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SubjectRe: x86 ptep_get_and_clear question
Jamie Lokier wrote:
>
> And how does that lose a dirty bit?
>
> For the other processor to not write a dirty bit, it must have a dirty
^^^^^^^^^^^
> TLB entry already which, along with the locked cycle in
> ptep_get_and_clear, means that `entry' will have _PAGE_DIRTY set. The
> dirty bit is not lost.
>
The other cpu writes the dirty bit - we just overwrite it ;-)
After the ptep_get_and_clear(), before the set_pte().

The current assumption about the page dirty logic is:
A cpu that has a writable, non-dirty pte cached in its tlb it may
unconditionally set the dirty bit - without honoring present or write
protected bits.

--> set_pte() can either loose a dirty bit or a 'pte_none() entry' could
suddenly become a swap entry unless it's guaranteed that no cpus has a
cached valid tlb entry.

Linus, does the proposed pte gather code handle the second part?
pte_none() suddenly becomes 0x0040.

Back to the current mprotect.c code:

pte is writable, not-dirty.

cpu1:
has a writable, non-dirty pte in it's tlb.
cpu 2: in mprotect.c
entry = ptep_get_and_clear(pte);
* pte now clear.
* entry contains the pte value without
the dirty bit
cpu decodes a write instruction, and dirties the pte.
lock; orl DIRTY_BIT, *pte
set_pte(pte, pte_modify(entry, newprot));
* pte overwritten with entry.

--> dirty bit lost.

--
Manfred
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