Hello Joe,
Thank you for your reply. I have an additional question.
Joseph A Knapka wrote:
> Martin Maletinsky wrote:
> > Hello,
> >
> > I am looking at the swapping mechanism in Linux. I have read the relevant chapter 16 in 'Understanding the Linux Kernel' from Bovet&Cesati, and looked at the 2.2.18 kernel
> > source code. I still have the follwing question:
> >
> > Function try_to_swap_out() [p. 481 in 'Understanding the Linux Kernel']:
> > If the page in question already belongs to the swap cache, the function performs no data transfer to the swap space on the disk (but only marks the page as swapped out).
> > The corresponding comment in the try_to_swap_out() functions states 'Is the page already in the swap cache? If so, ..... - it is already up-to-date on disk.
> > Understanding the Linux Kernel states on p. 482 'If the page belongs to the swap cache .... no memory transfer is performed'.
> > Now my question is, couldn't the page have been modified since it was added to the swap cache (and written to disk), and thus differ from the data in the swap space? In
> > this case shouldn't the page be written to disk (again)?
>
> If the page is in the swap cache, it's *effectively* up to date on disk,
> because the swap cache page is *the* authoritative image of the page.
> If it's dirty it will get written out by page_launder() in short
> order, because whomever dirtied it set the page_dirty bit in the
> page struct. That issue is unimportant to the process doing the
> swap_out, though - all it cares about is that the page is going
> to be taken care of by the cache machinery.
Assume a page P that is marked as clean (i.e. PG_dirty bit not set), and is in the page cache. Additionaly assume that P is mapped by a process A. Now let A perform a store
operation into the page P, which will mark A's page table entry mapping P as dirty (i.e. set the dirty bit).
Subsequently assume that try_to_swap_out() is called on A's page table entry that maps P. try_to_swap_out() will see that P is in the swap cache already, and thus drop the pte.
This leads to a situation, where P is in the swap cache, marked as clear (i.e. PG_dirty bit clear), while the disk image differs from the data that is in the main memory page
frame.
I would have expected try_to_swap_out() to check the page table entries dirty bit, and to mark the page dirty. However, I can't see any such code in the function (I pasted the
relevant lines of code from linux 2.2.18 below).
static int try_to_swap_out(struct task_struct * tsk, struct vm_area_struct* vma,
unsigned long address, pte_t * page_table, int gfp_mask)
{
pte_t pte;
unsigned long entry;
unsigned long page;
struct page * page_map;
pte = *page_table;
if (!pte_present(pte))
return 0;
page = pte_page(pte);
if (MAP_NR(page) >= max_mapnr)
return 0;
page_map = mem_map + MAP_NR(page);
if (pte_young(pte)) {
/*
* Transfer the "accessed" bit from the page
* tables to the global page map.
*/
set_pte(page_table, pte_mkold(pte));
flush_tlb_page(vma, address);
set_bit(PG_referenced, &page_map->flags);
return 0;
}
if (PageReserved(page_map)
|| PageLocked(page_map)
|| ((gfp_mask & __GFP_DMA) && !PageDMA(page_map)))
return 0;
/*
* Is the page already in the swap cache? If so, then
* we can just drop our reference to it without doing
* any IO - it's already up-to-date on disk.
*
* Return 0, as we didn't actually free any real
* memory, and we should just continue our scan.
*/
if (PageSwapCache(page_map)) {
entry = page_map->offset;
swap_duplicate(entry);
set_pte(page_table, __pte(entry));
drop_pte:
vma->vm_mm->rss--;
flush_tlb_page(vma, address);
__free_page(page_map);
return 0;
}
............
Thanks again, with best regards
Martin
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