The compiler can generate slightly smaller and simpler code when it
knows that "nbits" is non-negative. Since no-one passes a negative
bit-count, this shouldn't affect the semantics.
Signed-off-by: Rasmus Villemoes <linux@rasmusvillemoes.dk>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
extern int __bitmap_empty(const unsigned long *bitmap, unsigned int nbits);
extern int __bitmap_full(const unsigned long *bitmap, unsigned int nbits);
extern int __bitmap_equal(const unsigned long *bitmap1,
- const unsigned long *bitmap2, int bits);
+ const unsigned long *bitmap2, unsigned int nbits);
extern void __bitmap_complement(unsigned long *dst, const unsigned long *src,
int bits);
extern void __bitmap_shift_right(unsigned long *dst,
EXPORT_SYMBOL(__bitmap_full);
int __bitmap_equal(const unsigned long *bitmap1,
- const unsigned long *bitmap2, int bits)
+ const unsigned long *bitmap2, unsigned int bits)
{
- int k, lim = bits/BITS_PER_LONG;
+ unsigned int k, lim = bits/BITS_PER_LONG;
for (k = 0; k < lim; ++k)
if (bitmap1[k] != bitmap2[k])
return 0;