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x86: return conditional to mmu
Just return our allocation if we don't have an mmu. For i386, where this patch is being applied, we never have. So our goal is just to have the code to look like x86_64's. Signed-off-by: Glauber Costa <gcosta@redhat.com> Signed-off-by: Ingo Molnar <mingo@elte.hu> Signed-off-by: Thomas Gleixner <tglx@linutronix.de>
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@ -116,12 +116,42 @@ again:
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gfp = (gfp & ~GFP_DMA32) | GFP_DMA;
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goto again;
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}
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/* Let low level make its own zone decisions */
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gfp &= ~(GFP_DMA32|GFP_DMA);
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if (dma_ops->alloc_coherent)
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return dma_ops->alloc_coherent(dev, size,
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dma_handle, gfp);
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return NULL;
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}
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memset(ret, 0, size);
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*dma_handle = bus;
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if (!mmu) {
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*dma_handle = bus;
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return ret;
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}
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}
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return ret;
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if (dma_ops->alloc_coherent) {
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free_pages((unsigned long)ret, get_order(size));
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gfp &= ~(GFP_DMA|GFP_DMA32);
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return dma_ops->alloc_coherent(dev, size, dma_handle, gfp);
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}
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if (dma_ops->map_simple) {
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*dma_handle = dma_ops->map_simple(dev, virt_to_phys(ret),
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size,
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PCI_DMA_BIDIRECTIONAL);
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if (*dma_handle != bad_dma_address)
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return ret;
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}
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if (panic_on_overflow)
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panic("dma_alloc_coherent: IOMMU overflow by %lu bytes\n",
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(unsigned long)size);
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free_pages((unsigned long)ret, get_order(size));
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return NULL;
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}
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EXPORT_SYMBOL(dma_alloc_coherent);
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