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 <[email protected]>
---
 lib/bitmap.c | 9 ++-------
 1 file changed, 2 insertions(+), 7 deletions(-)

diff --git a/lib/bitmap.c b/lib/bitmap.c
index 2362601..acea8ae 100644
--- a/lib/bitmap.c
+++ b/lib/bitmap.c
@@ -157,7 +157,7 @@ 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;
@@ -170,13 +170,8 @@ 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));
-- 
2.0.4

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