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    SubjectRe: how does kernel get the "current" task struct?
    Hi,

    On Tue, 8 Feb 2000 13:01:44 +0100, Jamie Lokier
    <lkd@tantalophile.demon.co.uk> said:

    > Linus Torvalds wrote:
    >> The zones have to be aligned, there's no question about that. The
    >> preferred alignment is in the megabyte range, rather than in individual
    >> pages. In fact, I would suggest always making sure that it is aligned to
    >> the largest order that get_free_page() supports, and I think that's true
    >> of all current architectures..

    > I don't see why it's necessary though. The allocator should simply use
    > absolute address values for pairing instead of zone-relative
    > addresses.. is there anything more to it than that?

    Yes --- you need 8k allocations to be 8k-aligned physically, and 64k
    allocations to be 64k-aligned, etc. Just to think of one example, Intel
    MTRR absolutely requires that the address of a memory range is a
    multiple of the size of the range, and you need aligned zones to create
    such regions.

    This needn't be a problem --- you can easily create a zone aligned at a
    large (eg. 16MB) granularity, without filling it with free pages. As an
    example, the low-memory zone doesn't populate the free page list with
    pages in the 640k-1M hole. Just because your physical page region isn't
    aligned nicely doesn't mean you cannot align the zone itself.

    --Stephen

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