Will the GTK+ timeout callbacks be called in strict time order?

Question

When I add many different timeouts (with each intervall==0) in a thread, which is not the main thread (where gtk_main() resides)...

g_timeout_add(0, func, NULL);

... will then the different func() callbacks occur in the same order I called the corresponding g_timeout_add()'s?

The reason I'm asking is because GTK# is using internally timeouts to implement Application.Invoke() (see Application.cs and Timeout.cs).

EDIT: The concerning glib files are

score 4 · Answer 1 · answered Feb 10 '10 at 09:48

4

Internally, g_timeout_add calls g_hook_insert_sorted. If g_timeout_add_full is used, the priority determines the ordering, otherwise the hook is added at the end of the list. Hooks are executed in order, so when only g_timeout_add is used, the answer is yes.

Unfortunately, there is not no explicit guarantee, and to me, it looks like an implementation detail which might change in the future.

answered Feb 10 '10 at 09:48

Ondergetekende

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OK. But I cannot really find a call to `g_hook_insert_sorted()` inside *gmain.c* (I added a link in the question)? – ulrichb Feb 10 '10 at 15:28

score 0 · Answer 2 · answered Jan 26 '12 at 14:15

How about enforcing the order of the calls by storing your callbacks explicitly in a list, then using a single g_timeout_add() to call a function that iterates over that list?

static gboolean
call_in_order (GList* callbacks)
{
  for (; callbacks != NULL; callbacks = callbacks->next)
    {
      g_assert (callbacks->data != NULL);
      GSourceFunc callback = (GSourceFunc)(callbacks->data);

      callback(NULL);
    }
}


...

 GList *callbacks = NULL;
 callbacks = g_list_append(callbacks, func1);
 callbacks = g_list_append(callbacks, func2);
 callbacks = g_list_append(callbacks, func3);
 callbacks = g_list_append(callbacks, func4);

 g_timeout_add(0, (GSourceFunc)call_in_order, callbacks);

Will the GTK+ timeout callbacks be called in strict time order?

2 Answers2