Lines Matching refs:timeout
41 puts ("timeout");
58 // simple non-repeating 5.5 second timeout
96 file descriptor being readable or a timeout occurring), and it will manage
507 returning before the timeout value, resulting in additional iterations
523 because epoll returns immediately despite a nonzero timeout. And last
975 Likewise, by setting a higher I<timeout collect interval> you allow libev
991 Setting the I<timeout collect interval> can improve the opportunity for
998 Example: we only need 0.1s timeout granularity, and we wish not to poll
1495 other jobs, but since timers normally get invoked first, the timeout
1791 The callback is guaranteed to be invoked only I<after> its timeout has
1801 Many real-world problems involve some kind of timeout, usually for error
1808 In the following, a 60 second activity timeout is assumed - a timeout that
1866 you want to modify its timeout value, as libev does not have to completely
1879 but remember the time of last activity, and check for a real timeout only
1882 ev_tstamp timeout = 60.;
1889 // calculate when the timeout would happen
1890 ev_tstamp after = last_activity - ev_now (EV_A) + timeout;
1892 // if negative, it means we the timeout already occurred
1895 // timeout occurred, take action
1902 // the timeout can occur.
1909 timeout will occur (by calculating the absolute time when it would occur,
1910 C<last_activity + timeout>, and subtracting the current time, C<ev_now
1913 If this value is negative, then we are already past the timeout, i.e. we
1916 Otherwise, we now the earliest time at which the timeout would trigger,
1917 and simply start the timer with this timeout value.
1920 the timeout occurred. If not, it will simply reschedule itself to check
1925 libev to change the timeout.
1942 When your timeout value changes, then the timeout can be changed by simply
1946 timeout = new_value;
1956 employing some kind of timeout with the same timeout value, then one can
1959 When starting the timeout, calculate the timeout value and put the timeout
1962 Then use an C<ev_timer> to fire when the timeout at the I<beginning> of
1966 the timeout, append it to the end of the list again, and make sure to
1971 complication, and having to use a constant timeout. The constant timeout
2007 event loop will schedule your timeout to elapse at a system time of 500
2010 If an event library looks at the timeout 0.1s later, it will see "501 >=
2035 of the event triggering whatever timeout you are modifying/starting. If
2037 timeout on the current time, use something like the following to adjust
2119 is C<0.>, then it will automatically be stopped once the timeout is
2134 timeout to the C<repeat> value and calling C<ev_timer_start>.
2146 =item If the timer is repeating, make the C<repeat> value the new timeout
2158 the timeout value currently configured.
2169 or C<ev_timer_again> is called, and determines the next timeout (if any),
2188 Example: Create a timeout timer that times out after 10 seconds of
2203 // reset the timeout to start ticking again at 10 seconds
2225 it, as it uses a relative timeout).
3008 Method 1: Add IO watchers and a timeout watcher in a prepare handler,
3012 the callbacks for the IO/timeout watchers might not have been called yet.
3026 int timeout = 3600000;
3029 adns_beforepoll (ads, fds, &nfd, &timeout, timeval_from (ev_time ()));
3032 ev_timer_init (&tw, 0, timeout * 1e-3, 0.);
3105 event_poll_func (GPollFD *fds, guint nfds, gint timeout)
3112 if (timeout >= 0)
3119 if (timeout >= 0)
3522 =item ev_once (loop, int fd, int events, ev_tstamp timeout, callback)
3527 or timeout without having to allocate/configure/start/stop/free one or
3534 If C<timeout> is less than 0, then no timeout watcher will be
3535 started. Otherwise an C<ev_timer> watcher with after = C<timeout> (and
3536 repeat = 0) will be started. C<0> is a valid timeout.
3542 a timeout and an io event at the same time - you probably should give io