[ltt-dev] Problem in irq handling

[Mathieu Desnoyers]

* Kiran Kulkarni (kiran.1267 at gmail.com) wrote:
> Hello,
>
> I am using LTTng for my device which has arm processor. I have stuck at  
> a problem that after each timer interrupt
> my cpu spends 10ms time in serving irq (Graph representation also shows  
> same). So it actually spends time 10 ms for each timer interrupt,
> which I guess not correct. (Note: My timer interrupt update system timer  
> by 10ms).
>
> can you help me figure out what causing irq handlers to take 10ms....
>


You use an architecture with non-precise timestamp counter. Either you
have to implement your own trace clock, or accept that the precision is
at best 1 jiffy. See the include/asm-generic/trace-clock.h code for
details.

Mathieu

>
> Snippet example:
>
> kernel_sched_schedule: 98.730291747 (/home/kiran/Desktop/trace4/cpu_0),  
> 0, 0, swapper, , 0, 0x0, SYSCALL { prev_pid = 444, next_pid = 0,  
> prev_state = 2 }
> kernel_irq_entry: 98.730292968 (/home/kiran/Desktop/trace4/cpu_0), 0, 0,  
> swapper, , 0, 0x0, IRQ { irq_id = 8, kernel_mode = 1 }
> kernel_timer_update_time: 98.740001220  
> (/home/kiran/Desktop/trace4/cpu_0), 0, 0, swapper, , 0, 0x0, IRQ {  
> jiffies = 4294947170, xtime_sec = 69321, xtime_nsec = 732571667, wallto
> monotonic_sec = -69224, walltomonotonic_nsec = 687428333 }
> kernel_softirq_raise: 98.740002440 (/home/kiran/Desktop/trace4/cpu_0),  
> 0, 0, swapper, , 0, 0x0, IRQ { softirq_id = 1  
> [run_timer_softirq+0x0/0x208] }
> kernel_softirq_raise: 98.740003661 (/home/kiran/Desktop/trace4/cpu_0),  
> 0, 0, swapper, , 0, 0x0, IRQ { softirq_id = 8  
> [rcu_process_callbacks+0x0/0x38] }
> kernel_irq_exit: 98.740006102 (/home/kiran/Desktop/trace4/cpu_0), 0, 0,  
> swapper, , 0, 0x0, SYSCALL
>
>
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-- 
Mathieu Desnoyers
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