xref: /netbsd-src/sys/kern/subr_time_arith.c (revision 65195a4c92dbfea163b48af57736b67314fc5cbb)
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