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SubjectRe: [PATCH v2] f2fs: free_user_blocks should use reserved_segments instead
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Hi Yunlong,

I think you have changed original implication of the function, IMO, it would be
more accurate to use user_free_segment_blocks instead of free_user_blocks.

Thanks,

On 2017/8/18 18:02, Yunlong Song wrote:
> ping...
>
> On 2017/8/15 15:14, Yunlong Song wrote:
>> The part (overprovision_segments - reserved_segments) can still be used for LFS
>> in some case, e.g., there are lots of invalid block from dirty segments, then
>> the part (overprovision_segments - reserved_segments) can be safely used. So
>> free_use_blocks should use reserved_segments instead, rather than directly use
>> overprovision_segments. BTW, we also add the constraint of sbi->reserved_blocks.
>>
>> Signed-off-by: Yunlong Song <yunlong.song@huawei.com>
>> ---
>> fs/f2fs/gc.h | 18 +++++++++++++++---
>> 1 file changed, 15 insertions(+), 3 deletions(-)
>>
>> diff --git a/fs/f2fs/gc.h b/fs/f2fs/gc.h
>> index 9325191..542612a 100644
>> --- a/fs/f2fs/gc.h
>> +++ b/fs/f2fs/gc.h
>> @@ -49,11 +49,23 @@ struct gc_inode_list {
>> */
>> static inline block_t free_user_blocks(struct f2fs_sb_info *sbi)
>> {
>> - if (free_segments(sbi) < overprovision_segments(sbi))
>> + block_t avail_user_block_count, free_blocks, avail_free_blocks;
>> + block_t reserved_blocks;
>> +
>> + avail_user_block_count = sbi->user_block_count - sbi->reserved_blocks;
>> +
>> + if (unlikely(sbi->total_valid_block_count > avail_user_block_count))
>> + return 0;
>> +
>> + free_blocks = free_segments(sbi) << sbi->log_blocks_per_seg;
>> + avail_free_blocks = min(free_blocks, avail_user_block_count -
>> + sbi->total_valid_block_count);
>> + reserved_blocks = reserved_segments(sbi) << sbi->log_blocks_per_seg;
>> +
>> + if (avail_free_blocks < reserved_blocks)
>> return 0;
>> else
>> - return (free_segments(sbi) - overprovision_segments(sbi))
>> - << sbi->log_blocks_per_seg;
>> + return avail_free_blocks - reserved_blocks;
>> }
>> static inline block_t limit_invalid_user_blocks(struct f2fs_sb_info *sbi)
>

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