On Thu, Dec 13, 2012 at 02:01:55PM -0800, Yinghai Lu wrote:
> During debug load kernel above 4G, found one page if is not used in BRK
> and it should be with early page allocation.
What does that mean?
I see that this patch adds a change to not use the page at pgt_buf_top
anymore but why? Is pgt_buf_top the first invalid address we cannot
reserve anymore?
Generally, can we get this whole deal described in a bit more detail for
the mere mortals among us, maybe a short ascii art thing showing what
all those pgt_buf_{start,end,top} mean.
> Fix that checking and also add print out for every allocation from BRK
> page table allocation.
>
> Signed-off-by: Yinghai Lu <[email protected]>
> ---
> arch/x86/mm/init.c | 4 +++-
> 1 file changed, 3 insertions(+), 1 deletion(-)
>
> diff --git a/arch/x86/mm/init.c b/arch/x86/mm/init.c
> index 6f85de8..c4293cf 100644
> --- a/arch/x86/mm/init.c
> +++ b/arch/x86/mm/init.c
> @@ -47,7 +47,7 @@ __ref void *alloc_low_pages(unsigned int num)
> __GFP_ZERO, order);
> }
>
> - if ((pgt_buf_end + num) >= pgt_buf_top) {
> + if ((pgt_buf_end + num) > pgt_buf_top) {
> unsigned long ret;
> if (min_pfn_mapped >= max_pfn_mapped)
> panic("alloc_low_page: ran out of memory");
> @@ -61,6 +61,8 @@ __ref void *alloc_low_pages(unsigned int num)
> } else {
> pfn = pgt_buf_end;
> pgt_buf_end += num;
> + printk(KERN_DEBUG "BRK [%#010lx, %#010lx] PGTABLE\n",
pr_debug
> + pfn << PAGE_SHIFT, (pgt_buf_end << PAGE_SHIFT) - 1);
> }
>
> for (i = 0; i < num; i++) {
> --
> 1.7.10.4
Thanks.
--
Regards/Gruss,
Boris.
Sent from a fat crate under my desk. Formatting is fine.
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