Correct and simplify timeouts_readd()
William Ahern tells me that the intent here is that timeout_readd() should always reschedule the timeout at the first time in the future that is an even multiple of to->interval, as if we had called: do { to->expires += to->interval; } while (to->expires <= T->curtime); But of course, that's not efficient. The implementation strategy used in this patch simplifies the calculation down to a single % operation, plus a few additions and subtractions. To verify the correctness of the formula used here, note first that 0 <= r < to->interval, and so 0 < to->interval - r <= to->interval. Since expires' = curtime + (interval - r), curtime < expires' <= curtime + interval, and so the new expiration time is no more than one interval after curtime. Note second that since r = (curtime - expires) % interval, expires' = curtime + (interval - r), we have (expires' - expires) % interval = (curtime + (interval - r) - expires) % interval = (curtime - r - expires) % interval = (curtime - (curtime-expires) % interval - expires) % interval = (curtime - curtime + expires - expires) % interval = 0. And so the new expiration time is an even multiple of interval from the original expiration time. Since we have both properties we wanted, this formula should be right.
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