I am busy implementing some heartbeat monitoring code between two
machines. The spec calls for a 1 second recovery.
Basically if I get no heartbeats for 1 full second then I should
consider the peer system to have failed.
To cope with the leap-second scenario, one solution is to use a
timeout of 1 second longer than usual if the current time is close
to the turnover of the day. You can do this easily by checking
time(NULL) % 86400
and if we are at the turnover of the day use a 2 second timeout
instead of a 1 second timeout.
Now this seems like a nice and easy way of fixing old code. Here
is an example:
void process_event(Event e)
{
long long now = gettimeofday_in_millisecs();
if (now > last_recv_time + 1000) {
peer_has_failed();
} else if (e == EVENT_HEARTBEAT) {
last_recv_time = now;
}
}
Becomes:
int near_turnover_of_day(long long t)
{
#define FUDGE 2
if ((t + FUDGE) % 86400 <= FUDGE * 2)
return 1;
return 0;
}
void process_heartbeat(Event e)
{
long long now = gettimeofday_in_millisecs();
if (now > last_recv_time + 1000 +
1000 * near_turnover_of_day(now / 1000)) {
peer_has_failed();
} else if (e == EVENT_HEARTBEAT) {
last_recv_time = now;
}
}
Comments?
(This example is off the top of my head so please excuse any errors.)
-paul
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