Hi Arnd,
Sorry, I would say I made a logic mistake here. The schedule in the
return_from_int() will only be called when the process steps from the
user space into the kerner space( i.e. sys call). So, what you
described are all correct.
And, we have already had a solution to fix the issue you pointed out.
==================
inline static void default_idle(void)
{
int flag;
while (!need_resched()) {
leds_switch(LED_OFF);
local_irq_save(flag);
if ( likely(!need_resched()) {
#if defined (ANOMALY_05000244) && defined (CONFIG_BLKFIN_CACHE)
__asm__("nop; nop;\n");
#endif
__asm__(".align 64;\n STI %0; IDLE;\n"
: %0 (flag): :"cc");
}
local_irq_restore(flag);
leds_switch(LED_ON);
}
}======================================
Here, according to design, it's not possible that interrupt occurs
between "STI %0"(enable interrupt) and "IDLE".
__asm__(".align 64; STI %0; IDLE;" : %0 (x): :"cc");
Robin can explain more details.
-Aubrey
On 9/26/06, Arnd Bergmann <[email protected]> wrote:
On Monday 25 September 2006 17:39, Aubrey wrote:
> 1) Timer interrupt will call do_irq(), then return_from_int().
>
> 2) return_from_int() will check if there is interrupt pending or
> signal pending, if so, it will call schedule_and_signal_from_int().
>
> 3) schedule_and_signal_from_int() will jump to resume_userspace()
>
> 4) resume_userspace() will call _schedule to run the user task.
I have a little trouble reading your assembly code, but your
return_from_int() function should normally not call
schedule_and_signal_from_int() when the interrupt happened
in kernel context (like in the idle function):
+ /* if not return to user mode, get out */
+ p2.l = lo(IPEND);
+ p2.h = hi(IPEND);
+ r0 = [p2];
+ r1 = 0x17(Z);
+ r2 = ~r1;
+ r2.h = 0;
+ r0 = r2 & r0;
+ r1 = 1;
+ r1 = r0 - r1;
+ r2 = r0 & r1;
+ cc = r2 == 0;
+ if !cc jump 2f;
This looks a lot like you user_mode() function, so you jump
over schedule_and_signal_from_int() here.
What you described would be a preemptive kernel
(CONFIG_PREEMPT), but you clearly don't have that enabled.
Arnd <><
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