Re: x86 ptep_get_and_clear question

[Jamie Lokier <lk@tantalophile.demon.co.uk>]

Kanoj Sarcar wrote:
> > Is the sequence
> > << lock;
> > read pte
> > pte |= dirty
> > write pte
> > >> end lock;
> > or
> > << lock;
> > read pte
> > if (!present(pte))
> > 	do_page_fault();
> > pte |= dirty
> > write pte.
> > >> end lock;
> 
> No, it is a little more complicated. You also have to include in the
> tlb state into this algorithm. Since that is what we are talking about.
> Specifically, what does the processor do when it has a tlb entry allowing
> RW, the processor has only done reads using the translation, and the 
> in-memory pte is clear?

Yes (no to the no): Manfred's pseudo-code is exactly the question you're
asking.  Because when the TLB entry is non-dirty and you do a write, we
_know_ the processor will do a locked memory cycle to update the dirty
bit.  A locked memory cycle implies read-modify-write, not "write TLB
entry + dirty" (which would be a plain write) or anything like that.

Given you know it's a locked cycle, the only sensible design from Intel
is going to be one of Manfred's scenarios.

An interesting thought experiment though is this:

<< lock;
read pte
pte |= dirty
write pte
>> end lock;
if (!present(pte))
	do_page_fault();

It would have a mighty odd effect wouldn't it?

-- Jamie