tracepoint_synchronize_unregister() calls synchronize_sched(), but it is
between spin_lock() and spin_unlock(). Is it OK?
static int set_all_monitor_traces(int state)
{
int rc = 0;
struct dm_hw_stat_delta *new_stat = NULL;
struct dm_hw_stat_delta *temp;
spin_lock(&trace_state_lock);
switch (state) {
case TRACE_ON:
rc |= register_trace_kfree_skb(trace_kfree_skb_hit);
rc |= register_trace_napi_poll(trace_napi_poll_hit);
break;
case TRACE_OFF:
rc |= unregister_trace_kfree_skb(trace_kfree_skb_hit);
rc |= unregister_trace_napi_poll(trace_napi_poll_hit);
tracepoint_synchronize_unregister();
/*
* Clean the device list
*/
list_for_each_entry_safe(new_stat, temp, &hw_stats_list, list) {
if (new_stat->dev == NULL) {
list_del_rcu(&new_stat->list);
call_rcu(&new_stat->rcu, free_dm_hw_stat);
}
}
break;
default:
rc = 1;
break;
}
if (!rc)
trace_state = state;
spin_unlock(&trace_state_lock);
if (rc)
return -EINPROGRESS;
return rc;
}
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Calling synchronize_sched() while holding a spinlock would indeed be
very bad, but the code below seems to instead be invoking call_rcu(),
which is no problem.
Or am I missing something here?
Thanx, Paul
This has synchronize_sched() inside.
--
Dmitry
Ah! That would indeed be bad.
Thanx, Paul
set_all_monitor_traces() calls tracepoint_synchronize_unregister()
with a spinlock held, but tracepoint_synchronize_unregister() may sleep.