On Sat, Aug 14, 2010 at 08:57:09PM +0800, Parmenides wrote: > Hi, > > To acitvate a softirq, the raise_softirq() will be invorked. In this > function, in_interruput() function is called, and if we get value 1, > raise_softirq() will return simply. According to ULK3, this situation > indicates either that raise_softirq() has been invoked in interrupt > context, or that the softirqs are currently disabled. For the latter, > namely that the softirqs are currently disabled, I have no idea about > it. How can we know?
softirqs can be disabled using local_bh_disable(). This will "simulate" the fact you enter in a softirq context, which means you can't be interrupted by another level of softirq. This is simply a check of (preempt_count() & SOFTIRQ_MASK). When local_bh_disable() is called or when you enter softirqs, the softirq offset in preempt_count() is incremented. It is later decremented when you call local_bh_enable() or softirqs jobs finish. in_interrupt() returns 1 if you are on a hardirq or softirq. -- To unsubscribe from this list: send an email with "unsubscribe kernelnewbies" to [email protected] Please read the FAQ at http://kernelnewbies.org/FAQ
