Re: [PATCH v5 3/5] lib/bitmap: Introduce bitmap_scatter() and bitmap_gather() helpers

[Andy Shevchenko <andriy.shevchenko@linux.intel.com>]

On Thu, Feb 29, 2024 at 03:15:51PM +0100, Herve Codina wrote:
> From: Andy Shevchenko <andriy.shevchenko@linux.intel.com>
> 
> These helpers scatters or gathers a bitmap with the help of the mask
> position bits parameter.
> 
> bitmap_scatter() does the following:
>   src:  0000000001011010
>                   ||||||
>            +------+|||||
>            |  +----+||||
>            |  |+----+|||
>            |  ||   +-+||
>            |  ||   |  ||
>   mask: ...v..vv...v..vv
>         ...0..11...0..10
>   dst:  0000001100000010
> 
> and bitmap_gather() performs this one:
>    mask: ...v..vv...v..vv
>    src:  0000001100000010
>             ^  ^^   ^   0
>             |  ||   |  10
>             |  ||   > 010
>             |  |+--> 1010
>             |  +--> 11010
>             +----> 011010
>    dst:  0000000000011010
> 
> bitmap_gather() can the seen as the reverse bitmap_scatter() operation.

> The original work was done by Andy Shevchenko.

As I said, it's too much credit :-)

...

> + * A relationship exists between bitmap_gather() and bitmap_scatter(). See
> + * bitmap_scatter() for the bitmap scatter detailed operations.
> + * Suppose scattered computed using bitmap_scatter(scattered, src, mask, n).
> + * The operation bitmap_gather(result, scattered, mask, n) leads to a result
> + * equal or equivalent to src.
> + *
> + * The result can be 'equivalent' because bitmap_scatter() and bitmap_gather()
> + * are not bijective.
> + * The result and src values are equivalent in that sense that a call to

> + * bitmap_scatter(res, src, mask, n) and a call to bitmap_scatter(res, result,
> + * mask, n) will lead to the same res value.

For better readability I wouldn't break API calls, hence

 The result and src values are equivalent in that sense that a call to
 bitmap_scatter(res, src, mask, n) and a call to
 bitmap_scatter(res, result, mask, n) will lead to the same res value.

-- 
With Best Regards,
Andy Shevchenko