11-10-2003
thanx again
a ha, I see, thanx for clearify that, I have another little be query,
Quote:
This results in a string of 1's followed by a string of 0's. That will always be the case. Where we have 1's, the bits are the network number. Where we have 0's the bits are the host number.
how about the case, 11111111.11111111.11111111.11010100, so the rule for that is wether we count from the first left most 0 bit
or how many 0 bits or the position of 0 bits for the host number?
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LEARN ABOUT SUSE
bitmap_remap
BITMAP_REMAP(9) Basic Kernel Library Functions BITMAP_REMAP(9)
NAME
bitmap_remap - Apply map defined by a pair of bitmaps to another bitmap
SYNOPSIS
void bitmap_remap(unsigned long * dst, const unsigned long * src, const unsigned long * old, const unsigned long * new, int bits);
ARGUMENTS
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
DESCRIPTION
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.
COPYRIGHT
Kernel Hackers Manual 2.6. July 2010 BITMAP_REMAP(9)