[MEDIUM] scheduler: get rid of the 4 trees thanks and use ebtree v4.1
Since we're now able to search from a precise expiration date in
the timer tree using ebtree 4.1, we don't need to maintain 4 trees
anymore. Not only does this simplify the code a lot, but it also
ensures that we can always look 24 days back and ahead, which
doubles the ability of the previous scheduler. Indeed, while based
on absolute values, the timer tree is now relative to <now> as we
can always search from <now>-31 bits.
The run queue uses the exact same principle now, and is now simpler
and a bit faster to process. With these changes alone, an overall
0.5% performance gain was observed.
Tests were performed on the few wrapping cases and everything works
as expected.
diff --git a/include/common/ticks.h b/include/common/ticks.h
index 4587d56..de29b31 100644
--- a/include/common/ticks.h
+++ b/include/common/ticks.h
@@ -40,7 +40,7 @@
* in the past and as much in the future.
*
* We must both support absolute dates (well in fact, dates relative to now+/-
- * 12 days), and intervals (for timeouts). Both types need an "eternity" magic
+ * 24 days), and intervals (for timeouts). Both types need an "eternity" magic
* value. For optimal code generation, we'll use zero as the magic value
* indicating that an expiration timer or a timeout is not set. We have to
* check that we don't return this value when adding timeouts to <now>. If a
@@ -90,6 +90,18 @@
return tick_add(now, timeout);
}
+/* return 1 if timer <t1> is before <t2>, none of which can be infinite. */
+static inline int tick_is_lt(int t1, int t2)
+{
+ return (t1 - t2) < 0;
+}
+
+/* return 1 if timer <t1> is before or equal to <t2>, none of which can be infinite. */
+static inline int tick_is_le(int t1, int t2)
+{
+ return (t1 - t2) <= 0;
+}
+
/* return 1 if timer <timer> is expired at date <now>, otherwise zero */
static inline int tick_is_expired(int timer, int now)
{