Martin Schwidefsky wrote:
This code has me stumped. Does it mean that if a page already
has the PageWritable bit set (and count_ok stays 0), we will
always mark the page as volatile?
How does that work out on !s390?
> /**
> + * __page_check_writable() - check page state for new writable pte
> + *
> + * @page: the page the new writable pte refers to
> + * @pte: the new writable pte
> + */
> +void __page_check_writable(struct page *page, pte_t pte, unsigned int offset)
> +{
> + int count_ok = 0;
> +
> + preempt_disable();
> + while (page_test_set_state_change(page))
> + cpu_relax();
> +
> + if (!TestSetPageWritable(page)) {
> + count_ok = check_counts(page, offset);
> + if (check_bits(page) && count_ok)
> + page_set_volatile(page, 1);
> + else
> + /*
> + * If two processes create a write mapping at the
> + * same time check_counts will return false or if
> + * the page is currently isolated from the LRU
> + * check_bits will return false but the page might
> + * be in volatile state.
> + * We have to take care about the dirty bit so the
> + * only option left is to make the page stable but
> + * we can try to make it volatile a bit later.
> + */
> + page_set_stable_if_present(page);
> + }
> + page_clear_state_change(page);
> + if (!count_ok)
> + page_make_volatile(page, 1);
> + preempt_enable();
> +}
> +EXPORT_SYMBOL(__page_check_writable);
--
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