Re: [PATCH v4] btrfs: trim: fix underflow in trim length to prevent access beyond device boundary

[Qu Wenruo <quwenruo.btrfs@gmx.com>]

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On 2020/7/31 下午10:08, David Sterba wrote:
> On Fri, Jul 31, 2020 at 07:29:11PM +0800, Qu Wenruo wrote:
>> --- a/fs/btrfs/volumes.c
>> +++ b/fs/btrfs/volumes.c
>> @@ -4720,6 +4720,18 @@ int btrfs_shrink_device(struct btrfs_device *device, u64 new_size)
>>  	}
>>  
>>  	mutex_lock(&fs_info->chunk_mutex);
>> +	/*
>> +	 * Also clear any CHUNK_TRIMMED and CHUNK_ALLOCATED bits beyond the
>> +	 * current device boundary.
>> +	 * This shouldn't fail, as alloc_state should only utilize those two
>> +	 * bits, thus we shouldn't alloc new memory for clearing the status.
> 
> If this fails or not depends on implementation details of
> clear_extent_bits and this comment will get out of sync eventually, so I
> don't think it should be that specific.
> 
> If the new_size is somewhere in the middle of an existing state, it'll
> need to be split anyway, no?

Nope. Because in alloc_state we only have two bits utilized,
CHUNK_TRIMMED and CHUNK_ALLOCATED.

Thus what we're doing is to clear all utilized bits.

> 
> alloc_state |-----+++++|
> clear             |------------------------- ... (u64)-1|
> 
> So we'd need to keep the state "-" and unset bits only from "+", and
> this will require a split.

In this case, we would only reduce the the size of the existing status,
or just remove it completely.

> 
> But I still have doubts about just clearing the range, why are there any
> device->alloc_state entries at all after device is shrunk?

Because the alloc_state is mostly only utilized by trim facility, thus
existing functions won't bother clearing/setting it.

In this particular case, previous fstrim run would set the CHUNK_TRIMMED
bit for all unallocated range (except the super reserve).
Then shrink doesn't clear the exceed range, and cause problem.

Thus clearing the bit in btrfs_shrink_device() makes sense.

> Using
> clear_extent_bits here is not wrong if we look at the end result of
> clearing the range, but otherwise it leaves some state information
> and allocated memory behind.
> 
Not that complex case, just plain not fully considered corner case.

Thanks,
Qu


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