Thread (170 messages) 170 messages, 19 authors, 2012-09-16

[PATCH v2 14/31] arm64: DMA mapping API

From: catalin.marinas@arm.com (Catalin Marinas)
Date: 2012-08-21 13:00:30
Also in: linux-arch, lkml

On Wed, Aug 15, 2012 at 05:16:00PM +0100, Arnd Bergmann wrote:
On Tuesday 14 August 2012, Catalin Marinas wrote:
quoted
+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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