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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