Messages in this thread | | | From | Rasmus Villemoes <> | Subject | Re: [PATCH 2/3] Turn bitmap_set and bitmap_clear into memset when possible | Date | Wed, 07 Jun 2017 23:16:37 +0200 |
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On Wed, Jun 07 2017, Matthew Wilcox <willy@infradead.org> wrote:
> From: Matthew Wilcox <mawilcox@microsoft.com> > > 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 <mawilcox@microsoft.com> > --- > 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?
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