bitmap_onto man page on Oracle

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BITMAP_ONTO(9)		Basic Kernel Library Functions		BITMAP_ONTO(9)

       bitmap_onto - translate one bitmap relative to another

       void bitmap_onto(unsigned long * dst, const unsigned long * orig,
			const unsigned long * relmap, int bits);

	   resulting translated bitmap

	   original untranslated bitmap

	   bitmap relative to which translated

	   number of bits in each of these bitmaps

       Set the n-th bit of dst iff there exists some m such that the n-th bit
       of relmap is set, the m-th bit of orig is set, and the n-th bit of
       relmap is also the m-th _set_ bit of relmap. (If you understood the
       previous sentence the first time your read it, you're overqualified for
       your current job.)

       In other words, orig is mapped onto (surjectively) dst, using the the
       map { <n, m> | the n-th bit of relmap is the m-th set bit of relmap }.

       Any set bits in orig above bit number W, where W is the weight of
       (number of set bits in) relmap are mapped nowhere. In particular, if
       for all bits m set in orig, m >= W, then dst will end up empty. In
       situations where the possibility of such an empty result is not
       desired, one way to avoid it is to use the bitmap_fold operator, below,
       to first fold the orig bitmap over itself so that all its set bits x
       are in the range 0 <= x < W. The bitmap_fold operator does this by
       setting the bit (m % W) in dst, for each bit (m) set in orig.

       Example [1] for bitmap_onto: Let's say relmap has bits 30-39 set, and
       orig has bits 1, 3, 5, 7, 9 and 11 set. Then on return from this
       routine, dst will have bits 31, 33, 35, 37 and 39 set.

       When bit 0 is set in orig, it means turn on the bit in dst
       corresponding to whatever is the first bit (if any) that is turned on
       in relmap. Since bit 0 was off in the above example, we leave off that
       bit (bit 30) in dst.

       When bit 1 is set in orig (as in the above example), it means turn on
       the bit in dst corresponding to whatever is the second bit that is
       turned on in relmap. The second bit in relmap that was turned on in the
       above example was bit 31, so we turned on bit 31 in dst.

       Similarly, we turned on bits 33, 35, 37 and 39 in dst, because they
       were the 4th, 6th, 8th and 10th set bits set in relmap, and the 4th,
       6th, 8th and 10th bits of orig (i.e. bits 3, 5, 7 and 9) were also set.

       When bit 11 is set in orig, it means turn on the bit in dst
       corresponding to whatever is the twelfth bit that is turned on in
       relmap. In the above example, there were only ten bits turned on in
       relmap (30..39), so that bit 11 was set in orig had no affect on dst.

       Example [2] for bitmap_fold + bitmap_onto: Let's say relmap has these
       ten bits set: 40 41 42 43 45 48 53 61 74 95 (for the curious, that's 40
       plus the first ten terms of the Fibonacci sequence.)

       Further lets say we use the following code, invoking bitmap_fold then
       bitmap_onto, as suggested above to avoid the possitility of an empty
       dst result:

       unsigned long *tmp; // a temporary bitmap's bits

       bitmap_fold(tmp, orig, bitmap_weight(relmap, bits), bits);
       bitmap_onto(dst, tmp, relmap, bits);

       Then this table shows what various values of dst would be, for various
       orig's. I list the zero-based positions of each set bit. The tmp column
       shows the intermediate result, as computed by using bitmap_fold to fold
       the orig bitmap modulo ten (the weight of relmap).

       orig tmp dst 0 0 40 1 1 41 9 9 95 10 0 40 (*) 1 3 5 7 1 3 5 7 41 43 48
       61 0 1 2 3 4 0 1 2 3 4 40 41 42 43 45 0 9 18 27 0 9 8 7 40 61 74 95 0
       10 20 30 0 40 0 11 22 33 0 1 2 3 40 41 42 43 0 12 24 36 0 2 4 6 40 42
       45 53 78 102 211 1 2 8 41 42 74 (*)

       (*) For these marked lines, if we hadn't first done bitmap_fold into
       tmp, then the dst result would have been empty.

       If either of orig or relmap is empty (no set bits), then dst will be
       returned empty.

       If (as explained above) the only set bits in orig are in positions m
       where m >= W, (where W is the weight of relmap) then dst will once
       again be returned empty.

       All bits in dst not set by the above rule are cleared.

Kernel Hackers Manual 3.8.	   June 2014			BITMAP_ONTO(9)

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