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    SubjectRe: [PATCH v2 14/31] arm64: DMA mapping API
    On Wed, Aug 15, 2012 at 05:16:00PM +0100, Arnd Bergmann wrote:
    > On Tuesday 14 August 2012, Catalin Marinas wrote:
    > > +static struct dma_map_ops arm64_swiotlb_dma_ops = {
    > > + .alloc = arm64_swiotlb_alloc_coherent,
    > > + .free = arm64_swiotlb_free_coherent,
    > > + .map_page = arm64_swiotlb_map_page,
    > > + .unmap_page = arm64_swiotlb_unmap_page,
    > > + .map_sg = arm64_swiotlb_map_sg_attrs,
    > > + .unmap_sg = arm64_swiotlb_unmap_sg_attrs,
    > > + .sync_single_for_cpu = arm64_swiotlb_sync_single_for_cpu,
    > > + .sync_single_for_device = arm64_swiotlb_sync_single_for_device,
    > > + .sync_sg_for_cpu = arm64_swiotlb_sync_sg_for_cpu,
    > > + .sync_sg_for_device = arm64_swiotlb_sync_sg_for_device,
    > > + .dma_supported = swiotlb_dma_supported,
    > > + .mapping_error = swiotlb_dma_mapping_error,
    > > +};
    > > +
    > > +void __init swiotlb_init_with_default_size(size_t default_size, int verbose);
    > > +
    > > +void __init arm64_swiotlb_init(size_t max_size)
    > > +{
    > > + dma_ops = &arm64_swiotlb_dma_ops;
    > > + swiotlb_init_with_default_size(min((size_t)SZ_64M, max_size), 1);
    > > +}
    >
    > Why is swiotlb the default? I would expect that most devices can in fact
    > use the entire 64 bit address space, so you can use a simple linear
    > implementation for those.

    That was my worry, devices not capable of accessing the full 64-bit
    address space. We can hope that those SoCs would have an IOMMU but I
    can't tell for sure at this stage.

    The default implementation could be simpler. I can even drop it
    altogether from the initial patchset given that no SoC makes use of it
    yet.

    --
    Catalin


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