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: Rasmus Villemoes <linux@rasmusvillemoes.dk> Signed-off-by: Andrew Morton <akpm@linux-foundation.org> Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
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@ -159,7 +159,7 @@ void __bitmap_shift_left(unsigned long *dst, const unsigned long *src,
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unsigned int shift, unsigned int nbits)
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{
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int k;
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unsigned int lim = BITS_TO_LONGS(nbits), left = nbits % BITS_PER_LONG;
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unsigned int lim = BITS_TO_LONGS(nbits);
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unsigned int off = shift/BITS_PER_LONG, rem = shift % BITS_PER_LONG;
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for (k = lim - off - 1; k >= 0; --k) {
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unsigned long upper, lower;
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@ -172,13 +172,8 @@ void __bitmap_shift_left(unsigned long *dst, const unsigned long *src,
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lower = src[k - 1] >> (BITS_PER_LONG - rem);
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else
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lower = 0;
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upper = src[k];
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if (left && k == lim - 1)
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upper &= (1UL << left) - 1;
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upper <<= rem;
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upper = src[k] << rem;
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dst[k + off] = lower | upper;
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if (left && k + off == lim - 1)
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dst[k + off] &= (1UL << left) - 1;
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}
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if (off)
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memset(dst, 0, off*sizeof(unsigned long));
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