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]> Acked-by: Rasmus Villemoes <[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..c04c9d155e59 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((char *)map + start / 8, 0xff, nbits / 8); else __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((char *)map + start / 8, 0, nbits / 8); else __bitmap_clear(map, start, nbits); } -- 2.11.0
