Hi,
I am re-posting question posted by Daniel Rodrick <daniel.rodrick@gmai...> whose answer is still not very clear to me.

1) I read that an ISR cannot cause a page fault. Is the statement true?

2) In the case it is true, how are situations like this handled : an
ISR tries to use a data item that has been swapped out?

3) I have the following argument that convinces me that the statement
is false. When an ISR causes a page fault, the page fault handler is
invoked since its priority is higher. When it finishes the control
goes back to ISR, just like the normal case of nested ISR execution.
Hence an ISR can cause page fault. Is any thing wrong with this
argument?

My question is how Kernel assures that page fault doesn't happen during ISR? Is request_irq dues some miracle while registering the IRQ?

Thanks,
Shreshtha

Page faults are exceptions, and exceptions cannot interrupt ISRs.

I am not sure what would actually happen while inside an ISR if you tried to force access to memory that was paged out. I assume that on an Intel processor, some hardware mechanism would be triggered in the memory management unit which would eventually result in a kernel oops.

Once again, I am not sure what would actually happen while inside an ISR if you tried to force access to memory that was paged out. See my answer above.

There is everything wrong with this argument. Page faults are exceptions, and exceptions are lower priority than interrupts.

Driver writers assure that page faults will not happen during an ISR by only using memory that cannot be paged out. Memory allocated by kmalloc and its relatives cannot be paged out.

DEVIL'S ADVOCATE:
Even if page faults could occur during an ISR, they would bog the system down until it stopped. Memory is paged out to one of the slowest devices in the system: the hard drive. If ISRs had to wait for a page to be read from the hard drive, they would run too long and cause other interrupts to be mishandled, which would eventually lead to a slowdown and then a halt.

1 members found this post helpful.

Is this an existing thread here at LQ, or something from another forum? Can you post a link to the original post?

hi
this was posted in some other post
here is the link

http://www.nabble.com/Why-can%27t-an...td8517056.html

Hi David1357,
thanks a lot for your inputs.
here is my understanding now -

Double fault and triple fault is what happen if something nasty done in ISR or done forcibly (e.g. sometimes Linux does during shutdown).
In ISR, page fault cannot happen until we do it forcibly. Kernel pages are not page-able and all data are non page-able as allocated using kmalloc. So no page fault can happen during ISR.

I take a situation where code access some NULL pointer.
MMU (Memory Management Unit) will come in to action will generate page fault. But Page fault routine will also fail as no such page exists. NOTE: these all happened while ISR took a back seat. And then finally it will go to system crash without returning to executing ISR.
It is clear that without a NULL pointer reference its not possible to access a paged out data. Or is it possible? But how?

Looking in to DoubleFault and TripleFault also adds some light in to this topic.
http://en.wikipedia.org/wiki/Double_fault
http://en.wikipedia.org/wiki/Triple_fault

Thanks and Regards,
Shreshtha