Commit 53460a45 authored by Nikolay Borisov's avatar Nikolay Borisov Committed by David Sterba

btrfs: trim: make reserved device area adjustments more explicit

Currently the first megabyte on a device housing a btrfs filesystem is
exempt from allocation and trimming. Currently this is not a problem
since 'start' is set to 1M at the beginning of btrfs_trim_free_extents
and find_first_clear_extent_bit always returns a range that is >= start.

However, in a follow up patch find_first_clear_extent_bit will be
changed such that it will return a range containing 'start' and this
range may very well be 0...>=1M so 'start'.

Future proof the sole user of find_first_clear_extent_bit by setting
'start' after the function is called. No functional changes.
Reviewed-by: default avatarJohannes Thumshirn <jthumshirn@suse.de>
Signed-off-by: default avatarNikolay Borisov <nborisov@suse.com>
Signed-off-by: default avatarDavid Sterba <dsterba@suse.com>
parent 6f8e4fd4
......@@ -11179,12 +11179,17 @@ static int btrfs_trim_free_extents(struct btrfs_device *device, u64 *trimmed)
find_first_clear_extent_bit(&device->alloc_state, start,
&start, &end,
CHUNK_TRIMMED | CHUNK_ALLOCATED);
/* Ensure we skip the reserved area in the first 1M */
start = max_t(u64, start, SZ_1M);
/*
* If find_first_clear_extent_bit find a range that spans the
* end of the device it will set end to -1, in this case it's up
* to the caller to trim the value to the size of the device.
*/
end = min(end, device->total_bytes - 1);
len = end - start + 1;
/* We didn't find any extents */
......
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