On 2019/6/3 下午6:06, Nikolay Borisov wrote:
> 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.
>
> Signed-off-by: Nikolay Borisov <nbori...@suse.com>
Doesn't that previous patch already address this by:
+ u64 start = SZ_1M, len = 0, end = 0;
Thanks,
Qu
> ---
> fs/btrfs/extent-tree.c | 4 ++++
> 1 file changed, 4 insertions(+)
>
> diff --git a/fs/btrfs/extent-tree.c b/fs/btrfs/extent-tree.c
> index d8c5febf7636..5a11e4988243 100644
> --- a/fs/btrfs/extent-tree.c
> +++ b/fs/btrfs/extent-tree.c
> @@ -11183,6 +11183,10 @@ static int btrfs_trim_free_extents(struct
> btrfs_device *device, u64 *trimmed)
> * to the caller to trim the value to the size of the device.
> */
> end = min(end, device->total_bytes - 1);
> +
> + /* Ensure we skip first mb in case we have a bootloader there */
> + start = max_t(u64, start, SZ_1M);
> +
> len = end - start + 1;
>
> /* We didn't find any extents */
>
