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Message-ID: <B3C9C9EA-2B76-4AE5-8F1F-425FEB8560FD@nvidia.com>
Date: Thu, 29 May 2025 18:47:25 -0400
From: Zi Yan <ziy@...dia.com>
To: Dev Jain <dev.jain@....com>
Cc: akpm@...ux-foundation.org, willy@...radead.org,
 linux-fsdevel@...r.kernel.org, linux-mm@...ck.org,
 linux-kernel@...r.kernel.org, david@...hat.com, anshuman.khandual@....com,
 ryan.roberts@....com
Subject: Re: [PATCH] xarray: Add a BUG_ON() to ensure caller is not sibling

On 28 May 2025, at 23:17, Dev Jain wrote:

> On 28/05/25 10:42 pm, Zi Yan wrote:
>> On 28 May 2025, at 7:31, Dev Jain wrote:
>>
>>> Suppose xas is pointing somewhere near the end of the multi-entry batch.
>>> Then it may happen that the computed slot already falls beyond the batch,
>>> thus breaking the loop due to !xa_is_sibling(), and computing the wrong
>>> order. Thus ensure that the caller is aware of this by triggering a BUG
>>> when the entry is a sibling entry.
>> Is it possible to add a test case in lib/test_xarray.c for this?
>> You can compile the tests with “make -C tools/testing/radix-tree”
>> and run “./tools/testing/radix-tree/xarray”.
>
>
> Sorry forgot to Cc you.
> I can surely do that later, but does this patch look fine?

I am not sure the exact situation you are describing, so I asked you
to write a test case to demonstrate the issue. :)

>
>
>>
>>> This patch is motivated by code inspection and not a real bug report.
>>>
>>> Signed-off-by: Dev Jain <dev.jain@....com>
>>> ---
>>> The patch applies on 6.15 kernel.
>>>
>>>   lib/xarray.c | 2 ++
>>>   1 file changed, 2 insertions(+)
>>>
>>> diff --git a/lib/xarray.c b/lib/xarray.c
>>> index 9644b18af18d..0f699766c24f 100644
>>> --- a/lib/xarray.c
>>> +++ b/lib/xarray.c
>>> @@ -1917,6 +1917,8 @@ int xas_get_order(struct xa_state *xas)
>>>   	if (!xas->xa_node)
>>>   		return 0;
>>>
>>> +	XA_NODE_BUG_ON(xas->xa_node, xa_is_sibling(xa_entry(xas->xa,
>>> +		       xas->xa_node, xas->xa_offset)));
>>>   	for (;;) {
>>>   		unsigned int slot = xas->xa_offset + (1 << order);
>>>
>>> -- 
>>> 2.30.2
>>
>> Best Regards,
>> Yan, Zi


Best Regards,
Yan, Zi

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