bitmap_remap (9)

void bitmap_remap(unsigned long * dst, const unsigned long * src, const unsigned long * old, const unsigned long * new, int bits);

dst

remapped result

src

subset to be remapped

old

defines domain of map

new

defines range of map

bits

number of bits in each of these bitmaps

Let old and new define a mapping of bit positions, such that whatever position is held by the n-th set bit in old is mapped to the n-th set bit in new. In the more general case, allowing for the possibility that the weight 'w' of new is less than the weight of old, map the position of the n-th set bit in old to the position of the m-th set bit in new, where m == n % w.

If either of the old and new bitmaps are empty, or if src and dst point to the same location, then this routine copies src to dst.

The positions of unset bits in old are mapped to themselves (the identify map).

Apply the above specified mapping to src, placing the result in dst, clearing any bits previously set in dst.

For example, lets say that old has bits 4 through 7 set, and new has bits 12 through 15 set. This defines the mapping of bit position 4 to 12, 5 to 13, 6 to 14 and 7 to 15, and of all other bit positions unchanged. So if say src comes into this routine with bits 1, 5 and 7 set, then dst should leave with bits 1, 13 and 15 set.