linux-6.12.1/lib/bitmap.c

170  * Set the n-th bit of @dst iff the n-th bit of @src is set and
171  * n is less than @first, or the m-th bit of @src is set for any
453  * that bit 7 is the 3rd (starting with 0th) set bit in @buf.
474  * whatever position is held by the n-th set bit in @old is mapped
475  * to the n-th set bit in @new.  In the more general case, allowing
477  * weight of @old, map the position of the n-th set bit in @old to
478  * the position of the m-th set bit in @new, where m == n % w.
492  * position 4 to 12, 5 to 13, 6 to 14 and 7 to 15, and of all other
494  * with bits 1, 5 and 7 set, then @dst should leave with bits 1,
527  * whatever position is held by the n-th set bit in @old is mapped
528  * to the n-th set bit in @new.  In the more general case, allowing
530  * weight of @old, map the position of the n-th set bit in @old to
531  * the position of the m-th set bit in @new, where m == n % w.
541  * position 4 to 12, 5 to 13, 6 to 14 and 7 to 15, and of all other
542  * bit positions unchanged.  So if say @oldbit is 5, then this routine
565  * Set the n-th bit of @dst iff there exists some m such that the
566  * n-th bit of @relmap is set, the m-th bit of @orig is set, and
567  * the n-th bit of @relmap is also the m-th _set_ bit of @relmap.
572  * using the map { <n, m> | the n-th bit of @relmap is the
573  * m-th set bit of @relmap }.
587  *  1, 3, 5, 7, 9 and 11 set.  Then on return from this routine,
602  *  because they were the 4th, 6th, 8th and 10th set bits
603  *  set in @relmap, and the 4th, 6th, 8th and 10th bits of
604  *  @orig (i.e. bits 3, 5, 7 and 9) were also set.
641  *      1 3 5 7          1 3 5 7       41 43 48 61