If you need things to happen in a particular order, you need to sequence
your timeouts from each other. In an event based system, you cannot
guarantee the order that the events will be fired.
Instead of creating A B and C at the same time, if you need them to go in
that order, you will need to create A, when it fires let it create B, and
when it fires let it create C.
On Thu, Jun 25, 2009 at 1:28 PM, Denis Bilenko <denis.bile...@gmail.com>wrote:
> Hi,
>
> I've run into an issue that surprised me a bit. I'm installing a few
> timeouts to be
> executed on a following loop iteration, i.e. after zero seconds. I
> would expect the timeouts
> happen in the order they were installed, however, this is not the
> case. Here's the code
> that shows the issue:
>
> #include <stdio.h>
> #include "event.h"
>
> struct event e[3];
> struct timeval zero = {0, 0};
>
> void print(int fd, short events, char * text)
> {
> printf(text);
> }
>
> void callback(int fd, short events, void * arg)
> {
> timeout_set(&e[0], print, "A\n");
> timeout_add(&e[0], &zero);
> timeout_set(&e[1], print, "B\n");
> timeout_add(&e[1], &zero);
> timeout_set(&e[2], print, "C\n");
> timeout_add(&e[2], &zero);
> }
>
> int main()
> {
> struct event e;
> event_init();
> timeout_set(&e, callback, NULL);
> timeout_add(&e, &zero);
> event_dispatch();
> }
>
> The output of this program is
> libevent$ ./a.out
> A
> C
> B
>
> not ABC as I would like.
>
> If I do the same outside of the callback, the order is maintained as
> expected.
>
> Adding more timers suggests that all of them except the first one (A)
> are executed in the reverse order.
> Again, doing the same outside of a callback makes them run in the
> expected order.
>
> Is this a bug or it's supposed to work this way? Maybe I'm doing
> something wrong and there is a way
> of getting the expected behavior without resorting to increasing
> timeval for each new call?
>
> Thanks,
> Denis.
> _______________________________________________
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> Libevent-users@monkey.org
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>
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