I haven't tried to pass the timeout value but what do is with a global
variable.
Imagine that you have the reference of the timeout, as a global variable :
guint TimeOut_Ref;
then you have a first call to your callback function
TimeOut_Ref = g_timeout_add (20*1000, callback, (gpointer)&timeout);
you stop the timeout just when you enter the function :
gboolean callback (gpointer data) {
if (TimeOut_Ref > 0) {
g_source_remove(TimeOut_Ref);
TimeOut_Ref = 0;
}
....
Then, at the end of you called function, just before the return, you
reactivate the timeout :
.....
TimeOut_Ref = g_timeout_add (20*1000, callback, (gpointer)&timeout);
return TRUE;
}
Regards
Pascal
Mitko Haralanov a écrit :
> I have a question about g_timeout_add's timeout recalculation.
>
> Currently, the next call time is calculated based on <time of callback
> thread start> + <timeout given to g_timeout_add>. This means that if I
> call g_timeout_add with a timeout of 20*1000 msecs (20 seconds) and my
> callback takes 10 seconds to complete, my callback will be called 10
> seconds after it completes. Here is an example:
>
> #include <gtk/gtk.h>
> #include <glib.h>
> #include <time.h>
> #include <stdio.h>
>
> gboolean callback (gpointer data) {
> int timeout = *((int *)data);
> printf ("callback called at: %lu\n", time (NULL));
> sleep (timeout);
> printf ("callback returns at: %lu\n", time (NULL));
> return TRUE;
> }
>
> int main (int argc, char **argv) {
> int timeout = 10;
> gtk_init (&argc, &argv);
> g_timeout_add (20*1000, callback, (gpointer)&timeout);
> gtk_main ();
> }
>
>
> will produce this:
> callback called at: 1210893003
> callback returns at: 1210893013
> callback called at: 1210893023
> callback returns at: 1210893033
> callback called at: 1210893043
>
> What I would to happen is that my callback is called 20 seconds after
> it returns, not 20 seconds after it is started.
>
> How can I accomplish this?
>
>
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