On Wed, Jun 07 2017, Matthew Wilcox <[email protected]> wrote:
> From: Matthew Wilcox <[email protected]>
>
> Several callers have constant 'start' and an 'nbits' that is a multiple of
> 8, so we can turn them into calls to memset. We don't need the entirety
> of 'start' and 'nbits' to be constant, we just need to know whether
> they're divisible by 8.
>
> Signed-off-by: Matthew Wilcox <[email protected]>
> ---
> include/linux/bitmap.h | 6 ++++++
> 1 file changed, 6 insertions(+)
>
> diff --git a/include/linux/bitmap.h b/include/linux/bitmap.h
> index 4e0f0c8167af..0b3e4452b054 100644
> --- a/include/linux/bitmap.h
> +++ b/include/linux/bitmap.h
> @@ -319,6 +319,9 @@ static __always_inline void bitmap_set(unsigned long
> *map, unsigned int start,
> {
> if (__builtin_constant_p(nbits) && nbits == 1)
> __set_bit(start, map);
> + else if (__builtin_constant_p(start & 7) && IS_ALIGNED(start, 8) &&
> + __builtin_constant_p(nbits & 7) && IS_ALIGNED(nbits, 8))
> + memset(map + start / 8, 0xff, nbits / 8);
> else
Isn't the pointer arithmetic wrong here? I think you need to cast map to
(char*).
>
> __bitmap_set(map, start, nbits);
> }
> @@ -328,6 +331,9 @@ static __always_inline void bitmap_clear(unsigned long
> *map, unsigned int start,
> {
> if (__builtin_constant_p(nbits) && nbits == 1)
> __clear_bit(start, map);
> + else if (__builtin_constant_p(start & 7) && IS_ALIGNED(start, 8) &&
> + __builtin_constant_p(nbits & 7) && IS_ALIGNED(nbits, 8))
> + memset(map + start / 8, 0, nbits / 8);
> else
Ditto.
Do you have an example of how the generated code changes, both in the
case of actual constants and a case where gcc can see that start and
nbits are byte-aligned without knowing their actual values?
