How do you make a precise countdown timer using clock_gettime? - c

Question

Could somebody please explain how to make a countdown timer using clock_gettime, under Linux. I know you can use the clock() function to get cpu time, and multiply it by CLOCKS_PER_SEC to get actual time, but I'm told the clock() function is not well suited for this.
So far I have attempted this (a billion is to pause for one second)
#include <stdio.h>
#include <time.h>
#define BILLION 1000000000
int main()
{
struct timespec rawtime;
clock_gettime(CLOCK_MONOTONIC_RAW, &rawtime);
unsigned long int current = ( rawtime.tv_sec + rawtime.tv_nsec );
unsigned long int end = (( rawtime.tv_sec + rawtime.tv_nsec ) + BILLION );
while ( current < end )
{
clock_gettime(CLOCK_MONOTONIC_RAW, &rawtime);
current = ( rawtime.tv_sec + rawtime.tv_nsec );
}
return 0;
}
I know this wouldn't be very useful on its own, but once I've found out how to time correctly I can use this in my projects. I know that sleep() can be used for this purpose, but I want to code the timer myself so that I can better integrate it in my projects - such as the possibility of it returning the time left, as opposed to pausing the whole program.

Answer 1

Please, do not do that. You're burning CPU power for nothing in a busy loop.
Why not use the nanosleep() function instead? It's perfectly suited to the use case you outlined. Or, if you want an easier interface, perhaps something like
#define _POSIX_C_SOURCE 200809L
#include <time.h>
#include <errno.h>
/* Sleep for the specified number of seconds,
* and return the time left over.
*/
double dsleep(const double seconds)
{
struct timespec req, rem;
/* No sleep? */
if (seconds <= 0.0)
return 0.0;
/* Convert to seconds and nanoseconds. */
req.tv_sec = (time_t)seconds;
req.tv_nsec = (long)((seconds - (double)req.tv_sec) * 1000000000.0);
/* Take care of any rounding errors. */
if (req.tv_nsec < 0L)
req.tv_nsec = 0L;
else
if (req.tv_nsec > 999999999L)
req.tv_nsec = 999999999L;
/* Do the nanosleep. */
if (nanosleep(&req, &rem) != -1)
return 0.0;
/* Error? */
if (errno != EINTR)
return 0.0;
/* Return remainder. */
return (double)rem.tv_sec + (double)rem.tv_nsec / 1000000000.0;
}
The difference is that using this one the CPU is free to do something else, rather than spin like a crazed squirrel on speed.

Answer 2

This is not an answer, but an example of how to use signals and a POSIX timer to implement a timeout timer; intended as a response to the OP's followup question in a comment to the accepted answer.
#define _POSIX_C_SOURCE 200809L
#include <stdlib.h>
#include <signal.h>
#include <time.h>
#include <errno.h>
#include <string.h>
#include <stdio.h>
/* Timeout timer.
*/
static timer_t timeout_timer;
static volatile sig_atomic_t timeout_state = 0;
static volatile sig_atomic_t timeout_armed = 2;
static const int timeout_signo = SIGALRM;
#define TIMEDOUT() (timeout_state != 0)
/* Timeout signal handler.
*/
static void timeout_handler(int signo, siginfo_t *info, void *context __attribute__((unused)))
{
if (timeout_armed == 1)
if (signo == timeout_signo && info && info->si_code == SI_TIMER)
timeout_state = ~0;
}
/* Unset timeout.
* Returns nonzero if timeout had expired, zero otherwise.
*/
static int timeout_unset(void)
{
struct itimerspec t;
const int retval = timeout_state;
/* Not armed? */
if (timeout_armed != 1)
return retval;
/* Disarm. */
t.it_value.tv_sec = 0;
t.it_value.tv_nsec = 0;
t.it_interval.tv_sec = 0;
t.it_interval.tv_nsec = 0;
timer_settime(timeout_timer, 0, &t, NULL);
return retval;
}
/* Set timeout (in wall clock seconds).
* Cancels any pending timeouts.
*/
static int timeout_set(const double seconds)
{
struct itimerspec t;
/* Uninitialized yet? */
if (timeout_armed == 2) {
struct sigaction act;
struct sigevent evt;
/* Use timeout_handler() for timeout_signo signal. */
sigemptyset(&act.sa_mask);
act.sa_sigaction = timeout_handler;
act.sa_flags = SA_SIGINFO;
if (sigaction(timeout_signo, &act, NULL) == -1)
return errno;
/* Create a monotonic timer, delivering timeout_signo signal. */
evt.sigev_value.sival_ptr = NULL;
evt.sigev_signo = timeout_signo;
evt.sigev_notify = SIGEV_SIGNAL;
if (timer_create(CLOCK_MONOTONIC, &evt, &timeout_timer) == -1)
return errno;
/* Timeout is initialzied but unarmed. */
timeout_armed = 0;
}
/* Disarm timer, if armed. */
if (timeout_armed == 1) {
/* Set zero timeout, disarming the timer. */
t.it_value.tv_sec = 0;
t.it_value.tv_nsec = 0;
t.it_interval.tv_sec = 0;
t.it_interval.tv_nsec = 0;
if (timer_settime(timeout_timer, 0, &t, NULL) == -1)
return errno;
timeout_armed = 0;
}
/* Clear timeout state. It should be safe (no pending signals). */
timeout_state = 0;
/* Invalid timeout? */
if (seconds <= 0.0)
return errno = EINVAL;
/* Set new timeout. Check for underflow/overflow. */
t.it_value.tv_sec = (time_t)seconds;
t.it_value.tv_nsec = (long)((seconds - (double)t.it_value.tv_sec) * 1000000000.0);
if (t.it_value.tv_nsec < 0L)
t.it_value.tv_nsec = 0L;
else
if (t.it_value.tv_nsec > 999999999L)
t.it_value.tv_nsec = 999999999L;
/* Set it repeat once every millisecond, just in case the initial
* interrupt is missed. */
t.it_interval.tv_sec = 0;
t.it_interval.tv_nsec = 1000000L;
if (timer_settime(timeout_timer, 0, &t, NULL) == -1)
return errno;
timeout_armed = 1;
return 0;
}
int main(void)
{
char *line = NULL;
size_t size = 0;
ssize_t len;
fprintf(stderr, "Please supply input. The program will exit automatically if\n");
fprintf(stderr, "it takes more than five seconds for the next line to arrive.\n");
fflush(stderr);
while (1) {
if (timeout_set(5.0)) {
const char *const errmsg = strerror(errno);
fprintf(stderr, "Cannot set timeout: %s.\n", errmsg);
return 1;
}
len = getline(&line, &size, stdin);
if (len == (ssize_t)-1)
break;
if (len < (ssize_t)1) {
/* This should never occur (except for -1, of course). */
errno = EIO;
break;
}
/* We do not want *output* to be interrupted,
* so we cancel the timeout. */
timeout_unset();
if (fwrite(line, (size_t)len, 1, stdout) != 1) {
fprintf(stderr, "Error writing to standard output.\n");
fflush(stderr);
return 1;
}
fflush(stdout);
/* Next line. */
}
/* Remember to cancel the timeout. Also check it. */
if (timeout_unset())
fprintf(stderr, "Timed out.\n");
else
if (ferror(stdin) || !feof(stdin))
fprintf(stderr, "Error reading standard input.\n");
else
fprintf(stderr, "End of input.\n");
fflush(stderr);
/* Free line buffer. */
free(line);
line = NULL;
size = 0;
/* Done. */
return 0;
}
If you save the above as timer.c, you can compile it using e.g.
gcc -W -Wall -O3 -std=c99 -pedantic timer.c -lrt -o timer
and run it using ./timer.
If you read the code above carefully, you'll see that it is actually a periodic timer signal (at millisecond intervals), with a variable delay before the first signal. That is just a technique I like to use to make sure I don't miss the signal. (The signal repeats until the timeout is unset.)
Note that although you can do computation in an signal handler, you should only use functions that are async-signal-safe; see man 7 signal. Also, only the sig_atomic_t type is atomic wrt. normal single-threaded code and a signal handler. So, it is better to just use the signal as an indicator, and do the actual code in your own program.
If you wanted to e.g. update monster coordinates in a signal handler, it is possible but a bit tricky. I'd use three arrays containing the monster information, and use GCC __sync_bool_compare_and_swap() to update the array pointers -- very much the same technique as triple-buffering in graphics.
If you need more than one concurrent timeout, you could use multiple timers (there is a number of them available), but the best option is to define timeout slots. (You can use generation counters to detect "forgotten" timeouts, and so on.) Whenever a new timeout is set or unset, you update the timeout to reflect the next timeout that expires. It's a bit more code, but really a straightforward extension of the above.