linux-kernel - Re: linux-next: Tree for May 3

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Date:   Wed, 11 May 2022 17:37:50 +0300
From:   Mike Rapoport <rppt@...nel.org>
To:     Christoph Hellwig <hch@....de>
Cc:     Conor.Dooley@...rochip.com, sfr@...b.auug.org.au,
        linux-next@...r.kernel.org, linux-kernel@...r.kernel.org,
        linux-riscv@...ts.infradead.org
Subject: Re: linux-next: Tree for May 3

On Wed, May 11, 2022 at 04:10:34PM +0200, Christoph Hellwig wrote:
> On Wed, May 11, 2022 at 05:08:52PM +0300, Mike Rapoport wrote:
> > I guess the default to use memblock_alloc_low() backfires on system with
> > physical memory living at 0x1000200000:
> > 
> > [    0.000000] Early memory node ranges
> > [    0.000000]   node   0: [mem 0x0000001000200000-0x000000103fffffff]
> > 
> > The default limit for "low" memory is 0xffffffff and there is simply no
> > memory there.
> 
> Is there any way to ask memblock for a specific address limit?
> swiotlb just wants <= 32-bit by default.  With the little caveat
> that it should be 32-bit addressable for all devices, and we don't
> know the physical to dma address mapping at time of allocation.

There is 

void *memblock_alloc_try_nid(phys_addr_t size, phys_addr_t align,
			     phys_addr_t min_addr, phys_addr_t max_addr,
			     int nid);

that lets caller to specify min and max limits

Presuming that devices see [0x1000200000-0x103fffffff] as
[0x200000-0x3fffffff] we may try something like

	min = memblock_start_of_DRAM();
	max = min + 0xffffffff;

	if (flags & SWIOTLB_ANY)
		max = MEMBLOCK_ALLOC_ACCESSIBLE;

	tlb = memblock_alloc_try_nid(bytes, PAGE_SIZE, min, max, NUMA_NO_NODE);

-- 
Sincerely yours,
Mike.