Re: [2.6.32-rc5-git5] synchronize_sched() inside spin_lock()?

[Paul E. McKenney]

On Mon, Nov 02, 2009 at 09:00:06PM +0900, Tetsuo Handa wrote:
> Commit: 4ea7e38696c7e798c47ebbecadfd392f23f814f9
> 
> tracepoint_synchronize_unregister() calls synchronize_sched(), but it is
> between spin_lock() and spin_unlock(). Is it OK?

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

> 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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