On 8/27/21 8:32 AM, Juergen Gross wrote:
+static bool balloon_thread_cond(enum bp_state state, long credit)
+{
+ if (state != BP_EAGAIN)
+ credit = 0;
+
+ return current_credit() != credit || kthread_should_stop();
+}
+
+/*
+ * As this is a kthread it is guaranteed to run as a single instance only.
* We may of course race updates of the target counts (which are protected
* by the balloon lock), or with changes to the Xen hard limit, but we will
* recover from these in time.
*/
-static void balloon_process(struct work_struct *work)
+static int balloon_thread(void *unused)
{
enum bp_state state = BP_DONE;
long credit;
+ unsigned long timeout;
+
+ set_freezable();
+ for (;;) {
+ if (state == BP_EAGAIN)
+ timeout = balloon_stats.schedule_delay * HZ;
+ else
+ timeout = 3600 * HZ;
+ credit = current_credit();
+ wait_event_interruptible_timeout(balloon_thread_wq,
+ balloon_thread_cond(state, credit), timeout);
Given that wait_event_interruptible_timeout() is a bunch of nested macros do we need to worry here about overly aggressive compiler optimizing out 'credit = current_credit()'?
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