assert
     91 
     92 #endif
     93 
     94 /*
     95  * Compute number of ticks in the specified amount of time.
     96  */
     97 int
     98 tvtohz(const struct timeval *tv)
     99 {
    100 	unsigned long ticks;
    101 	long sec, usec;
    102 
    103 	/*
    104 	 * If the number of usecs in the whole seconds part of the time
    105 	 * difference fits in a long, then the total number of usecs will
    106 	 * fit in an unsigned long.  Compute the total and convert it to
    107 	 * ticks, rounding up and adding 1 to allow for the current tick
    108 	 * to expire.  Rounding also depends on unsigned long arithmetic
    109 	 * to avoid overflow.
    110 	 *
    111 	 * Otherwise, if the number of ticks in the whole seconds part of
    112 	 * the time difference fits in a long, then convert the parts to
    113 	 * ticks separately and add, using similar rounding methods and
    114 	 * overflow avoidance.  This method would work in the previous
    115 	 * case, but it is slightly slower and assumes that hz is integral.
    116 	 *
    117 	 * Otherwise, round the time difference down to the maximum
    118 	 * representable value.
    119 	 *
    120 	 * If ints are 32-bit, then the maximum value for any timeout in
    121 	 * 10ms ticks is 248 days.
    122 	 */
    123 	sec = tv->tv_sec;
    124 	usec = tv->tv_usec;
    125 
    126 	KASSERT(usec >= 0);
    127 	KASSERT(usec < 1000000);
    128 
    129 	/* catch overflows in conversion time_t->int */
    130 	if (tv->tv_sec > INT_MAX)
    131 		return INT_MAX;
    132 	if (tv->tv_sec < 0)
    133 		return 0;
    134 
    135 	if (sec < 0 || (sec == 0 && usec == 0)) {
    136 		/*
    137 		 * Would expire now or in the past.  Return 0 ticks.
    138 		 * This is different from the legacy tvhzto() interface,
    139 		 * and callers need to check for it.
    140 		 */
    141 		ticks = 0;
    142 	} else if (sec <= (LONG_MAX / 1000000))
    143 		ticks = (((sec * 1000000) + (unsigned long)usec + (tick - 1))
    144 		    / tick) + 1;
    145 	else if (sec <= (LONG_MAX / hz))
    146 		ticks = (sec * hz) +
    147 		    (((unsigned long)usec + (tick - 1)) / tick) + 1;
    148 	else
    149 		ticks = LONG_MAX;
    150 
    151 	if (ticks > INT_MAX)
    152 		ticks = INT_MAX;
    153 
    154 	return ((int)ticks);
    155 }
    156 
    157 /*
    158  * Check that a proposed value to load into the .it_value or
    159  * .it_interval part of an interval timer is acceptable, and
    160  * fix it to have at least minimal value (i.e. if it is less
    161  * than the resolution of the clock, round it up.). We don't
    162  * timeout the 0,0 value because this means to disable the
    163  * timer or the interval.
    164  */
    165 int
    166 itimerfix(struct timeval *tv)
    167 {
    168 
    169 	if (tv->tv_usec < 0 || tv->tv_usec >= 1000000)
    170 		return EINVAL;
    171 	if (tv->tv_sec < 0)
    172 		return ETIMEDOUT;
    173 	if (tv->tv_sec == 0 && tv->tv_usec != 0 && tv->tv_usec < tick)
    174 		tv->tv_usec = tick;
    175 	return 0;
    176 }
    177 
    178 int
    179 itimespecfix(struct timespec *ts)
    180 {
    181 
    182 	if (ts->tv_nsec < 0 || ts->tv_nsec >= 1000000000)
    183 		return EINVAL;
    184 	if (ts->tv_sec < 0)
    185 		return ETIMEDOUT;
    186 	if (ts->tv_sec == 0 && ts->tv_nsec != 0 && ts->tv_nsec < tick * 1000)
    187 		ts->tv_nsec = tick * 1000;
    188 	return 0;
    189 }
    190 
    191 /*
    192  * timespecaddok(tsp, usp)
    193  *
    194  *	True if tsp + usp can be computed without overflow, i.e., if it
    195  *	is OK to do timespecadd(tsp, usp, ...).
    196  */
    197 bool
    198 timespecaddok(const struct timespec *tsp, const struct timespec *usp)
    199 {
    200 	enum { TIME_MIN = __type_min(time_t), TIME_MAX = __type_max(time_t) };
    201 	time_t a = tsp->tv_sec;
    202 	time_t b = usp->tv_sec;
    203 	bool carry;
    204 
    205 	/*
    206 	 * Caller is responsible for guaranteeing valid timespec
    207 	 * inputs.  Any user-controlled inputs must be validated or
    208 	 * adjusted.
    209 	 */
    210 	KASSERT(tsp->tv_nsec >= 0);
    211 	KASSERT(usp->tv_nsec >= 0);
    212 	KASSERT(tsp->tv_nsec < 1000000000L);
    213 	KASSERT(usp->tv_nsec < 1000000000L);
    214 	__CTASSERT(1000000000L <= __type_max(long) - 1000000000L);
    215 
    216 	/*
    217 	 * Fail if a + b + carry overflows TIME_MAX, or if a + b
    218 	 * overflows TIME_MIN because timespecadd adds the carry after
    219 	 * computing a + b.
    220 	 *
    221 	 * Break it into two mutually exclusive and exhaustive cases:
    222 	 * I. a >= 0
    223 	 * II. a < 0
    224 	 */
    225 	carry = (tsp->tv_nsec + usp->tv_nsec >= 1000000000L);
    226 	if (a >= 0) {
    227 		/*
    228 		 * Case I: a >= 0.  If b < 0, then b + 1 <= 0, so
    229 		 *
    230 		 *	a + b + 1 <= a + 0 <= TIME_MAX,
    231 		 *
    232 		 * and
    233 		 *
    234 		 *	a + b >= 0 + b = b >= TIME_MIN,
    235 		 *
    236 		 * so this can't overflow.
    237 		 *
    238 		 * If b >= 0, then a + b + carry >= a + b >= 0, so
    239 		 * negative results and thus results below TIME_MIN are
    240 		 * impossible; we need only avoid
    241 		 *
    242 		 *	a + b + carry > TIME_MAX,
    243 		 *
    244 		 * which we will do by rejecting if
    245 		 *
    246 		 *	b > TIME_MAX - a - carry,
    247 		 *
    248 		 * which in turn is incidentally always false if b < 0
    249 		 * so we don't need extra logic to discriminate on the
    250 		 * b >= 0 and b < 0 cases.
    251 		 *
    252 		 * Since 0 <= a <= TIME_MAX, we know
    253 		 *
    254 		 *	0 <= TIME_MAX - a <= TIME_MAX,
    255 		 *
    256 		 * and hence
    257 		 *
    258 		 *	-1 <= TIME_MAX - a - 1 < TIME_MAX.
    259 		 *
    260 		 * So we can compute TIME_MAX - a - carry (i.e., either
    261 		 * TIME_MAX - a or TIME_MAX - a - 1) safely without
    262 		 * overflow.
    263 		 */
    264 		if (b > TIME_MAX - a - carry)
    265 			return false;
    266 	} else {
    267 		/*
    268 		 * Case II: a < 0.  If b >= 0, then since a + 1 <= 0,
    269 		 * we have
    270 		 *
    271 		 *	a + b + 1 <= b <= TIME_MAX,
    272 		 *
    273 		 * and
    274 		 *
    275 		 *	a + b >= a >= TIME_MIN,
    276 		 *
    277 		 * so this can't overflow.
    278 		 *
    279 		 * If b < 0, then the intermediate a + b is negative
    280 		 * and the outcome a + b + 1 is nonpositive, so we need
    281 		 * only avoid
    282 		 *
    283 		 *	a + b < TIME_MIN,
    284 		 *
    285 		 * which we will do by rejecting if
    286 		 *
    287 		 *	a < TIME_MIN - b.
    288 		 *
    289 		 * (Reminder: The carry is added afterward in
    290 		 * timespecadd, so to avoid overflow it is not enough
    291 		 * to merely reject a + b + carry < TIME_MIN.)
    292 		 *
    293 		 * It is safe to compute the difference TIME_MIN - b
    294 		 * because b is negative, so the result lies in
    295 		 * (TIME_MIN, 0].
    296 		 */
    297 		if (b < 0 && a < TIME_MIN - b)
    298 			return false;
    299 	}
    300 
    301 	return true;
    302 }
    303 
    304 /*
    305  * timespecsubok(tsp, usp)
    306  *
    307  *	True if tsp - usp can be computed without overflow, i.e., if it
    308  *	is OK to do timespecsub(tsp, usp, ...).
    309  */
    310 bool
    311 timespecsubok(const struct timespec *tsp, const struct timespec *usp)
    312 {
    313 	enum { TIME_MIN = __type_min(time_t), TIME_MAX = __type_max(time_t) };
    314 	time_t a = tsp->tv_sec, b = usp->tv_sec;
    315 	bool borrow;
    316 
    317 	/*
    318 	 * Caller is responsible for guaranteeing valid timespec
    319 	 * inputs.  Any user-controlled inputs must be validated or
    320 	 * adjusted.
    321 	 */
    322 	KASSERT(tsp->tv_nsec >= 0);
    323 	KASSERT(usp->tv_nsec >= 0);
    324 	KASSERT(tsp->tv_nsec < 1000000000L);
    325 	KASSERT(usp->tv_nsec < 1000000000L);
    326 	__CTASSERT(1000000000L <= __type_max(long) - 1000000000L);
    327 
    328 	/*
    329 	 * Fail if a - b - borrow overflows TIME_MIN, or if a - b
    330 	 * overflows TIME_MAX because timespecsub subtracts the borrow
    331 	 * after computing a - b.
    332 	 *
    333 	 * Break it into two mutually exclusive and exhaustive cases:
    334 	 * I. a < 0
    335 	 * II. a >= 0
    336 	 */
    337 	borrow = (tsp->tv_nsec - usp->tv_nsec < 0);
    338 	if (a < 0) {
    339 		/*
    340 		 * Case I: a < 0.  If b < 0, then -b - 1 >= 0, so
    341 		 *
    342 		 *	a - b - 1 >= a + 0 >= TIME_MIN,
    343 		 *
    344 		 * and, since a <= -1, provided that TIME_MIN <=
    345 		 * -TIME_MAX - 1 so that TIME_MAX <= -TIME_MIN - 1 (in
    346 		 * fact, equality holds, under the assumption of
    347 		 * two's-complement arithmetic),
    348 		 *
    349 		 *	a - b <= -1 - b = -b - 1 <= TIME_MAX,
    350 		 *
    351 		 * so this can't overflow.
    352 		 */
    353 		__CTASSERT(TIME_MIN <= -TIME_MAX - 1);
    354 
    355 		/*
    356 		 * If b >= 0, then a - b - borrow <= a - b < 0, so
    357 		 * positive results and thus results above TIME_MAX are
    358 		 * impossible; we need only avoid
    359 		 *
    360 		 *	a - b - borrow < TIME_MIN,
    361 		 *
    362 		 * which we will do by rejecting if
    363 		 *
    364 		 *	a < TIME_MIN + b + borrow.
    365 		 *
    366 		 * The right-hand side is safe to evaluate for any
    367 		 * values of b and borrow as long as TIME_MIN +
    368 		 * TIME_MAX + 1 <= TIME_MAX, i.e., TIME_MIN <= -1.
    369 		 * (Note: If time_t were unsigned, this would fail!)
    370 		 *
    371 		 * Note: Unlike Case I in timespecaddok, this criterion
    372 		 * does not work for b < 0, nor can the roles of a and
    373 		 * b in the inequality be reversed (e.g., -b < TIME_MIN
    374 		 * - a + borrow) without extra cases like checking for
    375 		 * b = TEST_MIN.
    376 		 */
    377 		__CTASSERT(TIME_MIN < -1);
    378 		if (b >= 0 && a < TIME_MIN + b + borrow)
    379 			return false;
    380 	} else {
    381 		/*
    382 		 * Case II: a >= 0.  If b >= 0, then
    383 		 *
    384 		 *	a - b <= a <= TIME_MAX,
    385 		 *
    386 		 * and, provided TIME_MIN <= -TIME_MAX - 1 (in fact,
    387 		 * equality holds, under the assumption of
    388 		 * two's-complement arithmetic)
    389 		 *
    390 		 *	a - b - 1 >= -b - 1 >= -TIME_MAX - 1 >= TIME_MIN,
    391 		 *
    392 		 * so this can't overflow.
    393 		 */
    394 		__CTASSERT(TIME_MIN <= -TIME_MAX - 1);
    395 
    396 		/*
    397 		 * If b < 0, then a - b >= a >= 0, so negative results
    398 		 * and thus results below TIME_MIN are impossible; we
    399 		 * need only avoid
    400 		 *
    401 		 *	a - b > TIME_MAX,
    402 		 *
    403 		 * which we will do by rejecting if
    404 		 *
    405 		 *	a > TIME_MAX + b.
    406 		 *
    407 		 * (Reminder: The borrow is subtracted afterward in
    408 		 * timespecsub, so to avoid overflow it is not enough
    409 		 * to merely reject a - b - borrow > TIME_MAX.)
    410 		 *
    411 		 * It is safe to compute the sum TIME_MAX + b because b
    412 		 * is negative, so the result lies in [0, TIME_MAX).
    413 		 */
    414 		if (b < 0 && a > TIME_MAX + b)
    415 			return false;
    416 	}
    417 
    418 	return true;
    419 }
    420 
    421 /*
    422  * itimer_transition(it, now, next, &overruns)
    423  *
    424  *	Given:
    425  *
    426  *	- it: the current state of an itimer (it_value = last expiry
    427  *	  time, it_interval = periodic rescheduling interval), and
    428  *
    429  *	- now: the current time on the itimer's clock;
    430  *
    431  *	compute:
    432  *
    433  *	- next: the next time the itimer should be scheduled for, and
    434  *	- overruns: the number of overruns if we're firing late.
    435  *
    436  *	XXX This should maybe also say whether the itimer should expire
    437  *	at all.
    438  */
    439 void
    440 itimer_transition(const struct itimerspec *restrict it,
    441     const struct timespec *restrict now,
    442     struct timespec *restrict next,
    443     int *restrict overrunsp)
    444 {
    445 	uint64_t last_val, next_val, interval, now_ns;
    446 	int backwards;
    447 
    448 	/*
    449 	 * Zero the outputs so we can test assertions in userland
    450 	 * without undefined behaviour.
    451 	 */
    452 	timespecclear(next);
    453 	*overrunsp = 0;
    454 
    455 	/*
    456 	 * Paranoia: Caller should guarantee this.
    457 	 */
    458 	if (!timespecisset(&it->it_interval)) {
    459 		timespecclear(next);
    460 		return;
    461 	}
    462 
    463 	backwards = (timespeccmp(&it->it_value, now, >));
    464 
    465 	/* Nonnegative interval guaranteed by itimerfix.  */
    466 	KASSERT(it->it_interval.tv_sec >= 0);
    467 	KASSERT(it->it_interval.tv_nsec >= 0);
    468 
    469 	/* Handle the easy case of non-overflown timers first. */
    470 	if (!backwards &&
    471 	    timespecaddok(&it->it_value, &it->it_interval)) {
    472 		timespecadd(&it->it_value, &it->it_interval,
    473 		    next);
    474 	} else {
    475 		now_ns = timespec2ns(now);
    476 		last_val = timespec2ns(&it->it_value);
    477 		interval = timespec2ns(&it->it_interval);
    478 
    479 		next_val = now_ns +
    480 		    (now_ns - last_val + interval - 1) % interval;
    481 
    482 		if (backwards)
    483 			next_val += interval;
    484 		else
    485 			*overrunsp = (now_ns - last_val) / interval;
    486 
    487 		next->tv_sec = next_val / 1000000000;
    488 		next->tv_nsec = next_val % 1000000000;
    489 	}
    490 }
    491