Re: [patch 4/6] Guest page hinting: writable page table entries.

[Rik van Riel <riel@redhat.com>]

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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