Re: [PATCH V4 2/3] powerpc/swiotlb: Enable at early stage and disable if not necessary

On Aug 3, 2012, at 9:42 AM, Li Yang wrote:

On Fri, Aug 3, 2012 at 8:38 PM, Kumar Gala [off-list ref] =

wrote:

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On Aug 2, 2012, at 9:21 PM, Jia Hongtao-B38951 wrote:
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-----Original Message-----
From: Kumar Gala [mailto:galak@kernel.crashing.org]
Sent: Thursday, August 02, 2012 8:55 PM
To: Jia Hongtao-B38951
Cc: linuxppc-dev@lists.ozlabs.org; Wood Scott-B07421; Li =

Yang-R58472

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Subject: Re: [PATCH V4 2/3] powerpc/swiotlb: Enable at early stage =

and

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disable if not necessary
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On Aug 2, 2012, at 6:42 AM, Jia Hongtao wrote:
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Remove the dependency on PCI initialization for SWIOTLB =

initialization.

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So that PCI can be initialized at proper time.
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SWIOTLB is partly determined by PCI inbound/outbound map which is

assigned

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in PCI initialization. But swiotlb_init() should be done at the =

stage

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of

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mem_init() which is much earlier than PCI initialization. So we =

reserve

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the

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memory for SWIOTLB first and free it if not necessary.
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All boards are converted to fit this change.
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Signed-off-by: Jia Hongtao <redacted>
Signed-off-by: Li Yang <redacted>
---

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This doesn't seem like it addresses our issue w/regards to not =

being able

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to map all of memory from PCI.

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PCI init will determine ppc_swiotlb_enable due to PCI map. =

swiotlb_late_init

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will handle all swiotlb things depend on the result of pci init.

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Think about the case that we have 4095M of memory & 1G of PCI memory =

mapped space.  The old code would enable swiotlb for this case since we =
would NOT be able to DMA to all 4095M of memory.  The patch does not =
handle this case correctly.

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The patch can handle it.  The ppc_swiotlb_enable is still being set in
fsl_pci_init() if there is 1G of PCI memory mapped space.  It is
after next patch that the ppc_swiotlb_enable is being set in the PCI
probe() routine.
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Leo

Gotcha.  I was thinking the swiotlb_init(0) was:

if (ppc_swiotlb_enable)
	swiotlb_init(0)

Now I see, we call it unconditionally, than fixup in swiotlb_late_init

- k=