Re: [PATCH 2/3] block: use lcm_not_zero() when stacking chunk_sectors

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On Fri, Sep 11, 2020 at 05:53:37PM -0400, Mike Snitzer wrote:
> Like 'io_opt', blk_stack_limits() should stack 'chunk_sectors' using
> lcm_not_zero() rather than min_not_zero() -- otherwise the final
> 'chunk_sectors' could result in sub-optimal alignment of IO to
> component devices in the IO stack.
> 
> Also, if 'chunk_sectors' isn't a multiple of 'physical_block_size'
> then it is a bug in the driver and the device should be flagged as
> 'misaligned'.
> 
> Signed-off-by: Mike Snitzer <snitzer@xxxxxxxxxx>
> ---
>  block/blk-settings.c | 12 ++++++++----
>  1 file changed, 8 insertions(+), 4 deletions(-)
> 
> diff --git a/block/blk-settings.c b/block/blk-settings.c
> index 76a7e03bcd6c..b09642d5f15e 100644
> --- a/block/blk-settings.c
> +++ b/block/blk-settings.c
> @@ -534,6 +534,7 @@ int blk_stack_limits(struct queue_limits *t, struct queue_limits *b,
>  
>  	t->io_min = max(t->io_min, b->io_min);
>  	t->io_opt = lcm_not_zero(t->io_opt, b->io_opt);
> +	t->chunk_sectors = lcm_not_zero(t->chunk_sectors, b->chunk_sectors);
>  
>  	/* Physical block size a multiple of the logical block size? */
>  	if (t->physical_block_size & (t->logical_block_size - 1)) {
> @@ -556,6 +557,13 @@ int blk_stack_limits(struct queue_limits *t, struct queue_limits *b,
>  		ret = -1;
>  	}
>  
> +	/* chunk_sectors a multiple of the physical block size? */
> +	if (t->chunk_sectors & (t->physical_block_size - 1)) {
> +		t->chunk_sectors = 0;
> +		t->misaligned = 1;
> +		ret = -1;
> +	}
> +
>  	t->raid_partial_stripes_expensive =
>  		max(t->raid_partial_stripes_expensive,
>  		    b->raid_partial_stripes_expensive);
> @@ -594,10 +602,6 @@ int blk_stack_limits(struct queue_limits *t, struct queue_limits *b,
>  			t->discard_granularity;
>  	}
>  
> -	if (b->chunk_sectors)
> -		t->chunk_sectors = min_not_zero(t->chunk_sectors,
> -						b->chunk_sectors);
> -
>  	t->zoned = max(t->zoned, b->zoned);
>  	return ret;
>  }

Looks fine:

Reviewed-by: Ming Lei <ming.lei@xxxxxxxxxx>


Thanks,
Ming




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