[PATCH] drivers/base: export gpl (un)register_memory_notifier

Dave Hansen haveblue at us.ibm.com
Sat Feb 16 03:55:38 EST 2008


On Fri, 2008-02-15 at 14:22 +0100, Christoph Raisch wrote:
> A translation from kernel to ehea_bmap space should be fast and
> predictable
> (ruling out hashes).
> If a driver doesn't know anything else about the mapping structure,
> the normal solution in kernel for this type of problem is a multi
> level
> look up table
> like pgd->pud->pmd->pte
> This doesn't sound right to be implemented in a device driver.
> 
> We didn't see from the existing code that such a mapping to a
> contiguous
> space already exists.
> Maybe we've missed it.

I've been thinking about that, and I don't think you really *need* to
keep a comprehensive map like that.  

When the memory is in a particular configuration (range of memory
present along with unique set of holes) you get a unique ehea_bmap
configuration.  That layout is completely predictable.

So, if at any time you want to figure out what the ehea_bmap address for
a particular *Linux* virtual address is, you just need to pretend that
you're creating the entire ehea_bmap, use the same algorithm and figure
out host you would have placed things, and use that result.

Now, that's going to be a slow, crappy linear search (but maybe not as
slow as recreating the silly thing).  So, you might eventually run into
some scalability problems with a lot of packets going around.  But, I'd
be curious if you do in practice.

The other idea is that you create a mapping that is precisely 1:1 with
kernel memory.  Let's say you have two sections present, 0 and 100.  You
have a high_section_index of 100, and you vmalloc() a 100 entry array.

You need to create a *CONTIGUOUS* ehea map?  Create one like this:

EHEA_VADDR->Linux Section
0->0
1->0
2->0
3->0
...
100->100

It's contiguous.  Each area points to a valid Linux memory address.
It's also discernable in O(1) to what EHEA address a given Linux address
is mapped.  You just have a couple of duplicate entries.  

-- Dave




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