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    SubjectRe: [PATCH 1/5] staging: zsmalloc: zsmalloc memory allocation library
    On 01/26/2012 01:12 PM, Dave Hansen wrote:
    > On 01/09/2012 02:51 PM, Seth Jennings wrote:
    >> + area = &get_cpu_var(zs_map_area);
    >> + if (off + class->size <= PAGE_SIZE) {
    >> + /* this object is contained entirely within a page */
    >> + area->vm_addr = kmap_atomic(page);
    >> + } else {
    >> + /* this object spans two pages */
    >> + struct page *nextp;
    >> +
    >> + nextp = get_next_page(page);
    >> + BUG_ON(!nextp);
    >> +
    >> +
    >> + set_pte(area->vm_ptes[0], mk_pte(page, PAGE_KERNEL));
    >> + set_pte(area->vm_ptes[1], mk_pte(nextp, PAGE_KERNEL));
    >> +
    >> + /* We pre-allocated VM area so mapping can never fail */
    >> + area->vm_addr = area->vm->addr;
    >> + }
    >
    > This bit appears to be trying to make kmap_atomic() variant that can map
    > two pages in to contigious virtual addresses. Instead of open-coding it
    > in a non-portable way like this, should we just make a new kmap_atomic()
    > variant that does this?
    >
    > From the way it's implemented, I _think_ you're guaranteed to get two
    > contiguous addresses if you do two adjacent kmap_atomics() on the same CPU:
    >
    > void *kmap_atomic_prot(struct page *page, pgprot_t prot)
    > {
    > ...
    > type = kmap_atomic_idx_push();
    > idx = type + KM_TYPE_NR*smp_processor_id();
    > vaddr = __fix_to_virt(FIX_KMAP_BEGIN + idx);
    >
    > I think if you do a get_cpu()/put_cpu() or just a preempt_disable()
    > across the operations you'll be guaranteed to get two contiguous addresses.

    I'm not quite following here. kmap_atomic() only does this for highmem pages.
    For normal pages (all pages for 64-bit), it doesn't do any mapping at all. It
    just returns the virtual address of the page since it is in the kernel's address
    space.

    For this design, the pages _must_ be mapped, even if the pages are directly
    reachable in the address space, because they must be virtually contiguous.

    --
    Seth



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