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Subject: Re: [PATCH v2] kernel/resource: Fix locking in
 request_free_mem_region
To:     Alistair Popple <apopple@nvidia.com>, linux-kernel@vger.kernel.org
Cc:     akpm@linux-foundation.org, daniel.vetter@ffwll.ch,
        dan.j.williams@intel.com, gregkh@linuxfoundation.org,
        jhubbard@nvidia.com, jglisse@redhat.com, linux-mm@kvack.org
References: <20210326012035.3853-1-apopple@nvidia.com>
From:   David Hildenbrand <david@redhat.com>
Organization: Red Hat GmbH
Message-ID: <9eef1283-28a3-845e-0e3e-80b763c9ec59@redhat.com>
Date:   Fri, 26 Mar 2021 09:57:51 +0100
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On 26.03.21 02:20, Alistair Popple wrote:
> request_free_mem_region() is used to find an empty range of physical
> addresses for hotplugging ZONE_DEVICE memory. It does this by iterating
> over the range of possible addresses using region_intersects() to see if
> the range is free.

Just a high-level question: how does this iteract with memory 
hot(un)plug? IOW, how defines and manages the "range of possible 
addresses" ?

> 
> region_intersects() obtains a read lock before walking the resource tree
> to protect against concurrent changes. However it drops the lock prior
> to returning. This means by the time request_mem_region() is called in
> request_free_mem_region() another thread may have already reserved the
> requested region resulting in unexpected failures and a message in the
> kernel log from hitting this condition:

I am confused. Why can't we return an error to the caller and let the 
caller continue searching? This feels much simpler than what you propose 
here. What am I missing?


-- 
Thanks,

David / dhildenb