提交 7f590657 编写于 作者: R Rasmus Villemoes 提交者: Linus Torvalds

lib: bitmap: remove redundant code from __bitmap_shift_left

The first of these conditionals is completely redundant: If k == lim-1, we
must have off==0, so the second conditional will also trigger and then it
wouldn't matter if upper had some high bits set.  But the second
conditional is in fact also redundant, since it only serves to clear out
some high-order "don't care" bits of dst, about which no guarantee is
made.
Signed-off-by: NRasmus Villemoes <linux@rasmusvillemoes.dk>
Signed-off-by: NAndrew Morton <akpm@linux-foundation.org>
Signed-off-by: NLinus Torvalds <torvalds@linux-foundation.org>
上级 6d874eca
......@@ -159,7 +159,7 @@ void __bitmap_shift_left(unsigned long *dst, const unsigned long *src,
unsigned int shift, unsigned int nbits)
{
int k;
unsigned int lim = BITS_TO_LONGS(nbits), left = nbits % BITS_PER_LONG;
unsigned int lim = BITS_TO_LONGS(nbits);
unsigned int off = shift/BITS_PER_LONG, rem = shift % BITS_PER_LONG;
for (k = lim - off - 1; k >= 0; --k) {
unsigned long upper, lower;
......@@ -172,13 +172,8 @@ void __bitmap_shift_left(unsigned long *dst, const unsigned long *src,
lower = src[k - 1] >> (BITS_PER_LONG - rem);
else
lower = 0;
upper = src[k];
if (left && k == lim - 1)
upper &= (1UL << left) - 1;
upper <<= rem;
upper = src[k] << rem;
dst[k + off] = lower | upper;
if (left && k + off == lim - 1)
dst[k + off] &= (1UL << left) - 1;
}
if (off)
memset(dst, 0, off*sizeof(unsigned long));
......
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